Find an orthonormal basis for the span of the following set of vectors.
\(\left [ \begin{array}{r} 3 \\ -4 \\ 0 \end{array} \right ] ,\left [ \begin{array}{r} 7 \\ -1 \\ 0 \end{array} \right ] ,\left [ \begin{array}{r} 1 \\ 7 \\ 1 \end{array} \right ] \)
Click the arrow to see answer.
We will use Gram-Schmidt orthogonalization
algorithm.
\[\vec {f}_1=\begin{bmatrix}3\\-4\\0\end{bmatrix}\]
\[\vec {f}_2=\begin{bmatrix}7\\-1\\0\end{bmatrix}-\frac {25}{25}\begin{bmatrix}3\\-4\\0\end{bmatrix}=\begin{bmatrix}4\\3\\0\end{bmatrix}\]
\[\vec {f}_3=\begin{bmatrix}1\\7\\1\end{bmatrix}-\frac {-25}{25}\begin{bmatrix}3\\-4\\0\end{bmatrix}-\frac {25}{25}\begin{bmatrix}4\\3\\0\end{bmatrix}=\begin{bmatrix}0\\0\\1\end{bmatrix}\]
Normalizing each vector we get:
\[ \left [ \begin{array}{c} \frac {3}{5} \\ -\frac {4}{5} \\ 0 \end{array} \right ] ,\left [ \begin{array}{c} \frac {4}{5} \\ \frac {3}{5} \\ 0 \end{array} \right ] ,\left [ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right ] \]
Using the Gram Schmidt process find an orthonormal basis for the following span:
\[ \mbox {span} \left \{ \left [ \begin{array}{r} 1 \\ 2 \\ 1 \end{array} \right ] ,\left [ \begin{array}{r} 2 \\ -1 \\ 3 \end{array} \right ] , \left [ \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right ] \right \} \]
Click the arrow to see the answer.
\[ \left [ \begin{array}{c} \frac {1}{6}\sqrt {6} \\ \frac {1}{3}\sqrt {6} \\ \frac {1}{6}\sqrt {6} \end{array} \right ] ,\left [ \begin{array}{c} \frac {3}{10}\sqrt {2} \\ -\frac {2}{5}\sqrt {2} \\ \frac {1}{2}\sqrt {2} \end{array} \right ] ,\left [ \begin{array}{c} \frac {7}{15}\sqrt {3} \\ -\frac {1}{15}\sqrt {3} \\ -\frac {1}{3}\sqrt {3} \end{array} \right ] \]
Write
\(\vec {x}\) as the sum of a vector in
\(W\) and a vector in
\(W^\perp \) if
\[\vec {x} = \begin{bmatrix}1\\ 5\\ 7\end{bmatrix},\quad W = \mbox {span}\left (\begin{bmatrix}1\\ -2\\ 3\end{bmatrix}, \begin{bmatrix}-1\\ 1\\ 1\end{bmatrix}\right )\]
Click the arrow to see the answer.
Observe that the two vectors that span
\(W\) are
orthogonal. We will refer to these vectors as
\(\vec {f}_1\) and
\(\vec {f}_2\) .
\[\vec {w}=\text {proj}_W(\vec {x})=\text {proj}_{\vec {f}_1}(\vec {x})+\text {proj}_{\vec {f}_2}(\vec {x})=\frac {12}{14}\begin{bmatrix}1\\-2\\3\end{bmatrix}+\frac {11}{3}\begin{bmatrix}-1\\1\\1\end{bmatrix}=\frac {1}{21}\begin{bmatrix}-59\\41\\131\end{bmatrix}\]
\[\vec {w}^{\perp }=\vec {x}-\vec {w}=\frac {1}{21}\begin{bmatrix}80\\64\\16\end{bmatrix}\]
\[\vec {x}=\vec {w}+\vec {w}^{\perp }\]
Write
\(\vec {x}\) as the sum of a vector in
\(W\) and a vector in
\(W^\perp \) if
\[\vec {x} = \begin{bmatrix}3\\ 1\\ 5\\ 9\end{bmatrix}, \quad W = \mbox {span}\left (\begin{bmatrix}1\\ 0\\ 1\\ 1\end{bmatrix}, \begin{bmatrix}0\\ 1\\ -1\\ 1\end{bmatrix}, \begin{bmatrix}-2\\ 0\\ 1\\ 1\end{bmatrix}\right )\]
Click the arrow to see the answer.
Observe that the three vectors that span
\(W\) form
an orthogonal set. We will refer to these vectors as
\(\vec {f}_1\) ,
\(\vec {f}_2\) and
\(\vec {f}_3\) .
\[\vec {w}=\text {proj}_W(\vec {x})=\text {proj}_{\vec {f}_1}(\vec {x})+\text {proj}_{\vec {f}_2}(\vec {x})+\text {proj}_{\vec {f}_3}(\vec {x})=\]
\[=\frac {17}{3}\begin{bmatrix}1\\0\\1\\1\end{bmatrix}+\frac {5}{3}\begin{bmatrix}0\\1\\-1\\1\end{bmatrix}+\frac {8}{6}\begin{bmatrix}-2\\0\\1\\1\end{bmatrix}=\frac {1}{3}\begin{bmatrix}9\\5\\16\\26\end{bmatrix}\]
\[\vec {w}^{\perp }=\vec {x}-\vec {w}=\frac {1}{3}\begin{bmatrix}0\\-2\\-1\\1\end{bmatrix}\]
\[\vec {x}=\vec {w}+\vec {w}^{\perp }\]
Some of the problems come from the end of Chapter 7 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.
433–438.