Additional Exercises for Ch 9

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Additional Exercises for Chapter 9: Orthogonality

Find an orthonormal basis for the span of each of the following sets of vectors.

(a): $\left [ \begin {array}{r} 3 \\ -4 \\ 0 \end {array} \right ] ,\left [ \begin {array}{r} 7 \\ -1 \\ 0 \end {array} \right ] ,\left [ \begin {array}{r} 1 \\ 7 \\ 1 \end {array} \right ]$
Click the arrow to see the answer.

$\left [ \begin {array}{c} \frac {3}{5} \\ -\frac {4}{5} \\ 0 \end {array} \right ] ,\left [ \begin {array}{c} \frac {4}{5} \\ \frac {3}{5} \\ 0 \end {array} \right ] ,\left [ \begin {array}{c} 0 \\ 0 \\ 1 \end {array} \right ]$
(b): $\left [ \begin {array}{r} 3 \\ 0 \\ -4 \end {array} \right ] ,\left [ \begin {array}{r} 11 \\ 0 \\ 2 \end {array} \right ] ,\left [ \begin {array}{r} 1 \\ 1 \\ 7 \end {array} \right ]$
Click the arrow to see the answer.

$\left [ \begin {array}{c} \frac {3}{5} \\ 0 \\ -\frac {4}{5} \end {array} \right ] ,\left [ \begin {array}{c} \frac {4}{5} \\ 0 \\ \frac {3}{5} \end {array} \right ] ,\left [ \begin {array}{c} 0 \\ 1 \\ 0 \end {array} \right ]$
(c): $\left [ \begin {array}{r} 3 \\ 0 \\ -4 \end {array} \right ] ,\left [ \begin {array}{r} 5 \\ 0 \\ 10 \end {array} \right ] ,\left [ \begin {array}{r} -7 \\ 1 \\ 1 \end {array} \right ]$

Click the arrow to see the answer.

$\left [ \begin {array}{c} \frac {3}{5} \\ 0 \\ -\frac {4}{5} \end {array} \right ] ,\left [ \begin {array}{c} \frac {4}{5} \\ 0 \\ \frac {3}{5} \end {array} \right ] ,\left [ \begin {array}{c} 0 \\ 1 \\ 0 \end {array} \right ]$

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.

A solution is $\left [ \begin {array}{c} \frac {1}{6}\sqrt {6} \\ \frac {1}{3}\sqrt {6} \\ \frac {1}{6}\sqrt {6} \end {array} \right ] ,\left [ \begin {array}{c} \frac {3}{10}\sqrt {2} \\ -\frac {2}{5}\sqrt {2} \\ \frac {1}{2}\sqrt {2} \end {array} \right ] ,\left [ \begin {array}{c} \frac {7}{15}\sqrt {3} \\ -\frac {1}{15}\sqrt {3} \\ -\frac {1}{3}\sqrt {3} \end {array} \right ]$

Using the Gram Schmidt process find an orthonormal basis for the following span: $\mbox {span}\left \{ \left [ \begin {array}{r} 1 \\ 2 \\ 1 \\ 0 \end {array} \right ] ,\left [ \begin {array}{r} 2 \\ -1 \\ 3 \\ 1 \end {array} \right ] , \left [ \begin {array}{r} 1 \\ 0 \\ 0 \\ 1 \end {array} \right ] \right \}$

Click the arrow to see the answer.

A solution is $\left [ \begin {array}{c} \frac {1}{6}\sqrt {6} \\ \frac {1}{3}\sqrt {6} \\ \frac {1}{6}\sqrt {6} \\ 0 \end {array} \right ] ,\left [ \begin {array}{c} \frac {1}{6}\sqrt {2}\sqrt {3} \\ -\frac {2}{9}\sqrt {2}\sqrt {3} \\ \frac {5}{18}\sqrt {2}\sqrt {3} \\ \frac {1}{9}\sqrt {2}\sqrt {3} \end {array} \right ] ,\left [ \begin {array}{c} \frac {5}{111}\sqrt {3}\sqrt {37} \\ \frac {1}{333}\sqrt {3}\sqrt {37} \\ -\frac {17}{333}\sqrt {3}\sqrt {37} \\ \frac {22}{333}\sqrt {3}\sqrt {37} \end {array} \right ]$

The set $V = \left \{ \left [ \begin {array}{r} x \\ y \\ z \end {array} \right ] :2x+3y-z=0\right \}$ is a subspace of $\mathbb {R}^{3}.$ Find an orthonormal basis for this subspace.

