Additional Exercises for Ch 10

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tdplotsinandcos }[3]{\pgfmathsetmacro {#1}{sin(#3)}\pgfmathsetmacro {#2}{cos(#3)}} \newcommand {\tdplotmult }[3]{\pgfmathsetmacro {#1}{#2*#3}} \newcommand {\tdplotdiv }[3]{\pgfmathsetmacro {#1}{#2/#3}} \newcommand {\tdplotcheckdiff }[5]{\par \par \pgfmathparse { abs(#2 -#1)<#3 } \par \ifthenelse {\equal {\pgfmathresult }{1}}{#4}{#5} } \newcommand {\tdplotsetmaincoords }[2]{\pgfmathsetmacro {\tdplotmaintheta }{#1} \pgfmathsetmacro {\tdplotmainphi }{#2} \tdplotcalctransformmainscreen \tikzset {tdplot_main_coords/.style={x={(\raarot cm,\rbarot cm)},y={(\rabrot cm,\rbbrot cm)},z={(\racrot cm,\rbcrot cm)}}}} \newcommand {\tdplotcalctransformmainscreen }[0]{\tdplotsinandcos {\sintheta }{\costheta }{\tdplotmaintheta }\tdplotsinandcos {\sinphi }{\cosphi }{\tdplotmainphi }\tdplotmult {\stsp }{\sintheta }{\sinphi }\tdplotmult {\stcp }{\sintheta }{\cosphi }\tdplotmult {\ctsp }{\costheta }{\sinphi }\tdplotmult {\ctcp }{\costheta }{\cosphi }\pgfmathsetmacro {\raarot }{\cosphi }\pgfmathsetmacro {\rabrot }{\sinphi }\pgfmathsetmacro {\racrot }{0}\pgfmathsetmacro {\rbarot }{-\ctsp }\pgfmathsetmacro {\rbbrot }{\ctcp }\pgfmathsetmacro {\rbcrot }{\sintheta }\pgfmathsetmacro {\rcarot }{\stsp }\pgfmathsetmacro {\rcbrot }{-\stcp }\pgfmathsetmacro {\rccrot }{\costheta }} \newcommand {\tdplotcalctransformrotmain }[0]{\tdplotsinandcos {\sinalpha }{\cosalpha }{\tdplotalpha } \tdplotsinandcos {\sinbeta }{\cosbeta }{\tdplotbeta } \tdplotsinandcos {\singamma }{\cosgamma }{\tdplotgamma } \tdplotmult {\sasb }{\sinalpha }{\sinbeta } \tdplotmult {\sbsg }{\sinbeta }{\singamma } \tdplotmult {\sasg }{\sinalpha }{\singamma } \tdplotmult {\sasbsg }{\sasb }{\singamma } \tdplotmult {\sacb }{\sinalpha }{\cosbeta } \tdplotmult {\sacg }{\sinalpha }{\cosgamma } \tdplotmult {\sbcg }{\sinbeta }{\cosgamma } \tdplotmult {\sacbsg }{\sacb }{\singamma } \tdplotmult {\sacbcg }{\sacb }{\cosgamma } \tdplotmult {\casb }{\cosalpha }{\sinbeta } \tdplotmult {\cacb }{\cosalpha }{\cosbeta } \tdplotmult {\cacg }{\cosalpha }{\cosgamma } \tdplotmult {\casg }{\cosalpha }{\singamma } \tdplotmult {\cacbsg }{\cacb }{\singamma } \tdplotmult {\cacbcg }{\cacb }{\cosgamma } \pgfmathsetmacro {\raaeul }{\cacbcg -\sasg } \pgfmathsetmacro {\rabeul }{-\cacbsg -\sacg } \pgfmathsetmacro {\raceul }{\casb } \pgfmathsetmacro {\rbaeul }{\sacbcg + \casg } \pgfmathsetmacro {\rbbeul }{-\sacbsg + \cacg } \pgfmathsetmacro {\rbceul }{\sasb } \pgfmathsetmacro {\rcaeul }{-\sbcg } \pgfmathsetmacro {\rcbeul }{\sbsg } \pgfmathsetmacro {\rcceul }{\cosbeta } } \newcommand {\tdplotcalctransformmainrot }[0]{\tdplotsinandcos {\sinalpha }{\cosalpha }{\tdplotalpha } \tdplotsinandcos {\sinbeta }{\cosbeta }{\tdplotbeta } \tdplotsinandcos {\singamma }{\cosgamma }{\tdplotgamma } \tdplotmult {\sasb }{\sinalpha }{\sinbeta } \tdplotmult {\sbsg }{\sinbeta }{\singamma } \tdplotmult {\sasg }{\sinalpha }{\singamma } \tdplotmult {\sasbsg }{\sasb }{\singamma } \tdplotmult {\sacb }{\sinalpha }{\cosbeta } \tdplotmult {\sacg }{\sinalpha }{\cosgamma } \tdplotmult {\sbcg }{\sinbeta }{\cosgamma } \tdplotmult {\sacbsg }{\sacb }{\singamma } \tdplotmult {\sacbcg }{\sacb }{\cosgamma } \tdplotmult {\casb }{\cosalpha }{\sinbeta } \tdplotmult {\cacb }{\cosalpha }{\cosbeta } \tdplotmult {\cacg }{\cosalpha }{\cosgamma } \tdplotmult {\casg }{\cosalpha }{\singamma } \tdplotmult {\cacbsg }{\cacb }{\singamma } \tdplotmult {\cacbcg }{\cacb }{\cosgamma } \pgfmathsetmacro {\raaeul }{\cacbcg -\sasg } \pgfmathsetmacro {\rabeul }{\sacbcg + \casg } \pgfmathsetmacro {\raceul }{-\sbcg } \pgfmathsetmacro {\rbaeul }{-\cacbsg -\sacg } \pgfmathsetmacro {\rbbeul }{-\sacbsg + \cacg } \pgfmathsetmacro {\rbceul }{\sbsg } \pgfmathsetmacro {\rcaeul }{\casb } \pgfmathsetmacro {\rcbeul }{\sasb } \pgfmathsetmacro {\rcceul }{\cosbeta } } \newcommand {\tdplottransformmainrot }[3]{\tdplotcalctransformmainrot \par \pgfmathsetmacro {\tdplotresx }{\raaeul * #1 + \rabeul * #2 + \raceul * #3} \pgfmathsetmacro {\tdplotresy }{\rbaeul * #1 + \rbbeul * #2 + \rbceul * #3} \pgfmathsetmacro {\tdplotresz }{\rcaeul * #1 + \rcbeul * #2 + \rcceul * #3} } \newcommand {\tdplottransformrotmain }[3]{\tdplotcalctransformrotmain \par \pgfmathsetmacro {\tdplotresx }{\raaeul * #1 + \rabeul * #2 + \raceul * #3} \pgfmathsetmacro {\tdplotresy }{\rbaeul * #1 + \rbbeul * #2 + \rbceul * #3} \pgfmathsetmacro {\tdplotresz }{\rcaeul * #1 + \rcbeul * #2 + \rcceul * #3} } \newcommand {\tdplottransformmainscreen }[3]{\tdplotcalctransformmainscreen \par \pgfmathsetmacro {\tdplotresx }{\raarot * #1 + \rabrot * #2 + \racrot * #3} \pgfmathsetmacro {\tdplotresy }{\rbarot * #1 + \rbbrot * #2 + \rbcrot * #3} } \newcommand {\tdplotsetrotatedcoords }[3]{\pgfmathsetmacro {\tdplotalpha }{#1} \pgfmathsetmacro {\tdplotbeta }{#2} \pgfmathsetmacro {\tdplotgamma }{#3} \tdplotcalctransformrotmain \par \tdplotmult {\raaeaa }{\raarot }{\raaeul } \tdplotmult {\rabeba }{\rabrot }{\rbaeul } \tdplotmult {\raceca }{\racrot }{\rcaeul } \tdplotmult {\raaeab }{\raarot }{\rabeul } \tdplotmult {\rabebb }{\rabrot }{\rbbeul } \tdplotmult {\racecb }{\racrot }{\rcbeul } \tdplotmult {\raaeac }{\raarot }{\raceul } \tdplotmult {\rabebc }{\rabrot }{\rbceul } \tdplotmult {\racecc }{\racrot }{\rcceul } \tdplotmult {\rbaeaa }{\rbarot }{\raaeul } \tdplotmult {\rbbeba }{\rbbrot }{\rbaeul } \tdplotmult {\rbceca }{\rbcrot }{\rcaeul } \tdplotmult {\rbaeab }{\rbarot }{\rabeul } \tdplotmult {\rbbebb }{\rbbrot }{\rbbeul } \tdplotmult {\rbcecb }{\rbcrot }{\rcbeul } \tdplotmult {\rbaeac }{\rbarot }{\raceul } \tdplotmult {\rbbebc }{\rbbrot }{\rbceul } \tdplotmult {\rbcecc }{\rbcrot }{\rcceul } \pgfmathsetmacro {\raarc }{\raaeaa + \rabeba + \raceca } \pgfmathsetmacro {\rabrc }{\raaeab + \rabebb + \racecb } \pgfmathsetmacro {\racrc }{\raaeac + \rabebc + \racecc } \pgfmathsetmacro {\rbarc }{\rbaeaa + \rbbeba + \rbceca } \pgfmathsetmacro {\rbbrc }{\rbaeab + \rbbebb + \rbcecb } \pgfmathsetmacro {\rbcrc }{\rbaeac + \rbbebc + \rbcecc } \tikzset {tdplot_rotated_coords/.append style={x={(\raarc cm,\rbarc cm)},y={(\rabrc cm,\rbbrc cm)},z={(\racrc cm,\rbcrc cm)}}}} \newcommand {\tdplotsetrotatedcoordsorigin }[1]{\tikzset {tdplot_rotated_coords/.append style={shift=#1}}} \newcommand {\tdplotresetrotatedcoordsorigin }[0]{\tikzset {tdplot_rotated_coords/.append style={shift={(0,0,0)}}}} \newcommand {\tdplotsetthetaplanecoords }[1]{\tdplotresetrotatedcoordsorigin \tdplotsetrotatedcoords {270 + #1}{270}{0}} \newcommand {\tdplotsetrotatedthetaplanecoords }[1]{\tdplotsetrotatedcoords {\tdplotalpha }{\tdplotbeta }{\tdplotgamma + #1}\tikzset {tdplot_rotated_coords/.append style={y={(\raarc cm,\rbarc cm)},z={(\rabrc cm,\rbbrc cm)},x={(\racrc cm,\rbcrc cm)}}}} \newcommand {\tdplotsetcoord }[4]{\tdplotsinandcos {\sinthetavec }{\costhetavec }{#3}\tdplotsinandcos {\sinphivec }{\cosphivec }{#4}\tdplotmult {\stcpv }{\sinthetavec }{\cosphivec }\tdplotmult {\stspv }{\sinthetavec }{\sinphivec }\coordinate (#1) at ($#2*(\stcpv ,\stspv ,\costhetavec )$); \coordinate (#1xy) at ($#2*(\stcpv ,\stspv ,0)$); \coordinate (#1xz) at ($#2*(\stcpv ,0,\costhetavec )$); \coordinate (#1yz) at ($#2*(0,\stspv ,\costhetavec )$); \coordinate (#1x) at ($#2*(\stcpv ,0,0)$); \coordinate (#1y) at ($#2*(0,\stspv ,0)$); \coordinate (#1z) at ($#2*(0,0,\costhetavec )$); } \newcommand {\tdplotsimplesetcoord }[4]{\tdplotsinandcos {\sinthetavec }{\costhetavec }{#3}\tdplotsinandcos {\sinphivec }{\cosphivec }{#4}\tdplotmult {\stcpv }{\sinthetavec }{\cosphivec }\tdplotmult {\stspv }{\sinthetavec }{\sinphivec }\coordinate (#1) at ($#2*(\stcpv ,\stspv ,\costhetavec )$); } \newcommand {\tdplotsetpolarplotrange }[4]{\pgfmathsetmacro {\tdplotlowerphi }{#3} \pgfmathsetmacro {\tdplotupperphi }{#4} \pgfmathsetmacro {\tdplotlowertheta }{#1} \pgfmathsetmacro {\tdplotuppertheta }{#2} } \newcommand {\tdplotresetpolarplotrange }[0]{\pgfmathsetmacro {\tdplotlowerphi }{0} \pgfmathsetmacro {\tdplotupperphi }{360} \pgfmathsetmacro {\tdplotlowertheta }{0} \pgfmathsetmacro {\tdplotuppertheta }{180} } \newcommand {\tdplotdosurfaceplot }[6]{\par \pgfmathsetmacro {\nextphi }{\curphi + \tdplotsuperfudge *\viewphistep } \par \begin {scope}[opacity=1] \par \par \tdplotcheckdiff {\nextphi }{360}{\origviewphistep }{#2}{} \tdplotcheckdiff {\nextphi }{0}{\origviewphistep }{#2}{} \par \tdplotcheckdiff {\nextphi }{90}{\origviewphistep }{#3}{} \tdplotcheckdiff {\nextphi }{450}{\origviewphistep }{#3}{} \end {scope} \par \foreach \curtheta in{\viewthetastart ,\viewthetainc ,...,\viewthetaend } { \par \pgfmathsetmacro {\curlongitude }{90 -\curphi } \pgfmathsetmacro {\curlatitude }{90 -\curtheta } \par \ifthenelse {\equal {\leftright }{-1.0}}{\pgfmathsetmacro {\curphi }{\curphi -\origviewphistep } }{} \par \pgfmathsetmacro {\tdplottheta }{mod(\curtheta ,360)} \pgfmathsetmacro {\tdplotphi }{mod(\curphi ,360)} \par \pgfmathparse {\tdplotphi <0} \ifthenelse {\equal {\pgfmathresult }{1}}{ \pgfmathsetmacro {\tdplotphi }{\tdplotphi + 360} }{}\par \pgfmathparse {\tdplottheta >\tdplotuppertheta } \pgfmathsetmacro {\logictest }{1 -\pgfmathresult } \par \pgfmathparse {\tdplottheta <\tdplotlowertheta } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathsetmacro {\tdplottheta }{\tdplottheta + \viewthetastep } \pgfmathparse {\tdplottheta >\tdplotuppertheta } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplottheta <\tdplotlowertheta } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplotphi >\tdplotupperphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplotphi <\tdplotlowerphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathsetmacro {\tdplotphi }{\tdplotphi + \viewphistep } \par \pgfmathparse {\tdplotphi <0} \ifthenelse {\equal {\pgfmathresult }{1}}{ \pgfmathsetmacro {\tdplotphi }{\tdplotphi + 360} }{}\par \pgfmathparse {\tdplotphi >\tdplotupperphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \pgfmathparse {\tdplotphi <\tdplotlowerphi } \pgfmathsetmacro {\logictest }{\logictest * (1 -\pgfmathresult )} \par \par \pgfmathsetmacro {\tdplottheta }{\curtheta } \pgfmathsetmacro {\tdplotphi }{\curphi } \par \ifthenelse {\equal {#6}{parametricfill}}{\ifthenelse {\equal {\logictest }{1.0}}{\pgfmathsetmacro {\radius }{#1} \pgfmathsetmacro {\tdplotr }{\radius *360} \par \pgfmathlessthan {\radius }{0} \pgfmathsetmacro {\phaseshift }{180 * \pgfmathresult } \par \pgfmathsetmacro {\colorarg }{#5} \pgfmathsetmacro {\colorarg }{\colorarg + \phaseshift } \pgfmathsetmacro {\colorarg }{mod(\colorarg ,360)} \par \pgfmathlessthan {\colorarg }{0} \pgfmathsetmacro {\colorarg }{\colorarg + 360*\pgfmathresult } \par \pgfmathdivide {\colorarg }{360} \definecolor {tdplotfillcolor}{hsb}{\pgfmathresult ,1,1} \color {tdplotfillcolor} }{}}{\pgfsetfillcolor {#5} } \pgfsetstrokecolor {#4} \par \ifthenelse {\equal {\leftright }{-1.0}}{\pgfmathsetmacro {\curphi }{\curphi + \origviewphistep } }{} \par \ifthenelse {\equal {\logictest }{1.0}}{\pgfmathsetmacro {\radius }{abs(#1)} \pgfpathmoveto {\pgfpointspherical {\curlongitude }{\curlatitude }{\radius }} \par \pgfmathsetmacro {\tdplotphi }{\curphi + \viewphistep } \pgfmathsetmacro {\radius }{abs(#1)} \pgfpathlineto {\pgfpointspherical {\curlongitude -\viewphistep }{\curlatitude }{\radius }} \par \pgfmathsetmacro {\tdplottheta }{\curtheta + \viewthetastep } \pgfmathsetmacro {\radius }{abs(#1)} \pgfpathlineto {\pgfpointspherical {\curlongitude -\viewphistep }{\curlatitude -\viewthetastep }{\radius }} \par \pgfmathsetmacro {\tdplotphi }{\curphi } \pgfmathsetmacro {\radius }{abs(#1)} \pgfpathlineto {\pgfpointspherical {\curlongitude }{\curlatitude -\viewthetastep }{\radius }} \pgfpathclose \par \pgfusepath {fill,stroke} }{} } } \newcommand {\tdplotshowargcolorguide }[4]{ \par \pgfmathsetmacro {\tdplotx }{#1} \pgfmathsetmacro {\tdploty }{#2} \pgfmathsetmacro {\tdplothuestep }{5} \pgfmathsetmacro {\tdplotxsize }{#3} \pgfmathsetmacro {\tdplotysize }{#4} \par \pgfmathsetmacro {\tdplotyscale }{\tdplotysize /360} \par \foreach \tdplotphi in {0,\tdplothuestep ,...,360} { \pgfmathdivide {\tdplotphi }{360} \definecolor {tdplotfillcolor}{hsb}{\pgfmathresult ,1,1} \color {tdplotfillcolor} \par \pgfmathsetmacro {\tdplotstarty }{\tdploty + \tdplotphi * \tdplotyscale } \pgfmathsetmacro {\tdplotstopy }{\tdplotstarty + \tdplothuestep * \tdplotyscale } \pgfmathsetmacro {\tdplotstartx }{\tdplotx } \pgfmathsetmacro {\tdplotstopx }{\tdplotx + \tdplotxsize } \filldraw [tdplot_screen_coords] (\tdplotstartx ,\tdplotstarty ) rectangle (\tdplotstopx ,\tdplotstopy ); } \par \pgfmathsetmacro {\tdplotstopy }{\tdploty + (360+\tdplothuestep )*\tdplotyscale } \pgfmathsetmacro {\tdplotstopx }{\tdplotx + \tdplotxsize } \par \draw [tdplot_screen_coords] (\tdplotx ,\tdploty ) rectangle (\tdplotstopx ,\tdplotstopy ); \par \node [tdplot_screen_coords,anchor=west,xshift=5pt] at (\tdplotstopx ,\tdploty ) {$0$}; \node [tdplot_screen_coords,anchor=west,xshift=5pt] at (\tdplotstopx ,\tdplotstopy ) {$2\pi $}; \par \pgfmathsetmacro {\tdplotstopy }{\tdploty + (360+\tdplothuestep )/2*\tdplotyscale } \node [tdplot_screen_coords,anchor=west,xshift=5pt] at (\tdplotstopx ,\tdplotstopy ) {$\pi $}; } \newcommand {\tdplotgetpolarcoords }[3]{\pgfmathsetmacro {\vxcalc }{#1} \pgfmathsetmacro {\vycalc }{#2} \pgfmathsetmacro {\vzcalc }{#3} \pgfmathsetmacro {\vcalc }{ sqrt((\vxcalc )^2 + (\vycalc )^2 + (\vzcalc )^2) } \par \pgfmathsetmacro {\vxycalc }{ sqrt((\vxcalc )^2 + (\vycalc )^2) } \par \pgfmathsetmacro {\tdplotrestheta }{asin(\vxycalc /\vcalc )} \pgfmathparse {\vzcalc <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotrestheta }{180 -\tdplotrestheta } } {} \ifthenelse {\equal {\vxcalc }{0.0}}{\pgfmathparse {\vycalc <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{270} } {\pgfmathparse {\vycalc >0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{90} } {\pgfmathsetmacro {\tdplotresphi }{0} } } } {\pgfmathsetmacro {\tdplotresphi }{atan(\vycalc /\vxcalc )} \pgfmathparse {\vxcalc <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{\tdplotresphi +180} } { } \par \pgfmathparse {\tdplotresphi <0} \ifthenelse {\equal {\pgfmathresult }{1}}{\pgfmathsetmacro {\tdplotresphi }{\tdplotresphi +360} } {} } } \newcommand {\vec }[0]{\mathbf } \newcommand {\RR }[0]{\mathbb {R}} \newcommand {\dfn }[0]{\textit } \newcommand {\dotp }[0]{\cdot } \newcommand {\id }[0]{\text {id}} \newcommand {\norm }[1]{\left \lVert #1\right \rVert } \newcommand {\mathtoolsset }[1]{\setkeys {\MT_options_name: }{#1}} \newcommand {\refeq }[1]{\textup {\ref {#1}}} \newcommand {\lparen }[0]{(} \newcommand {\rparen }[0]{)} \newcommand {\ordinarycolon }[0]{:} \newcommand {\MT_test_for_tcb_other:nnnnn }[1]{\if:w t#1\relax \expandafter \MH_use_choice_i:nnnn \else: \if:w c#1\relax \expandafter \expandafter \expandafter \MH_use_choice_ii:nnnn \else: \if:w b#1\relax \expandafter \expandafter \expandafter \expandafter \expandafter \expandafter \expandafter \MH_use_choice_iii:nnnn \else: \expandafter \expandafter \expandafter \expandafter \expandafter \expandafter \expandafter \MH_use_choice_iv:nnnn \fi: \fi: \fi: } \newcommand {\newcases }[6]{\newenvironment {#1}{\MT_start_cases:nnnn {#2}{#3}{#4}{#5}}{\MH_end_cases: \right #6}} \newcommand {\renewcases }[6]{\renewenvironment {#1}{\MT_start_cases:nnnn {#2}{#3}{#4}{#5}}{\MH_end_cases: \right #6}} \newcommand {\SwapAboveDisplaySkip }[0]{\noalign {\vskip -\abovedisplayskip \vskip \abovedisplayshortskip }} \newcommand {\vdotswithin }[1]{{\mathmakebox [\widthof {\ensuremath {{}#1{}}}][c]{{\vdots }}}} \newcommand {\MTFlushSpaceBelow }[0]{\\\noalign {\nobreak \vskip -\lineskip \vskip -\l_MT_shortvdotswithinadjustbelow_dim \vskip -\origjot \vskip \jot }} \newcommand {\mathmbox }[0]{\mathpalette \MT_mathmbox:nn } \newcommand {\prescript }[3]{\mathchoice {\MT_prescript_inner: {#1}{#2}{#3}{\scriptstyle }}{\MT_prescript_inner: {#1}{#2}{#3}{\scriptstyle }}{\MT_prescript_inner: {#1}{#2}{#3}{\scriptscriptstyle }}{\MT_prescript_inner: {#1}{#2}{#3}{\scriptscriptstyle }}} \newcommand {\spreadlines }[1]{\setlength {\jot }{#1}\ignorespaces } \newcommand {\newgathered }[4]{\newenvironment {#1}{\def \MT_gathered_pre: {#2}\def \MT_gathered_post: {#3}\def \MT_gathered_env_end: {#4}\MT_gathered_env }{\endMT_gathered_env }} \newcommand {\renewgathered }[4]{\renewenvironment {#1}{\def \MT_gathered_pre: {#2}\def \MT_gathered_post: {#3}\def \MT_gathered_env_end: {#4}\MT_gathered_env }{\endMT_gathered_env }} \newcommand {\lgathered }[0]{\def \MT_gathered_pre: {}\def \MT_gathered_post: {\hfil }\def \MT_gathered_env_end: {}\MT_gathered_env } \newcommand {\rgathered }[0]{\def \MT_gathered_pre: {\hfil }\def \MT_gathered_post: {}\def \MT_gathered_env_end: {}\MT_gathered_env } \newcommand {\gathered }[0]{\def \MT_gathered_pre: {\hfil }\def \MT_gathered_post: {\hfil }\def \MT_gathered_env_end: {}\MT_gathered_env } \newcommand {\splitfrac }[2]{\genfrac {}{}{0pt}{1}{\textstyle #1\quad \hfill }{\textstyle \hfill \quad \mathstrut #2}} \newcommand {\splitdfrac }[2]{\genfrac {}{}{0pt}{0}{#1\quad \hfill }{\hfill \quad \mathstrut #2}} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\dblcolon }[0]{\vcentcolon \mathrel {\mkern -.9mu}\vcentcolon } \newcommand {\coloneqq }[0]{\vcentcolon \mathrel {\mkern -1.2mu}=} \newcommand {\Coloneqq }[0]{\dblcolon \mathrel {\mkern -1.2mu}=} \newcommand {\coloneq }[0]{\vcentcolon \mathrel {\mkern -1.2mu}\mathrel {-}} \newcommand {\Coloneq }[0]{\dblcolon \mathrel {\mkern -1.2mu}\mathrel {-}} \newcommand {\eqqcolon }[0]{=\mathrel {\mkern -1.2mu}\vcentcolon } \newcommand {\Eqqcolon }[0]{=\mathrel {\mkern -1.2mu}\dblcolon } \newcommand {\eqcolon }[0]{\mathrel {-}\mathrel {\mkern -1.2mu}\vcentcolon } \newcommand {\Eqcolon }[0]{\mathrel {-}\mathrel {\mkern -1.2mu}\dblcolon } \newcommand {\colonapprox }[0]{\vcentcolon \mathrel {\mkern -1.2mu}\approx } \newcommand {\Colonapprox }[0]{\dblcolon \mathrel {\mkern -1.2mu}\approx } \newcommand {\colonsim }[0]{\vcentcolon \mathrel {\mkern -1.2mu}\sim } \newcommand {\Colonsim }[0]{\dblcolon \mathrel {\mkern -1.2mu}\sim } \newcommand {\nuparrow }[0]{\MH_nuparrow: } \newcommand {\ndownarrow }[0]{\MH_ndownarrow: } \newcommand {\bigtimes }[0]{\MH_csym_bigtimes: }$

