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Q51E

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Found in: Page 325

### Linear Algebra With Applications

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

# Find an eigenbasis for the given matrice and diagonalize:${\mathbit{A}}{\mathbf{=}}\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

The eigenbasis for the given matrice is $\left[\begin{array}{cc}0& 0\\ 0& 2\end{array}\right]$.

See the step by step solution

## Step 1: Solving the given matrices

We solve:

$det\left(A-\lambda l\right)=0\phantom{\rule{0ex}{0ex}}\left|\begin{array}{cc}1-\lambda & 1\\ 1& 1-\lambda \end{array}\right|=0\phantom{\rule{0ex}{0ex}}{\left(1-\lambda \right)}^{2}-1=0\phantom{\rule{0ex}{0ex}}{\lambda }^{2}-2\lambda =0\phantom{\rule{0ex}{0ex}}\lambda \left(\lambda -2\right)=0\phantom{\rule{0ex}{0ex}}{\lambda }_{1}=0,{\lambda }_{2}=2$

## Step 2: Solving by different values of λ

For $\lambda =0$, we solve:

$\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}{x}_{1}+{y}_{1}=0$

We can choose an eigenvector:

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

For $\lambda =2$, we solve:

$\left[\begin{array}{cc}-1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}{x}_{1}-{y}_{1}=0$

We can choose an eigenvector:

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

Now, localid="1659533233518" $\left\{{v}_{1},{v}_{2}\right\}$ is an eigenbasis for ${R}^{2}$, therefore the diagonalization of A in the eigenbasis is $\left[\begin{array}{cc}0& 0\\ 0& 2\end{array}\right]$.

Hence the final answer is $\left[\begin{array}{cc}0& 0\\ 0& 2\end{array}\right]$.