The subspace is of the form $\left [ \begin {array}{c} x \\ y \\ 2x+3y \end {array} \right ]$

So a basis is $\left [ \begin {array}{c} 1 \\ 0 \\ 2 \end {array} \right ] ,\left [ \begin {array}{c} 0 \\ 1 \\ 3 \end {array} \right ]$ . Therefore, an orthonormal basis is $\left [ \begin {array}{c} \frac {1}{5}\sqrt {5} \\ 0 \\ \frac {2}{5}\sqrt {5} \end {array} \right ] ,\left [ \begin {array}{c} -\frac {3}{35}\sqrt {5}\sqrt {14} \\ \frac {1}{14}\sqrt {5}\sqrt {14} \\ \frac {3}{70}\sqrt {5}\sqrt {14} \end {array} \right ]$

In each case, write $\vec {x}$ as the sum of a vector in $W$ and a vector in $W^\perp$ .

$\vec {x} = \begin {bmatrix}1\\ 5\\ 7\end {bmatrix}$ , $W = \mbox {span}\left (\begin {bmatrix}1\\ -2\\ 3\end {bmatrix}, \begin {bmatrix}-1\\ 1\\ 1\end {bmatrix}\right )$

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$\vec {x} = \frac {1}{182}\begin {bmatrix}271\\-221\\1030\end {bmatrix} + \frac {1}{182}\begin {bmatrix}93\\403\\62\end {bmatrix}$

$\vec {x} = \begin {bmatrix}3\\ 1\\ 5\\ 9\end {bmatrix}$

$W = \mbox {span}\left (\begin {bmatrix}1\\ 0\\ 1\\ 1\end {bmatrix}, \begin {bmatrix}0\\ 1\\ -1\\ 1\end {bmatrix}, \begin {bmatrix}-2\\ 0\\ 1\\ 1\end {bmatrix}\right )$

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$\vec {x}= \frac {1}{4}\begin {bmatrix}1\\ 7\\ 11\\ 17\end {bmatrix} + \frac {1}{4}\begin {bmatrix}7\\ -7\\ -7\\ 7\end {bmatrix}$

$\vec {x} = \begin {bmatrix}a\\ b\\ c\\ d\end {bmatrix}$

$W = \mbox {span}\left (\begin {bmatrix}1\\ 0\\ 0\\ 0\end {bmatrix}, \begin {bmatrix}0\\ 1\\ 0\\ 0\end {bmatrix}, \begin {bmatrix}0\\ 0\\ 1\\ 0\end {bmatrix}\right )$

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$\vec {x} = \frac {1}{12}\begin {bmatrix}5a - 5b + c - 3d\\ -5a + 5b - c + 3d\\ a - b + 11c + 3d\\ -3a + 3b + 3c + 3d\end {bmatrix} + \frac {1}{12}\begin {bmatrix}7a + 5b - c + 3d\\ 5a + 7b + c - 3d\\ -a + b + c -3d\\ 3a - 3b - 3c + 9d\end {bmatrix}$

Normalize the rows to make each of the following matrices orthogonal.