Additional Exercises for Chapter 10: Abstract Vector Spaces

Suppose you have $\mathbb {R}^{2}$ and the $+$ operation is as follows: $\begin{equation*} \left( a,b\right) +\left( c,d\right) =\left( a+d,b+c\right) . \end{equation*}$ Scalar multiplication is defined in the usual way. Is this a vector space? Explain why or why not.

Suppose you have $\mathbb {R}^{2}$ and the $+$ operation is defined as follows. $\begin{equation*} \left( a,b\right) +\left( c,d\right) =\left( 0,b+d\right) \end{equation*}$ Scalar multiplication is defined in the usual way. Is this a vector space? Explain why or why not.

Suppose you have $\mathbb {R}^{2}$ and scalar multiplication is defined as $c\left ( a,b\right ) =\left ( a,cb\right )$ while vector addition is defined as usual. Is this a vector space? Explain why or why not.

Suppose you have $\mathbb {R}^{2}$ and the $+$ operation is defined as follows. $\begin{equation*} \left( a,b\right) +\left( c,d\right) =\left( a-c,b-d\right) \end{equation*}$ Scalar multiplication is same as usual. Is this a vector space? Explain why or why not.

Consider all the functions defined on a non empty set which have values in $\mathbb {R}$ . Is this a vector space? Explain. The operations are defined as follows. Here $f,g$ signify functions and $a$ is a scalar.

$\begin{eqnarray*} \left( f+g\right) \left( x\right) &=&f\left( x\right) +g\left( x\right) \\ \left( af\right) \left( x\right) &=&a\left( f\left( x\right) \right) \end{eqnarray*}$

Consider the set of $n\times n$ symmetric matrices. That is, $A=A^{T}.$ In other words, $A_{ij}=A_{ji}$ . Show that this set of symmetric matrices is a vector space and a subspace of the vector space of $n\times n$ matrices.

Consider the set of all vectors $\begin {bmatrix}x\\y\end {bmatrix}$ in $\mathbb {R}^{2}$ such that $x+y\geq 0.$ Let the vector space operations be the usual ones. Is this a vector space? Is it a subspace of $\mathbb {R}^{2}$ ?

Consider the vectors $\begin {bmatrix}x\\y\end {bmatrix}$ in $\mathbb {R}^{2}$ such that $xy=0$ . Is this a subspace of $\mathbb {R}^{2}?$ Is it a vector space? The addition and scalar multiplication are the usual operations.

Define the operation of vector addition on $\mathbb {R}^{2}$ by $\left ( x,y\right ) +\left ( u,v\right ) =\left ( x+u,y+v+1\right ) .$ Let scalar multiplication be the usual operation. Is this a vector space with these operations? Explain.

Let the vectors be real numbers. Define vector space operations in the usual way. That is $x+y$ means to add the two numbers and $xy$ means to multiply them. Is $\mathbb {R}$ with these operations a vector space? Explain.

Let the scalars be the rational numbers and let the vectors be real numbers which are the form $a+b\sqrt {2}$ for $a,b$ rational numbers. Show that with the usual operations, this is a vector space.

Let $\mathbb {P}_{2}$ be the set of all polynomials of degree 2 or less. That is, these are of the form $a+bx+cx^{2}$ . Addition is defined as $\begin{equation*} \left( a+bx+cx^{2}\right) +\left( \hat{a}+\hat{b}x+\hat{c}x^{2}\right) =\left( a+\hat{a}\right) +\left( b+\hat{b}\right) x+\left( c+\hat{c}\right) x^{2} \end{equation*}$ and scalar multiplication is defined as $\begin{equation*} d\left( a+bx+cx^{2}\right) =da+dbx+cdx^{2} \end{equation*}$ Show that, with this definition of the vector space operations that $\mathbb { P}_{2}$ is a vector space. Now let $V$ denote those polynomials $a+bx+cx^{2}$ such that $a+b+c=0$ . Is $V$ a subspace of $\mathbb {P}_{2}?$ Explain.

Let $M,N$ be subspaces of a vector space $V$ and consider $M+N$ defined as the set of all $m+n$ where $m\in M$ and $n\in N$ . Show that $M+N$ is a subspace of $V$ .

Let $M,N$ be subspaces of a vector space $V$ . Then $M\cap N$ consists of all vectors which are in both $M$ and $N$ . Show that $M\cap N$ is a subspace of $V$ .

Let $M,N$ be subspaces of a vector space $\mathbb {R}^{2}.$ Then $N\cup M$ consists of all vectors which are in either $M$ or $N$ . Show that $N\cup M$ is not necessarily a subspace of $\mathbb {R}^{2}$ by giving an example where $N\cup M$ fails to be a subspace.

Let $X$ consist of the real valued functions which are defined on an interval $\left [ a,b\right ] .$ For $f,g\in X,\;f+g$ is the name of the function which satisfies $\left ( f+g\right ) \left ( x\right ) =f\left ( x\right ) +g\left ( x\right )$ . For $s$ a real number, $\left ( s f\right ) \left ( x\right ) = s \left ( f\left ( x\right ) \right )$ . Show this is a vector space.

The axioms of a vector space all hold because they hold for a vector space. The only thing left to verify is the assertions about the things which are supposed to exist. $0$ would be the zero function which sends everything to $0$ . This is an additive identity. Now if $f$ is a function, $-f\left ( x\right ) \equiv \left ( -f\left ( x\right ) \right )$ . Then $\left ( f+\left ( -f\right ) \right ) \left ( x\right ) \equiv f\left ( x\right ) +\left ( -f\right ) \left ( x\right ) \equiv f\left ( x\right ) +\left ( -f\left ( x\right ) \right ) =0$ Hence $f+-f=\mathbf {0}$ . For each $x\in \left [ a,b\right ] ,$ let $f_{x}\left ( x\right ) =1$ and $f_{x}\left ( y\right ) =0$ if $y\neq x.$ Then these vectors are obviously linearly independent.

