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Q42E

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Found in: Page 393

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Find a symmetric 2x2 matrix B such that ${{\mathbit{B}}}^{{\mathbf{3}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{5}}\left[\begin{array}{cc}12& 14\\ 14& 33\end{array}\right]$

A symmetric 2x2 matrix $B=\frac{1}{5}\left[\begin{array}{cc}6& 2\\ 2& 9\end{array}\right]$

See the step by step solution

## Step 1: Symmetric matrix:

In linear algebra, a symmetric matrix is a square matrix that stays unchanged when its transpose is calculated. In that instance, a symmetric matrix is one whose transpose equals the matrix itself.

## Step 2: To determine the eigenvalues of the matrix A :

Let $A={B}^{3}$ , determine the eigenvalues of the matrix A

$det\left(A-\lambda {l}_{n}\right)=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}\frac{12}{5}-\lambda & \frac{14}{5}\\ \frac{14}{5}& \frac{33}{5}-\lambda \end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left(\frac{12}{5}-\lambda \right)\left(\frac{33}{5}-\lambda \right)-\left(\frac{14}{5}\right)\left(\frac{14}{5}\right)=0\phantom{\rule{0ex}{0ex}}25{\lambda }^{2}-225\lambda +200=0\phantom{\rule{0ex}{0ex}}\left(\lambda -8\right)\left(\lambda -1\right)=0$

The eigenvalues are $\lambda =1$ , and $\lambda =8$ . Now we obtain the eigenvectors.

CASE 1: When $\lambda =1$

$\left[A-l\right]\stackrel{~}{x}=0⇒\left[\begin{array}{cc}\frac{12}{5}-1& \frac{14}{5}\\ \frac{14}{5}& \frac{33}{5}-1\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}\frac{7}{5}& \frac{14}{5}\\ \frac{14}{5}& \frac{28}{5}\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0$

apply row operation ${R}_{2}\to 2{R}_{1}-{R}_{2}$ :

$\left[\begin{array}{cc}\frac{7}{5}& \frac{14}{5}\\ 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0$

apply row operation ${R}_{1}\to 5{R}_{1}$

$\left[\begin{array}{cc}7& 14\\ 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0$

## Step 3: Compute eigenvector corresponding:

so here we have $a=-2b$ , choosing $b=1$ yields $a=-2$ . The eigenvector corresponding to the eigenvalue $\lambda =1$ is

$\stackrel{\to }{v}=\left[\begin{array}{c}-2\\ 1\end{array}\right]$

therefore,

${E}_{1}=span\left(\frac{1}{\sqrt{5}}\left[\begin{array}{c}-2\\ 1\end{array}\right]\right)$

CASE 2: When $\lambda =8$

role="math" localid="1660104429366" $\left[A-l\right]\stackrel{~}{x}=0\to \left[\begin{array}{cc}\frac{12}{5}-8& \frac{14}{5}\\ \frac{14}{5}& \frac{33}{5}-8\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}-\frac{28}{5}& \frac{14}{5}\\ \frac{14}{5}& -\frac{7}{5}\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0$

apply row operation ${R}_{2}\to 2{R}_{2}+{R}_{1}$ and ${R}_{1}\to 5{R}_{1}$ :

$\left[\begin{array}{cc}-28& 14\\ 0& 0\end{array}\right]\left[\begin{array}{c}a\\ b\end{array}\right]=0$

so here we have $a=\frac{1}{2}b$ , choosing b = 2 yields a = 1 . The eigenvector corresponding to the eigenvalue $\lambda =8$ is

${\stackrel{\to }{v}}_{2}=\left[\begin{array}{c}1\\ 2\end{array}\right]$

therefore,

${E}_{8}=span\left(\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ 2\end{array}\right]\right)$

## Step 4: Find a symmetric 2x2 matrix B :

In decomposition $A=SD{S}^{-1}$ , the orthogonal matrix is given by

$S=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]$

and the diagonal matrix is given by

$D=\left[\begin{array}{cc}8& 0\\ 0& 1\end{array}\right]$

Now let

$P=\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]$

notice that ${P}^{3}=D$ so ${\left(SD{S}^{-1}\right)}^{3}=S{P}^{3}{S}^{-1}=A$. So, we compute B such that

${S}^{-1}BS=P\to B=SP{S}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]\left[\begin{array}{cc}2& 0\\ 0& 1\end{array}\right]{\left[\begin{array}{c}1\\ \sqrt{5}\end{array}\left[\begin{array}{cc}1& -2\\ 2& 1\end{array}\right]\right]}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{5}}\left[\begin{array}{cc}2& -2\\ 4& 1\end{array}\right]\frac{1}{\sqrt{5}}\left[\begin{array}{cc}1& 2\\ -2& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{5}\left[\begin{array}{cc}6& 2\\ 2& 9\end{array}\right]$

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