$A = \begin {bmatrix} 3 & -4 \\ 4 & 3 \end {bmatrix}$

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$\frac {1}{5}\begin {bmatrix} 3 & -4 \\ 4 & 3 \end {bmatrix}$

$A = \begin {bmatrix} a & b \\ -b & a \end {bmatrix}$ $a$ , $b$ not both zero

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$\frac {1}{\sqrt {a^2 + b^2}}\begin {bmatrix} a & b \\ -b & a \end {bmatrix}$

$A = \begin {bmatrix} 2 & 1 & -1 \\ 1 & -1 & 1 \\ 0 & 1 & 1 \end {bmatrix}$

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$\begin {bmatrix} \frac {2}{\sqrt {6}} & \frac {1}{\sqrt {6}} & -\frac {1}{\sqrt {6}}\\ \frac {1}{\sqrt {3}} & -\frac {1}{\sqrt {3}} & \frac {1}{\sqrt {3}} \\ 0 & \frac {1}{\sqrt {2}} & \frac {1}{\sqrt {2}} \end {bmatrix}$

$A = \begin {bmatrix} 2 & 6 & -3 \\ 3 & 2 & 6 \\ -6 & 3 & 2 \end {bmatrix}$

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$\frac {1}{7}\begin {bmatrix} 2 & 6 & -3 \\ 3 & 2 & 6 \\ -6 & 3 & 2 \end {bmatrix}$

Find the eigenvalues and an orthonormal basis of eigenvectors for $A.$ $\begin{equation*} A=\left[ \begin{array}{rrr} 11 & -1 & -4 \\ -1 & 11 & -4 \\ -4 & -4 & 14 \end{array} \right] \end{equation*}$

Two eigenvalues are 12 and 18.

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The eigenvectors and eigenvalues are: $\left \{ \frac {1}{\sqrt {3}}\left [ \begin {array}{c} 1 \\ 1 \\ 1 \end {array} \right ] \right \} \leftrightarrow 6,\left \{ \frac {1}{\sqrt {2}}\left [ \begin {array}{r} -1 \\ 1 \\ 0 \end {array} \right ] \right \} \leftrightarrow 12,\left \{ \frac {1}{\sqrt {6}}\left [ \begin {array}{r} -1 \\ -1 \\ 2 \end {array} \right ] \right \} \leftrightarrow 18$

Find the eigenvalues and an orthonormal basis of eigenvectors for $A.$ $\begin{equation*} A=\left[ \begin{array}{rrr} 4 & 1 & -2 \\ 1 & 4 & -2 \\ -2 & -2 & 7 \end{array} \right] \end{equation*}$

One eigenvalue is 3.

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The eigenvectors and eigenvalues are: $\left \{ \frac {1}{\sqrt {2}}\left [ \begin {array}{r} -1 \\ 1 \\ 0 \end {array} \right ] ,\frac {1}{\sqrt {3}}\left [ \begin {array}{c} 1 \\ 1 \\ 1 \end {array} \right ] \right \} \leftrightarrow 3,\left \{ \frac {1}{\sqrt {6}}\left [ \begin {array}{r} -1 \\ -1 \\ 2 \end {array} \right ] \right \} \leftrightarrow 9$

One eigenvalue is $-2$ .

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The eigenvectors and eigenvalues are: $\left \{ \left [ \begin {array}{c} \frac {1}{3}\sqrt {3} \\ \frac {1}{3}\sqrt {3} \\ \frac {1}{3}\sqrt {3} \end {array} \right ] \right \} \leftrightarrow 1,\left \{ \left [ \begin {array}{c} -\frac {1}{2}\sqrt {2} \\ \frac {1}{2}\sqrt {2} \\ 0 \end {array} \right ] , \left [ \begin {array}{c} -\frac {1}{6}\sqrt {6} \\ -\frac {1}{6}\sqrt {6} \\ \frac {1}{3}\sqrt {2}\sqrt {3} \end {array} \right ] \right \} \leftrightarrow -2$ $\left [ \begin {array}{ccc} \sqrt {3}/3 & -\sqrt {2}/2 & -\sqrt {6}/6 \\ \sqrt {3}/3 & \sqrt {2}/2 & -\sqrt {6}/6 \\ \sqrt {3}/3 & 0 & \frac {1}{3}\sqrt {2}\sqrt {3} \end {array} \right ]^{T}\left [ \begin {array}{rrr} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end {array} \right ] \left [ \begin {array}{ccc} \sqrt {3}/3 & -\sqrt {2}/2 & -\sqrt {6}/6 \\ \sqrt {3}/3 & \sqrt {2}/2 & -\sqrt {6}/6 \\ \sqrt {3}/3 & 0 & \frac {1}{3}\sqrt {2}\sqrt {3} \end {array} \right ]$ $=\left [ \begin {array}{rrr} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end {array} \right ]$

Two eigenvalues are $18$ and $24$ .