Let the vectors be polynomials of degree no more than 3. Show that with the usual definitions of scalar multiplication and addition wherein, for $p\left ( x\right )$ a polynomial, $\left ( a p\right ) \left ( x\right ) = a p\left ( x\right )$ and for $p,q$ polynomials $\left ( p+q\right ) \left ( x\right ) = p\left ( x\right ) +q\left ( x\right ) ,$ this is a vector space.

This is just a subspace of the vector space of functions because it is closed with respect to vector addition and scalar multiplication. Hence this is a vector space.

Let $p(x) = 4x^2-x$ . Is $p(x)$ in $\mbox {span} \left ( x^2+x, x^2-1, -x + 2 \right )$

Let $p(x) = - x^2 + x + 2$ . Is $p(x)$ in $\mbox {span} \left ( x^2 + x + 1, 2x^2 + x \right )$ ?

Let $A = \left [ \begin {array}{rr} 1 & 3 \\ 0 & 0 \end {array} \right ]$ . Is $A$ in $\mbox {span} \left ( \left [ \begin {array}{rr} 1 & 0 \\ 0 & 1 \end {array} \right ], \left [ \begin {array}{rr} 0 & 1 \\ 1 & 0 \end {array} \right ], \left [ \begin {array}{rr} 1 & 0 \\ 1 & 1 \end {array} \right ], \left [ \begin {array}{rr} 0 & 1 \\ 1 & 1 \end {array} \right ] \right )?$

Show that the spanning set in Question prb:10.26 is a spanning set for $\mathbb {M}_{22}$ , the vector space of all $2 \times 2$ matrices.

Consider the vector space of polynomials of degree at most $2,$ $\mathbb {P}_{2}$ . Determine whether the following is a basis for $\mathbb {P}_{2}$ . $\begin{equation*} \left\{ x^{2}+x+1,2x^{2}+2x+1,x+1\right\} \end{equation*}$ Hint: There is a isomorphism from $\mathbb {R}^{3}$ to $\mathbb {P} _{2}$ . It is defined as follows: $\begin{equation*} T(\vec{e}_{1})=1,\quad T(\vec{e}_{2})=x,\quad T(\vec{e}_{3})=x^{2} \end{equation*}$ Then extend $T$ linearly. Thus $\begin{equation*} T\left(\left[ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right]\right) =x^{2}+x+1,\quad T\left(\left[ \begin{array}{c} 1 \\ 2 \\ 2 \end{array} \right]\right) =2x^{2}+2x+1,\quad T\left(\left[ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right]\right) =1+x \end{equation*}$ It follows that if $\begin{equation*} \left\{ \left[ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 1 \\ 2 \\ 2 \end{array} \right] ,\left[ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right] \right\} \end{equation*}$ is a basis for $\mathbb {R}^{3},$ then the polynomials will be a basis for $\mathbb {P}_{2}$ because they will be independent. Recall that an isomorphism takes a linearly independent set to a linearly independent set. Also, since $T$ is an isomorphism, it preserves all linear relations.

Find a basis in $\mathbb {P}_{2}$ for the subspace $\begin{equation*} \mbox{span}\left( 1+x+x^{2},1+2x,1+5x-3x^{2}\right) \end{equation*}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

This is the situation in which you have a spanning set and you want to cut it down to form a linearly independent set which is also a spanning set. Use the same isomorphism above. Since $T$ is an isomorphism, it preserves all linear relations so if such can be found in $\mathbb {R}^{3},$ the same linear relations will be present in $\mathbb {P}_{2}$ .

Find a basis in $\mathbb {P}_{3}$ for the subspace $\begin{equation*} \mbox{span}\left( 1+x-x^{2}+x^{3},1+2x+3x^{3},-1+3x+5x^{2}+7x^{3},1+6x+4x^{2}+11x^{3}\right) \end{equation*}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Find a basis in $\mathbb {P}_{3}$ for the subspace $\begin{equation*} \mbox{span}\left( x^{3}-2x^{2}+x+2,3x^{3}-x^{2}+2x+2,7x^{3}+x^{2}+4x+2,5x^{3}+3x+2\right) \end{equation*}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Find a basis in $\mathbb {P}_{3}$ for the subspace $\begin{equation*} \mbox{span}\left( x^{3}+2x^{2}+x-2,3x^{3}+3x^{2}+2x-2,3x^{3}+x+2,3x^{3}+x+2\right) \end{equation*}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Find a basis in $\mathbb {P}_{3}$ for the subspace $\begin{equation*} \mbox{span}\left( x^{3}-5x^{2}+x+5,3x^{3}-4x^{2}+2x+5,5x^{3}+8x^{2}+2x-5,11x^{3}+6x+5\right) \end{equation*}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Find a basis in $\mathbb {P}_{3}$ for the subspace $\begin{equation*} \mbox{span}\left( x^{3}-3x^{2}+x+3,3x^{3}-2x^{2}+2x+3,7x^{3}+7x^{2}+3x-3,7x^{3}+4x+3\right) \end{equation*}$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Here are some vectors. $\begin{equation*} \left\{ x^{3}+x^{2}-x-1,3x^{3}+2x^{2}+2x-1\right\} \end{equation*}$ If these are linearly independent, extend to a basis for all of $\mathbb {P} _{3}$ .

Here are some vectors. $\begin{equation*} \left\{ x^{3}-2x^{2}-x+2,3x^{3}-x^{2}+2x+2\right\} \end{equation*}$ If these are linearly independent, extend to a basis for all of $\mathbb {P} _{3}$ .

Here are some vectors. $\begin{equation*} \left\{ x^{3}-3x^{2}-x+3,3x^{3}-2x^{2}+2x+3\right\} \end{equation*}$ If these are linearly independent, extend to a basis for all of $\mathbb {P} _{3}$ .

Here are some vectors. $\begin{equation*} \left\{ x^{3}-2x^{2}-3x+2,3x^{3}-x^{2}-6x+2,-8x^{3}+18x+10\right\} \end{equation*}$ If these are linearly independent, extend to a basis for all of $\mathbb {P} _{3}$ .

Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. $\left \{ x+1, x^2 + 2, x^2 - x -3 \right \}$

Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. $\left \{ x^2 + x, -2x^2 -4x -6 , 2x - 2 \right \}$

Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. $\left \{ \left [ \begin {array}{rr} 1 & 2 \\ 0 & 1 \end {array} \right ], \left [ \begin {array}{rr} -7 & 2 \\ -2 & -3 \end {array} \right ], \left [ \begin {array}{rr} 4 & 0 \\ 1 & 2 \end {array} \right ] \right \}$

Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. $\left \{ \left [ \begin {array}{rr} 1 & 0 \\ 0 & 1 \end {array} \right ], \left [ \begin {array}{rr} 0 & 1 \\ 0 & 1 \end {array} \right ], \left [ \begin {array}{rr} 1 & 0 \\ 1 & 0 \end {array} \right ], \left [ \begin {array}{rr} 0 & 0 \\ 1 & 1 \end {array} \right ] \right \}$

If you have $5$ vectors in $\mathbb {R}^{5}$ and the vectors are linearly independent, can it always be concluded they span $\mathbb {R}^{5}?$

Yes. If not, there would exist a vector not in the span. But then you could add in this vector and obtain a linearly independent set of vectors with more vectors than a basis.

If you have $6$ vectors in $\mathbb {R}^{5},$ is it possible they are linearly independent? Explain.

No. They can’t be.

Let $\mathbb {P}_3$ be the polynomials of degree no more than 3. Determine which of the following are bases for this vector space.

(a): $\left \{ x+1,x^{3}+x^{2}+2x,x^{2}+x,x^{3}+x^{2}+x\right \}$
(b): $\left \{ x^{3}+1,x^{2}+x,2x^{3}+x^{2},2x^{3}-x^{2}-3x+1\right \}$

(a)
(b): Suppose $c_{1}\left ( x^{3}+1\right ) +c_{2}\left ( x^{2}+x\right ) +c_{3}\left ( 2x^{3}+x^{2}\right ) +c_{4}\left ( 2x^{3}-x^{2}-3x+1\right ) =0$ Then combine the terms according to power of $x.$ $\left ( c_{1}+2c_{3}+2c_{4}\right ) x^{3}+\left ( c_{2}+c_{3}-c_{4}\right ) x^{2}+\left ( c_{2}-3c_{4}\right ) x+\left ( c_{1}+c_{4}\right ) =0$ Is there a non zero solution to the system $\begin {array}{c} c_{1}+2c_{3}+2c_{4}=0 \\ c_{2}+c_{3}-c_{4}=0 \\ c_{2}-3c_{4}=0 \\ c_{1}+c_{4}=0 \end {array}$ Solution is: $c_{1}=0,c_{2}=0,c_{3}=0,c_{4}=0$ Therefore, these are linearly independent.

Define $T:\mathbb {P}^2\rightarrow \mathbb {P}^3$ by $T(p(x))=xp(x)$ .

Find $T(-x^2+2x-4)$ .

Answer: $T(-x^2+2x-4)=\answer {-x^3+2x^2-4x}$

Is $T$ a linear transformation? If so, prove it. If not, give a counterexample.

Let $\vec {v}_1=\begin {bmatrix}1\\3\\0\end {bmatrix},\quad \vec {v}_2=\begin {bmatrix}0\\1\\-2\end {bmatrix}$ $\vec {w}_1=\begin {bmatrix}1\\-1\\4\end {bmatrix},\quad \vec {w}_2=\begin {bmatrix}0\\2\\-1\end {bmatrix}$

Let $V=\text {span}(\vec {v}_1, \vec {v}_2)$ and $W=\text {span}(\vec {w}_1, \vec {w}_2)$ .