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Two eigenvalues are $12$ and 18.

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The eigenvectors and eigenvalues are: $\left \{ \left [ \begin {array}{c} -\frac {1}{6}\sqrt {6} \\ -\frac {1}{6}\sqrt {6} \\ \frac {1}{3}\sqrt {2}\sqrt {3} \end {array} \right ] \right \} \leftrightarrow 6,\left \{ \left [ \begin {array}{c} -\frac {1}{2}\sqrt {2} \\ \frac {1}{2}\sqrt {2} \\ 0 \end {array} \right ] \right \} \leftrightarrow 12,\left \{ \left [ \begin {array}{c} \frac {1}{3}\sqrt {3} \\ \frac {1}{3}\sqrt {3} \\ \frac {1}{3}\sqrt {3} \end {array} \right ] \right \} \leftrightarrow 18.$ The matrix $Q$ has these as its columns.

Find the eigenvalues and an orthonormal basis of eigenvectors for $A.$ Diagonalize $A$ by finding an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $Q^{T}AQ=D$ . $\begin{equation*} A=\left[ \begin{array}{ccc} - \frac{5}{3} & \frac{1}{15}\sqrt{6}\sqrt{5} & \frac{8}{15}\sqrt{5} \\ \frac{1}{15}\sqrt{6}\sqrt{5} & - \frac{14}{5} & - \frac{1}{15}\sqrt{6} \\ \frac{8}{15}\sqrt{5} & - \frac{1}{15}\sqrt{6} & \frac{7}{15} \end{array} \right] \end{equation*}$

The eigenvalues are $-3,-2,1.$

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eigenvectors: $\left \{ \left [ \begin {array}{c} \frac {1}{6}\sqrt {6} \\ 0 \\ \frac {1}{6}\sqrt {5}\sqrt {6} \end {array} \right ] \right \} \leftrightarrow 1,\left \{ \left [ \begin {array}{c} -\frac {1}{3}\sqrt {2}\sqrt {3} \\ -\frac {1}{5}\sqrt {5} \\ \frac {1}{15}\sqrt {2}\sqrt {15} \end {array} \right ] \right \} \leftrightarrow -2,\left \{ \left [ \begin {array}{c} -\frac {1}{6}\sqrt {6} \\ \frac {2}{5}\sqrt {5} \\ \frac {1}{30}\sqrt {30} \end {array} \right ] \right \} \leftrightarrow -3$ These vectors are the columns of $Q$ .

Find the eigenvalues and an orthonormal basis of eigenvectors for $A.$ Diagonalize $A$ by finding an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $Q^{T}AQ=D$ . $\begin{equation*} A=\left[ \begin{array}{rrr} 3 & 0 & 0 \\ 0 & \frac{3}{2} & \frac{1}{2} \\ 0 & \frac{1}{2} & \frac{3}{2} \end{array} \right] \end{equation*}$

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The eigenvectors and eigenvalues are: $\left \{ \left [ \begin {array}{c} 0 \\ -\frac {1}{2}\sqrt {2} \\ \frac {1}{2}\sqrt {2} \end {array} \right ] \right \} \leftrightarrow 1,\left \{ \left [ \begin {array}{c} 0 \\ \frac {1}{2}\sqrt {2} \\ \frac {1}{2}\sqrt {2} \end {array} \right ] \right \} \leftrightarrow 2,\left \{ \left [ \begin {array}{c} 1 \\ 0 \\ 0 \end {array} \right ] \right \} \leftrightarrow 3.$ These vectors are the columns of the matrix $Q$ .

Click the arrow to see the answer.

The eigenvectors and eigenvalues are: $\left \{ \left [ \begin {array}{c} 1 \\ 0 \\ 0 \end {array} \right ] \right \} \leftrightarrow 2,\left \{ \left [ \begin {array}{c} 0 \\ -\frac {1}{2}\sqrt {2} \\ \frac {1}{2}\sqrt {2} \end {array} \right ] \right \} \leftrightarrow 4,\left \{ \left [ \begin {array}{c} 0 \\ \frac {1}{2}\sqrt {2} \\ \frac {1}{2}\sqrt {2} \end {array} \right ] \right \} \leftrightarrow 6.$ These vectors are the columns of $Q$ .

Find the eigenvalues and an orthonormal basis of eigenvectors for $A.$ Diagonalize $A$ by finding an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $Q^{T}AQ=D$ . $\begin{equation*} A=\left[ \begin{array}{ccc} \frac{4}{3} & \frac{1}{3}\sqrt{3}\sqrt{2} & \frac{1}{3}\sqrt{2} \\ \frac{1}{3}\sqrt{3}\sqrt{2} & 1 & - \frac{1}{3} \sqrt{3} \\ \frac{1}{3}\sqrt{2} & - \frac{1}{3}\sqrt{3} & \frac{5}{3} \end{array} \right] \end{equation*}$

The eigenvalues are $0,2,2$ where $2$ is listed twice because it is a root of multiplicity 2.

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The eigenvectors and eigenvalues are: $\left \{ \left [ \begin {array}{c} -\frac {1}{5}\sqrt {2}\sqrt {5} \\ \frac {1}{5}\sqrt {3}\sqrt {5} \\ \frac {1}{5}\sqrt {5} \end {array} \right ] \right \} \leftrightarrow 0,\left \{ \left [ \begin {array}{c} \frac {1}{3}\sqrt {3} \\ 0 \\ \frac {1}{3}\sqrt {2}\sqrt {3}\end {array} \right ] ,\left [ \begin {array}{c} \frac {1}{5}\sqrt {2}\sqrt {5} \\ \frac {1}{5}\sqrt {3}\sqrt {5} \\ -\frac {1}{5}\sqrt {5} \end {array} \right ] \right \} \leftrightarrow 2.$ The columns are these vectors.

Find the eigenvalues and an orthonormal basis of eigenvectors for $A.$ Diagonalize $A$ by finding an orthogonal matrix $Q$ and a diagonal matrix $D$ such that $Q^{T}AQ=D$ . $\begin{equation*} A=\left[ \begin{array}{ccc} 1 & \frac{1}{6}\sqrt{3}\sqrt{2} & \frac{1}{6} \sqrt{3}\sqrt{6} \\ \frac{1}{6}\sqrt{3}\sqrt{2} & \frac{3}{2} & \frac{1}{12}\sqrt{2}\sqrt{6} \\ \frac{1}{6}\sqrt{3}\sqrt{6} & \frac{1}{12} \sqrt{2}\sqrt{6} & \frac{1}{2} \end{array} \right] \end{equation*}$

The eigenvalues are $2,1,0.$

Click the arrow to see the answer.

The eigenvectors and eigenvalues are: $\left \{ \left [ \begin {array}{c} -\frac {1}{3}\sqrt {3} \\ 0 \\ \frac {1}{3}\sqrt {2}\sqrt {3} \end {array} \right ] \right \} \leftrightarrow 0,\left \{ \left [ \begin {array}{c} \frac {1}{3}\sqrt {3} \\ -\frac {1}{2}\sqrt {2} \\ \frac {1}{6}\sqrt {6} \end {array} \right ] \right \} \leftrightarrow 1, \left \{ \left [ \begin {array}{c} \frac {1}{3}\sqrt {3} \\ \frac {1}{2}\sqrt {2} \\ \frac {1}{6}\sqrt {6} \end {array} \right ] \right \} \leftrightarrow 2.$ The columns are these vectors.

Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix $\begin{equation*} A = \left[ \begin{array}{ccc} \frac{1}{3} & \frac{1}{6}\sqrt{3}\sqrt{2} & - \frac{7}{18}\sqrt{3}\sqrt{6} \\ \frac{1}{6}\sqrt{3}\sqrt{2} & \frac{3}{2} & - \frac{1}{12}\sqrt{2}\sqrt{6} \\ - \frac{7}{18}\sqrt{3}\sqrt{6} & - \frac{1}{12} \sqrt{2}\sqrt{6} & - \frac{5}{6} \end{array} \right] \end{equation*}$

The eigenvalues are $1,2,-2.$

Click the arrow to see the answer.