Suppose $T:V\rightarrow W$ is a linear transformation such that $T(\vec {v}_1)=\begin {bmatrix}2\\0\\7\end {bmatrix},\quad T(\vec {v}_2)=\begin {bmatrix}-1\\7\\1\end {bmatrix}$

Verify that vectors $\begin {bmatrix}2\\0\\7\end {bmatrix}$ and $\begin {bmatrix}-1\\7\\1\end {bmatrix}$ are in $W$ by expressing each as a linear combination of $\vec {w}_1$ and $\vec {w}_2$ . $\begin {bmatrix}2\\0\\7\end {bmatrix}=\answer {2}\vec {w}_1+\answer {1}\vec {w}_2$ $\begin {bmatrix}-1\\7\\1\end {bmatrix}=\answer {-1}\vec {w}_1+\answer {3}\vec {w}_2$

Show that $\vec {v}=\begin {bmatrix}1\\2\\2\end {bmatrix}$ is in $V$ by expressing it as a linear combination of $\vec {v}_1$ and $\vec {v}_2$ . $\vec {v}=\answer {1}\vec {v}_1+\answer {-1}\vec {v}_2$

Find $T(\vec {v})$ and express it as a linear combination of $\vec {w}_1$ and $\vec {w}_2$ . $T(\vec {v})=\answer {3}\vec {w}_1+\answer {-2}\vec {w}_2$

In the context of the above problem, consider polynomials $\begin{equation*} \left\{ a_{i}x^{3}+b_{i}x^{2}+c_{i}x+d_{i},\ i=1,2,3,4\right\} \end{equation*}$ Show that this collection of polynomials is linearly independent on an interval $\left [ s,t\right ]$ if and only if $\begin{equation*} \left[ \begin{array}{cccc} a_{1} & b_{1} & c_{1} & d_{1} \\ a_{2} & b_{2} & c_{2} & d_{2} \\ a_{3} & b_{3} & c_{3} & d_{3} \\ a_{4} & b_{4} & c_{4} & d_{4} \end{array} \right] \end{equation*}$ is an invertible matrix.

Let $p_{i}\left ( x\right )$ denote the $i^{th}$ of these polynomials. Suppose $\sum _{i}C_{i}p_{i}\left ( x\right ) =0.$ Then collecting terms according to the exponent of $x,$ you need to have

$\begin{eqnarray*} C_{1}a_{1}+C_{2}a_{2}+C_{3}a_{3}+C_{4}a_{4} &=&0 \\ C_{1}b_{1}+C_{2}b_{2}+C_{3}b_{3}+C_{4}b_{4} &=&0 \\ C_{1}c_{1}+C_{2}c_{2}+C_{3}c_{3}+C_{4}c_{4} &=&0 \\ C_{1}d_{1}+C_{2}d_{2}+C_{3}d_{3}+C_{4}d_{4} &=&0 \end{eqnarray*}$

The matrix of coefficients is just the transpose of the above matrix. There exists a non trivial solution if and only if the determinant of this matrix equals 0.

Let $M=\left \{ \vec {u}=\begin {bmatrix}u_{1}\\u_{2}\\u_{3}\\u_{4}\end {bmatrix} \in \mathbb {R}^{4}:\left | u_{1}\right | \leq 4\right \} .$ Is $M$ a subspace of $\mathbb {R}^4$ ?

This is not a subspace. $\left [ \begin {array}{r} 1 \\ 1 \\ 1 \\ 1 \end {array} \right ]$ is in it, but $20\left [ \begin {array}{r} 1 \\ 1 \\ 1 \\ 1 \end {array} \right ]$ is not.

Let $M=\left \{ \vec {u}=\begin {bmatrix} u_{1}\\u_{2}\\u_{3}\\u_{4}\end {bmatrix} \in \mathbb {R}^{4}:\sin \left ( u_{1}\right ) =1\right \} .$ Is $M$ a subspace of $\mathbb {R}^4$ ?

This is not a subspace.

Let $W$ be a subset of $\mathbb {M}_{22}$ given by $W = \left \{ A | A \in \mathbb {M}_{22}, A^T = A \right \}$ In words, $W$ is the set of all symmetric $2 \times 2$ matrices. Is $W$ a subspace of $\mathbb {M}_{22}$ ?

Let $W$ be a subset of $\mathbb {M}_{22}$ given by $W = \left \{ \left [ \begin {array}{rr} a & b \\ c & d \end {array} \right ] | a,b,c,d \in \mathbb {R}, a + b = c + d \right \}$ Is $W$ a subspace of $\mathbb {M}_{22}$ ?

Let $W$ be a subset of $P_3$ given by $W = \left \{ ax^3 + bx^2 + cx + d | a,b,c,d \in \mathbb {R}, d = 0 \right \}$ Is $W$ a subspace of $P_3$ ?

Let $W$ be a subset of $P_3$ given by $W = \left \{ p(x) = ax^3 + bx^2 + cx + d | a,b,c,d \in \mathbb {R}, p(2) = 1 \right \}$ Is $W$ a subspace of $P_3$ ?

Let $T:\mathbb {P}_2 \to \mathbb {R}$ be a linear transformation such that $T(x^2)=1; T(x^2+x)=5; T(x^2+x+1)=-1.$ Find $T(ax^2+bx+c)$ .

By linearity we have $T(x^2)=1$ , $T(x) = T(x^2+x - x^2)= T(x^2+x) - T(x^2)= 5-1=5$ , and $T(1) = T(x^2+x+1 -(x^2+x))=T(x^2+x+1) -T(x^2+x))= -1-5=-6$ .

Thus $T(ax^2+bx+c) = aT(x^2) + bT(x) + cT(1) = a+5b-6c$ .

Consider the following functions $T:\mathbb {R}^{3}\rightarrow \mathbb {R}^{2}.$ Explain why each of these functions $T$ is not linear.

(a): $T\left (\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ]\right ) =\left [ \begin {array}{c} x+2y+3z+1 \\ 2y-3x+z \end {array} \right ]$
(b): $T\left (\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ]\right ) =\left [ \begin {array}{c} x+2y^{2}+3z \\ 2y+3x+z \end {array} \right ]$
(c): $T\left (\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ]\right ) =\left [ \begin {array}{c} \sin x+2y+3z \\ 2y+3x+z \end {array} \right ]$
(d): $T\left (\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ]\right ) =\left [ \begin {array}{c} x+2y+3z \\ 2y+3x-\ln z \end {array} \right ]$

Suppose $T$ is a linear transformation such that

$\begin{eqnarray*} T\left(\left[ \begin{array}{r} 1 \\ 1 \\ -7 \end{array} \right]\right) &=&\left[ \begin{array}{r} 3 \\ 3 \\ 3 \end{array} \right] \\ T\left(\left[ \begin{array}{r} -1 \\ 0 \\ 6 \end{array} \right]\right) &=&\left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] \\ T\left(\left[ \begin{array}{r} 0 \\ -1 \\ 2 \end{array} \right]\right) &=&\left[ \begin{array}{r} 1 \\ 3 \\ -1 \end{array} \right] \end{eqnarray*}$

Find the matrix of $T$ . That is find $A$ such that $T(\vec {x})=A\vec {x}$ .

$\left [ \begin {array}{rrr} 3 & 1 & 1 \\ 3 & 2 & 3 \\ 3 & 3 & -1 \end {array} \right ] \left [ \begin {array}{ccc} 6 & 2 & 1 \\ 5 & 2 & 1 \\ 6 & 1 & 1 \end {array} \right ] =\left [ \begin {array}{ccc} 29 & 9 & 5 \\ 46 & 13 & 8 \\ 27 & 11 & 5 \end {array} \right ]$

Suppose $T$ is a linear transformation such that

$\begin{eqnarray*} T\left(\left[ \begin{array}{r} 1 \\ 2 \\ -18 \end{array} \right]\right) &=&\left[ \begin{array}{r} 5 \\ 2 \\ 5 \end{array} \right] \\ T\left(\left[ \begin{array}{r} -1 \\ -1 \\ 15 \end{array} \right]\right) &=&\left[ \begin{array}{r} 3 \\ 3 \\ 5 \end{array} \right] \\ T\left(\left[ \begin{array}{r} 0 \\ -1 \\ 4 \end{array} \right]\right) &=&\left[ \begin{array}{r} 2 \\ 5 \\ -2 \end{array} \right] \end{eqnarray*}$

Find the matrix of $T$ . That is find $A$ such that $T(\vec {x})=A\vec {x}$ .

$\left [ \begin {array}{rrr} 5 & 3 & 2 \\ 2 & 3 & 5 \\ 5 & 5 & -2 \end {array} \right ] \left [ \begin {array}{ccc} 11 & 4 & 1 \\ 10 & 4 & 1 \\ 12 & 3 & 1 \end {array} \right ] =\left [ \begin {array}{ccc} 109 & 38 & 10 \\ 112 & 35 & 10 \\ 81 & 34 & 8 \end {array} \right ]$

Consider the following functions $T:\mathbb {R}^{3}\rightarrow \mathbb {R}^{2}$ . Show that each is a linear transformation and determine for each the matrix $A$ such that $T(\vec {x})=A\vec {x}$ .

(a): $T\left (\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ]\right ) =\left [ \begin {array}{c} x+2y+3z \\ 2y-3x+z \end {array} \right ]$
(b): $T\left (\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ]\right ) =\left [ \begin {array}{c} 7x+2y+z \\ 3x-11y+2z \end {array} \right ]$
(c): $T\left (\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ]\right ) =\left [ \begin {array}{c} 3x+2y+z \\ x+2y+6z \end {array} \right ]$
(d): $T\left (\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ]\right ) =\left [ \begin {array}{c} 2y-5x+z \\ x+y+z \end {array} \right ]$

Let $V$ and $W$ be subspaces of $\mathbb {R}^{n}$ and $\mathbb {R}^{m}$ respectively and let $T:V\rightarrow W$ be a linear transformation. Suppose that $\left \{ T(\vec {v}_{1}),\cdots ,T(\vec {v}_{r})\right \}$ is linearly independent. Show that it must be the case that $\left \{ \vec {v}_{1},\cdots , \vec {v}_{r}\right \}$ is also linearly independent.

If $\sum _i^r a_i \vec {v}_r =0$ , then using linearity properties of $T$ we get $0 = T(0) = T\left (\sum _i^r a_i \vec {v}_r\right ) = \sum _i^r a_i T(\vec {v}_r).$ Since we assume that $\left \{ T\vec {v}_{1},\cdots ,T\vec {v}_{r}\right \}$ is linearly independent, we must have all $a_i=0$ , and therefore we conclude that $\left \{ \vec {v}_{1},\cdots , \vec {v}_{r}\right \}$ is also linearly independent.

Let $\begin{equation*} V=\mbox{span}\left\{ \left[ \begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 1 \\ 1 \\ 0 \\ 1 \end{array} \right] \right\} \end{equation*}$ Let $T\vec {x}=A\vec {x}$ where $A$ is the matrix $\begin{equation*} \left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 1 & 1 & 1 & 2 \end{array} \right] \end{equation*}$ Give a basis for $\mbox {im}\left ( T\right )$ .