The eigenvectors: $\left \{ \left [ \begin {array}{c} -\frac {1}{3}\sqrt {3} \\ \frac {1}{2}\sqrt {2} \\ \frac {1}{6}\sqrt {6} \end {array} \right ] \right \} \leftrightarrow 1,\left \{ \left [ \begin {array}{c} \frac {1}{3}\sqrt {3} \\ 0 \\ \frac {1}{3}\sqrt {2}\sqrt {3} \end {array} \right ] \right \} \leftrightarrow -2, \left \{ \left [ \begin {array}{c} \frac {1}{3}\sqrt {3} \\ \frac {1}{2}\sqrt {2} \\ -\frac {1}{6}\sqrt {6} \end {array} \right ] \right \} \leftrightarrow 2.$ Then the columns of $Q$ are these vectors

Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix $\begin{equation*} A = \left[ \begin{array}{ccc} - \frac{1}{2} & - \frac{1}{5}\sqrt{6}\sqrt{5} & \frac{1}{10}\sqrt{5} \\ - \frac{1}{5}\sqrt{6}\sqrt{5} & \frac{7}{5} & - \frac{1}{5}\sqrt{6} \\ \frac{1}{10}\sqrt{5} & - \frac{1}{5}\sqrt{6} & - \frac{9}{10} \end{array} \right] \end{equation*}$

The eigenvalues are $-1,2,-1$ where $-1$ is listed twice because it has multiplicity 2 as a zero of the characteristic equation.

Click the arrow to see the answer.

The eigenvectors and eigenvalues are: $\left \{ \left [ \begin {array}{c} -\frac {1}{6}\sqrt {6} \\ 0 \\ \frac {1}{6}\sqrt {5}\sqrt {6} \end {array} \right ] ,\left [ \begin {array}{c} \frac {1}{3}\sqrt {2}\sqrt {3} \\ \frac {1}{5}\sqrt {5} \\ \frac {1}{15}\sqrt {2}\sqrt {15} \end {array} \right ] \right \} \leftrightarrow -1,\left \{ \left [ \begin {array}{c} \frac {1}{6}\sqrt {6} \\ -\frac {2}{5}\sqrt {5} \\ \frac {1}{30}\sqrt {30} \end {array} \right ] \right \} \leftrightarrow 2 .$ The columns of $Q$ are these vectors. $\left [ \begin {array}{ccc} -\frac {1}{6}\sqrt {6} & \frac {1}{3}\sqrt {2}\sqrt {3} & \frac {1}{6}\sqrt {6} \\ 0 & \frac {1}{5}\sqrt {5} & -\frac {2}{5}\sqrt {5} \\ \frac {1}{6}\sqrt {5}\sqrt {6} & \frac {1}{15}\sqrt {2}\sqrt {15} & \frac {1}{30} \sqrt {30} \end {array} \right ] ^{T}\left [ \begin {array}{ccc} - \frac {1}{2} & - \frac {1}{5}\sqrt {6}\sqrt {5} & \frac {1}{10}\sqrt {5} \\ - \frac {1}{5}\sqrt {6}\sqrt {5} & \frac {7}{5} & - \frac {1}{5}\sqrt {6} \\ \frac {1}{10}\sqrt {5} & - \frac {1}{5}\sqrt {6} & - \frac {9}{10} \end {array} \right ] \cdot$ $\left [ \begin {array}{ccc} -\frac {1}{6}\sqrt {6} & \frac {1}{3}\sqrt {2}\sqrt {3} & \frac {1}{6}\sqrt {6} \\ 0 & \frac {1}{5}\sqrt {5} & -\frac {2}{5}\sqrt {5} \\ \frac {1}{6}\sqrt {5}\sqrt {6} & \frac {1}{15}\sqrt {2}\sqrt {15} & \frac {1}{30} \sqrt {30} \end {array} \right ] =\left [ \begin {array}{rrr} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end {array} \right ]$

Explain why a matrix $A$ is symmetric if and only if there exists an orthogonal matrix $Q$ such that $A=Q^{T}DQ$ for $D$ a diagonal matrix.

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If $A$ is given by the formula, then $A^{T}=Q^{T}D^{T}Q=Q^{T}DQ=A$ Next suppose $A=A^{T}.$ Then by the theorems on symmetric matrices, there exists an orthogonal matrix $Q$ such that $QAQ^{T}=D$ for $D$ diagonal. Hence $A=Q^{T}DQ$

Show that if $A$ is a real symmetric matrix and $\lambda$ and $\mu$ are two different eigenvalues, then if $\vec {x}$ is an eigenvector for $\lambda$ and $\vec {y}$ is an eigenvector for $\mu ,$ then $\vec {x}\dotp \vec {y}=0.$ Also all eigenvalues are real. Supply reasons for each step in the following argument. First $\begin{equation*} \lambda \vec{x}^{T}\overline{\vec{x}}=\left( A\vec{x}\right) ^{T} \overline{\vec{x}}=\vec{x}^{T}A\overline{\vec{x}}=\vec{x}^{T} \overline{A\vec{x}}=\vec{x}^{T}\overline{\lambda }\overline{\vec{x}} =\overline{\lambda }\vec{x}^{T}\overline{\vec{x}} \end{equation*}$ and so $\lambda =\overline {\lambda }.$ This shows that all eigenvalues are real. It follows all the eigenvectors are real. Why? Now let $\vec {x},\vec {y} ,\mu$ and $\lambda$ be given as above. $\begin{equation*} \lambda \left( \vec{x}\dotp \vec{y}\right) =\lambda \vec{x}\dotp \vec{y}=A\vec{x}\dotp \vec{y}=\vec{x}\dotp A\vec{y}=\vec{x}\dotp \mu \vec{y}=\mu \left( \vec{x}\dotp \vec{y}\right) =\mu \left( \vec{x}\dotp \vec{y}\right) \end{equation*}$ and so $\begin{equation*} \left( \lambda -\mu \right) \vec{x}\dotp \vec{y}=0 \end{equation*}$ Why does it follow that $\vec {x}\dotp \vec {y}=0?$

Click the arrow to see the answer.

Since $\lambda \neq \mu ,$ it follows $X \dotp \vec {y} =0.$

Challenge Exercises

If $W$ is a subspace of $\RR ^n$ , show that $\left (U^{\perp }\right )^\perp = U$ .

Show that $W \subseteq \left (U^{\perp }\right )^\perp$ , then use Theorem th:023783c twice.

If $W$ is a subspace of $\RR ^n$ , show how to find an $n \times n$ matrix $A$ such that $W = \{\vec {x} \mid A\vec {x} = \vec {0}\}$ .

Practice Problem prob:8_1_13.

Let $\{\vec {y}_{1}, \vec {y}_{2}, \dots , \vec {y}_{m}\}$ be a basis of $W^\perp$ , and let $A$ be the $n \times n$ matrix with rows $\vec {y}^T_1, \vec {y}^T_2, \dots , \vec {y}^T_m, 0, \dots , 0$ . Then $A\vec {x} = \vec {0}$ if and only if $\vec {y}_{i} \dotp \vec {x} = 0$ for each $i = 1, 2, \dots , m$ ; if and only if $\vec {x}$ is in $W^{\perp \perp } = U$ .

If $U$ and $W$ are subspaces, define $U+W$ to be the set of all possible sums of elements of $U$ with elements of $W$ . Is $U+W$ a subspace? Show that $(U + W)^\perp = U^\perp \cap W^\perp$ .

Think of $\RR ^n$ as consisting of row vectors.

Let $E$ be an $n \times n$ matrix, and let $W = \{\vec {x} E \mid \vec {x} \mbox { in } \RR ^n\}$ . Show that the following are equivalent.