Let $\begin{equation*} V=\mbox{span}\left\{ \left[ \begin{array}{c} 1 \\ 0 \\ 0 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array} \right] ,\left[ \begin{array}{c} 1 \\ 4 \\ 4 \\ 1 \end{array} \right] \right\} \end{equation*}$ Let $T\vec {x}=A\vec {x}$ where $A$ is the matrix $\begin{equation*} \left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 1 & 1 & 1 & 2 \end{array} \right] \end{equation*}$ Find a basis for $\mbox {im}\left ( T\right )$ . In this case, the original vectors do not form an independent set.

Since the third vector is a linear combinations of the first two, then the image of the third vector will also be a linear combinations of the image of the first two. However the image of the first two vectors are linearly independent (check!), and hence form a basis of the image.

Thus a basis for $\mbox {im}\left ( T\right )$ is:

$\begin{equation*} V=\mbox{span}\left\{ \left[ \begin{array}{c} 2 \\ 0 \\ 1 \\ 3 \end{array} \right] ,\left[ \begin{array}{c} 4 \\ 2 \\ 4 \\ 5 \end{array} \right] \right\} \end{equation*}$

If $\left \{ \vec {v}_{1},\cdots ,\vec {v}_{r}\right \}$ is linearly independent and $T$ is a one to one linear transformation, show that $\left \{ T(\vec {v}_{1}),\cdots ,T(\vec {v}_{r})\right \}$ is also linearly independent. Give an example which shows that if $T$ is only linear, it can happen that, although $\left \{ \vec {v}_{1},\cdots ,\vec {v}_{r}\right \}$ is linearly independent, $\left \{ T\vec {v}_{1},\cdots ,T\vec {v}_{r}\right \}$ is not. In fact, show that it can happen that each of the $T\vec {v}_{j}$ equals 0.

Let $V$ and $W$ be subspaces of $\mathbb {R}^{n}$ and $\mathbb {R}^{m}$ respectively and let $T:V\rightarrow W$ be a linear transformation. Show that if $T$ is onto $W$ and if $\left \{ \vec {v}_{1},\cdots ,\vec {v} _{r}\right \}$ is a basis for $V,$ then $\mbox {span}\left \{ T(\vec {v}) _{1},\cdots ,T(\vec {v}_{r})\right \} =W$ .

Define $T:\mathbb {R}^{4}\rightarrow \mathbb {R}^{3}$ as follows. $\begin{equation*} T(\vec{x})=\left[ \begin{array}{rrrr} 3 & 2 & 1 & 8 \\ 2 & 2 & -2 & 6 \\ 1 & 1 & -1 & 3 \end{array} \right] \vec{x} \end{equation*}$ Find a basis for $\mbox {im}\left ( T\right )$ . Also find a basis for $\ker \left ( T\right ) .$

Define $T:\mathbb {R}^{3}\rightarrow \mathbb {R}^{3}$ as follows. $\begin{equation*} T(\vec{x})=\left[ \begin{array}{ccc} 1 & 2 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right] \vec{x} \end{equation*}$ where on the right, it is just matrix multiplication of the vector $\vec {x}$ which is meant. Explain why $T$ is an isomorphism of $\mathbb {R}^{3}$ to $\mathbb {R}^{3}$ .

Suppose $T:\mathbb {R}^{3}\rightarrow \mathbb {R}^{3}$ is a linear transformation given by $\begin{equation*} T(\vec{x})=A\vec{x} \end{equation*}$ where $A$ is a $3\times 3$ matrix. Show that $T$ is an isomorphism if and only if $A$ is invertible.

Suppose $T:\mathbb {R}^{n}\rightarrow \mathbb {R}^{m}$ is a linear transformation given by $\begin{equation*} T(\vec{x})=A\vec{x} \end{equation*}$ where $A$ is an $m\times n$ matrix. Show that $T$ is never an isomorphism if $m\neq n$ . In particular, show that if $m>n,$ $T$ cannot be onto and if $m<n,$ then $T$ cannot be one to one.

Define $T:\mathbb {R}^{2}\rightarrow \mathbb {R}^{3}$ as follows. $\begin{equation*} T(\vec{x})=\left[ \begin{array}{cc} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{array} \right] \vec{x} \end{equation*}$ where on the right, it is just matrix multiplication of the vector $\vec {x}$ which is meant. Show that $T$ is one to one. Next let $W=\mbox {im}\left ( T\right ) .$ Show that $T$ is an isomorphism of $\mathbb {R}^{2}$ and $\mbox {im }\left ( T\right )$ .

In the above problem, find a $2\times 3$ matrix $A$ such that the restriction of $A$ to $\mbox {im}\left ( T\right )$ gives the same result as $T^{-1}$ on $\mbox {im}\left ( T\right )$ . Hint: You might let $A$ be such that $\begin{equation*} A\left[ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right] =\left[ \begin{array}{c} 1 \\ 0 \end{array} \right] ,\ A\left[ \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right] =\left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \end{equation*}$ now find another vector $\vec {v}\in \mathbb {R}^{3}$ such that $\begin{equation*} \left\{ \left[ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right] ,\vec{v}\right\} \end{equation*}$ is a basis. You could pick $\begin{equation*} \vec{v}=\left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] \end{equation*}$ for example. Explain why this one works or one of your choice works. Then you could define $A\vec {v}$ to equal some vector in $\mathbb {R}^{2}.$ Explain why there will be more than one such matrix $A$ which will deliver the inverse isomorphism $T^{-1}$ on $\mbox {im}\left ( T\right )$ .

Now let $V$ equal $\mbox {span}\left \{ \left [ \begin {array}{c} 1 \\ 0 \\ 1 \end {array} \right ] ,\left [ \begin {array}{c} 0 \\ 1 \\ 1 \end {array} \right ] \right \}$ and let $T:V\rightarrow W$ be a linear transformation where $\begin{equation*} W=\mbox{span}\left\{ \left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right] ,\left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 1 \end{array} \right] \right\} \end{equation*}$ $\$ and $\begin{equation*} T\left[ \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right] =\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 0 \end{array} \right] ,T\left[ \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right] =\left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 1 \end{array} \right] \end{equation*}$ Explain why $T$ is an isomorphism. Determine a matrix $A$ which, when multiplied on the left gives the same result as $T$ on $V$ and a matrix $B$ which delivers $T^{-1}$ on $W$ .

You need to have $\begin{equation*} A\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{array} \right] =\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 1 & 1 \\ 0 & 1 \end{array} \right] \end{equation*}$ Now enlarge $\left [ \begin {array}{c} 1 \\ 0 \\ 1 \end {array} \right ] ,\left [ \begin {array}{c} 0 \\ 1 \\ 1 \end {array} \right ]$ to obtain a basis for $\mathbb {R}^{3}$ . You could add in $\left [ \begin {array}{c} 0 \\ 0 \\ 1 \end {array} \right ]$ for example, and then pick another vector in $\mathbb {R}^{4}$ and let $A\left [ \begin {array}{c} 0 \\ 0 \\ 1 \end {array} \right ]$ equal this other vector. Then you would have $\begin{equation*} A\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{array} \right] =\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{array} \right] \end{equation*}$ This would involve picking for the new vector in $\mathbb {R}^{4}$ the vector $\left [ \begin {array}{cccc} 0 & 0 & 0 & 1 \end {array} \right ] ^{T}.$ Then you could find $A$ . You can do something similar to find a matrix for $T^{-1}$ denoted as $B$ .

Let $V=\mathbb {R}^{3}$ and let $\begin{equation*} W=\mbox{span} \left( S \right), \mbox{ where } S=\left\{ \left[ \begin{array}{r} 1 \\ -1 \\ 1 \end{array} \right] ,\left[ \begin{array}{r} -2 \\ 2 \\ -2 \end{array} \right],\left[ \begin{array}{r} -1 \\ 1 \\ 1 \end{array} \right],\left[ \begin{array}{r} 1 \\ -1 \\ 3 \end{array} \right] \right\} \end{equation*}$ Find a basis of $W$ consisting of vectors in $S$ .

In this case $\dim (W)=1$ and a basis for $W$ consisting of vectors in $S$ can be obtained by taking any (nonzero) vector from $S$ .

Let $T$ be a linear transformation given by $T\left ( \left [ \begin {array}{r} x\\ y \end {array}\right ]\right ) = \left [ \begin {array}{rrr} 1 &1 \\ 1 & 1 \end {array}\right ] \left [ \begin {array}{r} x\\ y \end {array}\right ]$ Find a basis for $\ker \left ( T\right )$ and $\mbox {im} \left ( T\right )$ .

A basis for $\ker \left ( T\right )$ is $\left \{ \left [ \begin {array}{r} 1 \\ -1 \end {array} \right ] \right \}$ and a basis for $\mbox {im} \left ( T\right )$ is $\left \{ \left [ \begin {array}{r} 1 \\ 1 \end {array} \right ] \right \}$ .
There are many other possibilities for the specific bases, but in this case $\dim \left ( \ker \left ( T\right ) \right )=1$ and $\dim \left ( \mbox {im} \left ( T\right ) \right )=1$ .

Let $T$ be a linear transformation given by $T\left ( \left [ \begin {array}{r} x\\ y \end {array}\right ]\right ) = \left [ \begin {array}{rrr} 1 & 0 \\ 1 & 1 \end {array}\right ] \left [ \begin {array}{r} x\\ y \end {array}\right ]$ Find a basis for $\ker \left ( T\right )$ and $\mbox {im} \left ( T\right )$ .

In this case $\ker \left ( T\right ) =\{0\}$ and $\mbox {im} \left ( T\right ) = \mathbb {R}^2$ (pick any basis of $\mathbb {R}^2$ ).

Let $V=\mathbb {R}^{3}$ and let $\begin{equation*} W=\mbox{span}\left\{ \left[ \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right] ,\left[ \begin{array}{r} -1 \\ 2 \\ -1 \end{array} \right] \right\} \end{equation*}$ Extend this basis of $W$ to a basis of $V$ .

There are many possible such extensions, one is (how do we know?): $\begin{equation*} \left\{ \left[ \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right] ,\left[ \begin{array}{r} -1 \\ 2 \\ -1 \end{array} \right] ,\left[ \begin{array}{r} 0 \\ 0\\ 1 \end{array} \right] \right\} \end{equation*}$

Let $T$ be a linear transformation given by $T \left [ \begin {array}{r} x\\ y \\ z \end {array}\right ] = \left [ \begin {array}{rrr} 1 & 1 & 1 \\ 1 & 1 & 1 \end {array}\right ] \left [ \begin {array}{r} x\\ y \\ z \end {array}\right ]$ What is $\dim ( \ker \left ( T \right ) )$ ?

We can easily see that $\dim ( \mbox {im} \left ( T \right ) ) =1$ , and thus $\dim ( \ker \left ( T \right ) ) = 3 - \dim ( \mbox {im} \left ( T \right ) ) = 3- 1 = 2$ .