(a): $E^{2} = E = E^{T}$ ( $E$ is a projection matrix).
(b): $(\vec {x} - \vec {x}E) \dotp (\vec {y}E) = 0$ for all $\vec {x}$ and $\vec {y}$ in $\RR ^n$ .
(c): $\mbox {proj}_W(\vec {x}) = \vec {x}E$ for all $\vec {x}$ in $\RR ^n$ .
For (ii) implies (iii): Write $\vec {x} = \vec {x}E + (\vec {x} - \vec {x}E)$ and use the uniqueness argument preceding the definition of $\mbox {proj}_W(\vec {x})$ . For (iii) implies (ii): $\vec {x} - \vec {x}E$ is in $W^\perp$ for all $\vec {x}$ in $\RR ^n$ .

If $E$ is a projection matrix, show that $I - E$ is also a projection matrix.

If $EF = 0 = FE$ and $E$ and $F$ are projection matrices, show that $E + F$ is also a projection matrix.

If $A$ is $m \times n$ and $AA^{T}$ is invertible, show that $E = A^{T}(AA^{T})^{-1}A$ is a projection matrix.

Consider $A = \begin {bmatrix} 0 & a & 0 \\ a & 0 & c \\ 0 & c & 0 \end {bmatrix}$ where one of $a, c \neq 0$ . Show that the characteristic polynomial (see Definition def:char_poly_complex) is given by $c_{A}(z) = z(z - k)(z + k)$ , where $k = \sqrt {a^2 + c^2}$ , and find an orthogonal matrix $Q$ such that $Q^{-1}AQ$ is diagonal.

Click the arrow to see the answer.

$Q = \frac {1}{\sqrt {2}k}\begin {bmatrix} c\sqrt {2} & a & a \\ 0 & k & -k \\ -a\sqrt {2} & c & c \end {bmatrix}$

Consider $A = \begin {bmatrix} 0 & 0 & a \\ 0 & b & 0 \\ a & 0 & 0 \end {bmatrix}$ . Show that the characteristic polynomial (see Definition def:char_poly_complex) is given by $c_{A}(z) = (z - b)(z - a)(z + a)$ , and find an orthogonal matrix $Q$ such that $Q^{-1}AQ$ is diagonal.

Given $A = \begin {bmatrix} b & a \\ a & b \end {bmatrix}$ , show that the characteristic polynomial (see Definition def:char_poly_complex) is given by $c_{A}(z) = (z - a - b)(z + a - b)$ , and find an orthogonal matrix $Q$ such that $Q^{-1}AQ$ is diagonal.

Consider $A = \begin {bmatrix} b & 0 & a \\ 0 & b & 0 \\ a & 0 & b \end {bmatrix}$ . Show that the characteristic polynomial (see Definition def:char_poly_complex) is given by $c_{A}(z) = (z - b)(z - b - a)(z - b + a)$ , and find an orthogonal matrix $Q$ such that $Q^{-1}AQ$ is diagonal.

Octave Exercises

Use Octave to check your work on Problems prb:9.1 to prb:9.12. The first steps of prb:9.2 are in the code below. See if you can finish the rest of the problem.

To use Octave, go to the Sage Math Cell Webpage, copy the code below into the cell, select OCTAVE as the language, and press EVALUATE.

A=[4 1 -2; 1 4 -2; -2 -2 7];
 

 
[Q,D]=eig(A)
 

 
% The eigenvalue 3 has algebraic multiplicity 2 and geometric multiplicity 2
 
% One way to see this is to compute:
 
rref(A-3*eye(size(A)))
 

                                                                  

                                                                  
 
% By hand, we will not get the same eigenvectors for 3 that Octave did, and we will need to
 
% orthogonalize them using Gram-Schmidt.  Octave does so automatically.  Observe:
 
transpose(Q)*Q
 

 
% And so we can check that the following is (approximately) the same as D:
 
transpose(Q)*A*Q

Bibliography

Some of these problems come from Section 7.4 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.

Other problems come from the second part of Section 8.1 of Keith Nicholson’s Linear Algebra with Applications. (CC-BY-NC-SA)

W. Keith Nicholson, Linear Algebra with Applications, Lyryx 2018, Open Edition, p. 422–423

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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 9: Orthogonality

Challenge Exercises

Octave Exercises

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