Consider the following functions which map $\mathbb {R}^{n}$ to $\mathbb {R}^{n}$ .

(a): $T$ multiplies the $j^{th}$ component of $\vec {x}$ by a nonzero number $b.$
(b): $T$ replaces the $i^{th}$ component of $\vec {x}$ with $b$ times the $j^{th}$ component added to the $i^{th}$ component.
(c): $T$ switches the $i^{th}$ and $j^{th}$ components.

Show these functions are linear transformations and describe their matrices $A$ such that $T\left (\vec {x}\right ) = A\vec {x}$ .

(a): The matrix of $T$ is the elementary matrix which multiplies the $j^{th}$ diagonal entry of the identity matrix by $b$ .
(b): The matrix of $T$ is the elementary matrix which takes $b$ times the $j^{th}$ row and adds to the $i^{th}$ row.
(c): The matrix of $T$ is the elementary matrix which switches the $i^{th}$ and the $j^{th}$ rows where the two components are in the $i^{th}$ and $j^{th}$ positions.

Suppose $T$ is a linear transformation such that

$\begin{eqnarray*} T\left[ \begin{array}{r} 1 \\ 2 \\ -6 \end{array} \right] &=&\left[ \begin{array}{r} 5 \\ 1 \\ 3 \end{array} \right] \\ T\left[ \begin{array}{r} -1 \\ -1 \\ 5 \end{array} \right] &=&\left[ \begin{array}{r} 1 \\ 1 \\ 5 \end{array} \right] \\ T\left[ \begin{array}{r} 0 \\ -1 \\ 2 \end{array} \right] &=&\left[ \begin{array}{r} 5 \\ 3 \\ -2 \end{array} \right] \end{eqnarray*}$

Find the matrix of $T$ . That is find $A$ such that $T(\vec {x})=A\vec {x}$ .

$\left [ \begin {array}{rrr} 5 & 1 & 5 \\ 1 & 1 & 3 \\ 3 & 5 & -2 \end {array} \right ] \left [ \begin {array}{ccc} 3 & 2 & 1 \\ 2 & 2 & 1 \\ 4 & 1 & 1 \end {array} \right ] =\left [ \begin {array}{ccc} 37 & 17 & 11 \\ 17 & 7 & 5 \\ 11 & 14 & 6 \end {array} \right ]$

Suppose $T$ is a linear transformation such that

$\begin{eqnarray*} T\left[ \begin{array}{r} 1 \\ 1 \\ -8 \end{array} \right] &=&\left[ \begin{array}{r} 1 \\ 3 \\ 1 \end{array} \right] \\ T\left[ \begin{array}{r} -1 \\ 0 \\ 6 \end{array} \right] &=&\left[ \begin{array}{r} 2 \\ 4 \\ 1 \end{array} \right] \\ T\left[ \begin{array}{r} 0 \\ -1 \\ 3 \end{array} \right] &=&\left[ \begin{array}{r} 6 \\ 1 \\ -1 \end{array} \right] \end{eqnarray*}$

Find the matrix of $T$ . That is find $A$ such that $T(\vec {x})=A\vec {x}$ .

$\left [ \begin {array}{rrr} 1 & 2 & 6 \\ 3 & 4 & 1 \\ 1 & 1 & -1 \end {array} \right ] \left [ \begin {array}{ccc} 6 & 3 & 1 \\ 5 & 3 & 1 \\ 6 & 2 & 1 \end {array} \right ] =\left [ \begin {array}{ccc} 52 & 21 & 9 \\ 44 & 23 & 8 \\ 5 & 4 & 1 \end {array} \right ]$

Suppose $T$ is a linear transformation such that

$\begin{eqnarray*} T\left[ \begin{array}{r} 1 \\ 3 \\ -7 \end{array} \right] &=&\left[ \begin{array}{r} -3 \\ 1 \\ 3 \end{array} \right] \\ T\left[ \begin{array}{r} -1 \\ -2 \\ 6 \end{array} \right] &=&\left[ \begin{array}{r} 1 \\ 3 \\ -3 \end{array} \right] \\ T\left[ \begin{array}{r} 0 \\ -1 \\ 2 \end{array} \right] &=&\left[ \begin{array}{r} 5 \\ 3 \\ -3 \end{array} \right] \end{eqnarray*}$

Find the matrix of $T$ . That is find $A$ such that $T(\vec {x})=A\vec {x}$ .

$\left [ \begin {array}{rrr} -3 & 1 & 5 \\ 1 & 3 & 3 \\ 3 & -3 & -3 \end {array} \right ] \left [ \begin {array}{ccc} 2 & 2 & 1 \\ 1 & 2 & 1 \\ 4 & 1 & 1 \end {array} \right ] = \left [ \begin {array}{rrr} 15 & 1 & 3 \\ 17 & 11 & 7 \\ -9 & -3 & -3 \end {array} \right ]$

Suppose $T$ is a linear transformation such that

$\begin{eqnarray*} T\left[ \begin{array}{r} 1 \\ 1 \\ -7 \end{array} \right] &=&\left[ \begin{array}{r} 3 \\ 3 \\ 3 \end{array} \right] \\ T\left[ \begin{array}{r} -1 \\ 0 \\ 6 \end{array} \right] &=&\left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] \\ T\left[ \begin{array}{r} 0 \\ -1 \\ 2 \end{array} \right] &=&\left[ \begin{array}{r} 1 \\ 3 \\ -1 \end{array} \right] \end{eqnarray*}$

Find the matrix of $T$ . That is find $A$ such that $T(\vec {x})=A\vec {x}$ .

Suppose $T$ is a linear transformation such that

$\begin{eqnarray*} T\left[ \begin{array}{r} 1 \\ 2 \\ -18 \end{array} \right] &=&\left[ \begin{array}{r} 5 \\ 2 \\ 5 \end{array} \right] \\ T\left[ \begin{array}{r} -1 \\ -1 \\ 15 \end{array} \right] &=&\left[ \begin{array}{r} 3 \\ 3 \\ 5 \end{array} \right] \\ T\left[ \begin{array}{r} 0 \\ -1 \\ 4 \end{array} \right] &=&\left[ \begin{array}{r} 2 \\ 5 \\ -2 \end{array} \right] \end{eqnarray*}$

Find the matrix of $T$ . That is find $A$ such that $T(\vec {x})=A\vec {x}$ .

Consider the following functions $T:\mathbb {R}^{3}\rightarrow \mathbb {R}^{2}$ . Show that each is a linear transformation and determine for each the matrix $A$ such that $T(\vec {x})=A\vec {x}$ .

(a): $T\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ] =\left [ \begin {array}{c} x+2y+3z \\ 2y-3x+z \end {array} \right ]$
(b): $T\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ] =\left [ \begin {array}{c} 7x+2y+z \\ 3x-11y+2z \end {array} \right ]$
(c): $T\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ] =\left [ \begin {array}{c} 3x+2y+z \\ x+2y+6z \end {array} \right ]$
(d): $T\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ] =\left [ \begin {array}{c} 2y-5x+z \\ x+y+z \end {array} \right ]$

Consider the following functions $T:\mathbb {R}^{3}\rightarrow \mathbb {R}^{2}.$ Explain why each of these functions $T$ is not linear.

(a): $T\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ] =\left [ \begin {array}{c} x+2y+3z+1 \\ 2y-3x+z \end {array} \right ]$
(b): $T\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ] =\left [ \begin {array}{c} x+2y^{2}+3z \\ 2y+3x+z \end {array} \right ]$
(c): $T\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ] =\left [ \begin {array}{c} \sin x+2y+3z \\ 2y+3x+z \end {array} \right ]$
(d): $T\left [ \begin {array}{c} x \\ y \\ z \end {array} \right ] =\left [ \begin {array}{c} x+2y+3z \\ 2y+3x-\ln z \end {array} \right ]$

Find the matrix for $T\left (\vec {w} \right ) = \mbox {proj}_{\vec {v}}\left ( \vec {w}\right )$ where $\vec {v}=\left [ \begin {array}{rrr} 1 & -2 & 3 \end {array} \right ] ^{T}.$

Recall that $\mbox {proj}_{\vec {u}}\left ( \vec {v}\right ) =\frac {\vec {v}\dotp \vec {u} }{\norm {\vec {u}}^{2}}\vec {u}$ and so the desired matrix has $i^{th}$ column equal to $\mbox {proj}_{\vec {u}}\left ( \vec {e}_{i}\right ) .$ Therefore, the matrix desired is $\frac {1}{14}\left [ \begin {array}{rrr} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 3 & -6 & 9 \end {array} \right ]$

Find the matrix for $T\left (\vec {w} \right ) = \mbox {proj}_{\vec {v}}\left ( \vec {w}\right )$ where $\vec {v}=\left [ \begin {array}{rrr} 1 & 5 & 3 \end {array} \right ] ^{T}.$

$\frac {1}{35}\left [ \begin {array}{rrr} 1 & 5 & 3 \\ 5 & 25 & 15 \\ 3 & 15 & 9 \end {array} \right ]$

Find the matrix for $T\left (\vec {w} \right ) = \mbox {proj}_{\vec {v}}\left ( \vec {w}\right )$ where $\vec {v}=\left [ \begin {array}{rrr} 1 & 0 & 3 \end {array} \right ] ^{T}.$

$\frac {1}{10}\left [ \begin {array}{ccc} 1 & 0 & 3 \\ 0 & 0 & 0 \\ 3 & 0 & 9 \end {array} \right ]$

Let $B = \left \{ \left [ \begin {array}{r} 2 \\ -1 \end {array} \right ], \left [ \begin {array}{r} 3 \\ 2 \end {array} \right ] \right \}$ be a basis of $\mathbb {R}^2$ and let $\vec {x} = \left [ \begin {array}{r} 5 \\ -7 \end {array} \right ]$ be a vector in $\mathbb {R}^2$ . Find $[\vec {x}]_B$ .

Let $B = \left \{ \left [ \begin {array}{r} 1 \\ -1 \\ 2 \end {array} \right ], \left [ \begin {array}{r} 2 \\ 1 \\ 2 \end {array} \right ], \left [ \begin {array}{r} -1 \\ 0 \\ 2 \end {array} \right ] \right \}$ be a basis of $\mathbb {R}^3$ and let $\vec {x} = \left [ \begin {array}{r} 5 \\ -1 \\ 4 \end {array} \right ]$ be a vector in $\mathbb {R}^2$ . Find $[\vec {x}]_B$ .

$[\vec {x}]_B = \left [ \begin {array}{r} 2 \\ 1 \\ -1 \end {array} \right ]$ .

Let $T: \mathbb {R}^2 \mapsto \mathbb {R}^2$ be a linear transformation defined by $T \left ( \left [ \begin {array}{r} a \\ b \end {array} \right ] \right ) = \left [ \begin {array}{r} a+b \\ a-b \end {array} \right ]$ .

Consider the two bases $B_1 = \left \{ \vec {v}_{1}, \vec {v}_{2} \right \} = \left \{ \left [ \begin {array}{r} 1 \\ 0 \end {array}\right ], \left [ \begin {array}{r} -1 \\ 1 \end {array} \right ] \right \}$ and $B_2 = \left \{ \left [ \begin {array}{r} 1 \\ 1 \end {array} \right ], \left [ \begin {array}{r} 1 \\ -1 \end {array} \right ] \right \}$

Find the matrix $M_{B_2,B_1}$ of $T$ with respect to the ordered bases $B_1$ and $B_2$ .

$M_{B_{2} B_{1}} = \left [ \begin {array}{rr} 1 & 0 \\ -1 & 1 \end {array} \right ]$

Challenge Exercises

Denote by $\mathbb {R}^{\mathbb {N}}$ the set of real valued sequences. For $\vec {a}\equiv \left \{ a_{n}\right \} _{n=1}^{\infty },\vec {b}\equiv \left \{ b_{n}\right \} _{n=1}^{\infty }$ two of these, define their sum to be given by $\begin{equation*} \vec{a}+\vec{b} = \left\{ a_{n}+b_{n}\right\} _{n=1}^{\infty } \end{equation*}$ and define scalar multiplication by $\begin{equation*} c\vec{a}=\left\{ ca_{n}\right\} _{n=1}^{\infty }\text{ where }\vec{a} =\left\{ a_{n}\right\} _{n=1}^{\infty } \end{equation*}$ Is this a special case of Problem prb:10.5? Is this a vector space?

Let $\mathbb {C}^{2}$ be the set of ordered pairs of complex numbers. Define addition and scalar multiplication in the usual way. $\begin{equation*} \left( z,w\right) +\left( \hat{z},\hat{w}\right) = \left( z+\hat{z},w+ \hat{w}\right) ,\ u\left( z,w\right) \equiv \left( uz,uw\right) \end{equation*}$ Here the scalars are from $\mathbb {C}$ . Show this is a vector space.

Let $V$ be the set of functions defined on a nonempty set which have values in a vector space $W.$ Is this a vector space? Explain.

Consider functions defined on $\left \{ 1,2,\cdots ,n\right \}$ having values in $\mathbb {R}$ . Explain how, if $V$ is the set of all such functions, $V$ can be considered as $\mathbb {R}^{n}$ .

Let $f\left ( i\right )$ be the $i^{th}$ component of a vector $\vec {x}\in \mathbb {R}^{n}$ . Thus a typical element in $\mathbb {R}^{n}$ is $\left ( f\left ( 1\right ) ,\cdots ,f\left ( n\right ) \right )$ .

Let the field of scalars be $\mathbb {Q}$ , the rational numbers and let the vectors be of the form $a+b\sqrt {2}$ where $a,b$ are rational numbers. Show that this collection of vectors is a vector space with field of scalars $\mathbb {Q}$ and give a basis for this vector space.

When you add two of these you get one and when you multiply one of these by a scalar, you get another one. A basis is $\left \{ 1,\sqrt {2}\right \}$ . By definition, the span of these gives the collection of vectors. Are they independent? Say $a+b\sqrt {2}=0$ where $a,b$ are rational numbers. If $a\neq 0,$ then $b\sqrt {2}=-a$ which can’t happen since $a$ is rational. If $b\neq 0,$ then $-a=b\sqrt {2}$ which again can’t happen because on the left is a rational number and on the right is an irrational. Hence both $a,b=0$ and so this is a basis.

Suppose $V$ is a finite dimensional vector space. Based on the exchange theorem above, it was shown that any two bases have the same number of vectors in them. Give a different proof of this fact using the earlier material in the book.

Suppose $\left \{ \vec {x}_{1}\vec { ,\cdots ,x}_{n}\right \}$ and $\left \{ \vec {y}_{1}\vec {,\cdots ,y} _{m}\right \}$ are two bases with $m<n.$ Then define $\begin{equation*} \phi :\mathbb{R}^{n}\mapsto V,\psi :\mathbb{R}^{m}\mapsto V \end{equation*}$ by $\begin{equation*} \phi \left( \vec{a}\right) = \sum_{k=1}^{n}a_{k}\vec{x} _{k},\;\psi \left( \vec{b}\right) = \sum_{j=1}^{m}b_{j}\vec{y}_{j} \end{equation*}$ Consider the linear transformation, $\psi ^{-1}\circ \phi .$ Argue it is a one to one and onto mapping from $\mathbb {R}^{n}$ to $\mathbb {R}^{m}.$ Now consider a matrix of this linear transformation and its reduced row echelon form.

This is obvious because when you add two of these you get one and when you multiply one of these by a scalar, you get another one. A basis is $\left \{ 1,\sqrt {2}\right \}$ . By definition, the span of these gives the collection of vectors. Are they independent? Say $a+b\sqrt {2}=0$ where $a,b$ are rational numbers. If $a\neq 0,$ then $b\sqrt {2}=-a$ which can’t happen since $a$ is rational. If $b\neq 0,$ then $-a=b\sqrt {2}$ which again can’t happen because on the left is a rational number and on the right is an irrational. Hence both $a,b=0$ and so this is a basis.

If $p = p(x)$ and $q = q(x)$ are polynomials in $\mathbb {P}^{n}$ , define $\begin{equation*} \langle p, q \rangle = p(0)q(0) + p(1)q(1) + \dots + p(n)q(n) \end{equation*}$ Show that this is an inner product on $\mathbb {P}^{n}$ .

Let $\mathcal {D}_{n}$ denote the space of all functions from the set $\{1, 2, 3, \dots , n\}$ to $\RR$ with pointwise addition and scalar multiplication. Show that $\langle \ , \rangle$ is an inner product on $\mathcal {D}_{n}$ if
$\langle \vec {f}, \vec {g}\rangle = f(1)g(1) + f(2)g(2) + \dots + f(n)g(n)$ .

P1 and P2 are clear since $f(i)$ and $g(i)$ are real numbers.

P3: ⟨f + g,h⟩	= ∑ _i(f + g)(i) ⋅ h(i)
	= ∑ _i(f(i) + g(i)) ⋅ h(i)
	= ∑ _i[f(i)h(i) + g(i)h(i)]
	= ∑ _if(i)h(i) + ∑ _ig(i)h(i)
	= ⟨f,h⟩ + ⟨g,h⟩.
P4: ⟨rf,g⟩	= ∑ _i(rf)(i) ⋅ g(i)
	= ∑ _irf(i) ⋅ g(i)
	= r∑ _if(i) ⋅ g(i)
	= r⟨f,g⟩

P5: If $f \neq 0$ , then $\langle f, f \rangle = \displaystyle \sum _{i}f(i)^2 > 0$ because some $f(i) \neq 0$ .

Practice Problem Source

Problems prb:10.1 to prb:10.106 come from Chapter 9 of Ken Kuttler’s A First Course in Linear Algebra. (CC-BY)

Ken Kuttler, A First Course in Linear Algebra, Lyryx 2017, Open Edition, pp. 469–535.

Problems prob:inner_prod_7 to come from Section 10.1 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, pp. 528–530.

Press...	...to do
left/right arrows	Move cursor
shift+left/right arrows	Select region
ctrl+a	Select all
ctrl+x/c/v	Cut/copy/paste
ctrl+z/y	Undo/redo
ctrl+left/right	Add entry to list or column to matrix
shift+ctrl+left/right	Add copy of current entry/column to to list/matrix
ctrl+up/down	Add row to matrix
shift+ctrl+up/down	Add copy of current row to matrix
ctrl+backspace	Delete current entry in list or column in matrix
ctrl+shift+backspace	Delete current row in matrix

Type...	...to get
norm	$\|\|\blue{[?]}\|\|$
text	$\text{\blue{[?]}}$
sym_name	$\backslash\texttt{\blue{[?]}}$
abs	$\left\|\blue{[?]}\right\|$
sqrt	$\sqrt{\blue{[?]}}$
paren	$\left(\blue{[?]}\right)$
floor	$\lfloor \blue{[?]} \rfloor$
factorial	$\blue{[?]}!$
exp	${\blue{[?]}}^{\blue{[?]}}$
sub	${\blue{[?]}}_{\blue{[?]}}$
frac	$\dfrac{\blue{[?]}}{\blue{[?]}}$
int	$\displaystyle\int{\blue{[?]}}d\blue{[?]}$
defi	$\displaystyle\int_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}d\blue{[?]}$
deriv	$\displaystyle\frac{d}{d\blue{[?]}}\blue{[?]}$
sum	$\displaystyle\sum_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
prod	$\displaystyle\prod_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
root	$\sqrt[\blue{[?]}]{\blue{[?]}}$
vec	$\left\langle \blue{[?]} \right\rangle$
mat	$\left(\begin{matrix} \blue{[?]} \end{matrix}\right)$
*	$\cdot$
infinity	$\infty$
arcsin	$\arcsin\left(\blue{[?]}\right)$
arccos	$\arccos\left(\blue{[?]}\right)$
arctan	$\arctan\left(\blue{[?]}\right)$
sin	$\sin\left(\blue{[?]}\right)$
cos	$\cos\left(\blue{[?]}\right)$
tan	$\tan\left(\blue{[?]}\right)$
sec	$\sec\left(\blue{[?]}\right)$
csc	$\csc\left(\blue{[?]}\right)$
cot	$\cot\left(\blue{[?]}\right)$
log	$\log\left(\blue{[?]}\right)$
ln	$\ln\left(\blue{[?]}\right)$
alpha	$\alpha$
beta	$\beta$
gamma	$\gamma$
delta	$\delta$
epsilon	$\epsilon$
zeta	$\zeta$
eta	$\eta$
theta	$\theta$
iota	$\iota$
kappa	$\kappa$
lambda	$\lambda$
mu	$\mu$
nu	$\nu$
xi	$\xi$
omicron	$\omicron$
pi	$\pi$
rho	$\rho$
sigma	$\sigma$
tau	$\tau$
upsilon	$\upsilon$
phi	$\phi$
chi	$\chi$
psi	$\psi$
omega	$\omega$
Gamma	$\Gamma$
Delta	$\Delta$
Theta	$\Theta$
Lambda	$\Lambda$
Xi	$\Xi$
Pi	$\Pi$
Sigma	$\Sigma$
Phi	$\Phi$
Psi	$\Psi$
Omega	$\Omega$

Additional Exercises for Chapter 10: Abstract Vector Spaces

Challenge Exercises

Practice Problem Source

Controls

Symbols

Settings