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Expert-verified 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:${\mathbit{A}}{\mathbf{=}}\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 6\\ 3& 6& 9\end{array}\right]$

The eigenbasis for the given matrice is $\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 14\end{array}\right]$.

See the step by step solution

## Step 1: Solving the given matrices

We solve:

$\mathrm{det}\left(\mathrm{A}-\mathrm{\lambda l}\right)=0\phantom{\rule{0ex}{0ex}}\left|\begin{array}{ccc}1-\mathrm{\lambda }& 2& 3\\ 2& 4-\mathrm{\lambda }& 6\\ 3& 6& 9-\mathrm{\lambda }\end{array}\right|=0\phantom{\rule{0ex}{0ex}}\left(1-\mathrm{\lambda }\right)\left(4-\mathrm{\lambda }\right)\left(9-\mathrm{\lambda }\right)+36+36-36\left(1-\mathrm{\lambda }\right)-9\left(4-\mathrm{\lambda }\right)-4\left(9-\mathrm{\lambda }\right)=0\phantom{\rule{0ex}{0ex}}-{\mathrm{\lambda }}^{3}+14{\mathrm{\lambda }}^{2}=0\phantom{\rule{0ex}{0ex}}{\mathrm{\lambda }}^{2}\left(-\mathrm{\lambda }+14\mathrm{\lambda }\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{\lambda }}_{1,2}=0,{\mathrm{\lambda }}_{3}=14$

## Step 2: Solving by different values of λ

For $\lambda =0$, we solve:

$\left[\begin{array}{ccc}1& 2& 3\\ 2& 4& 6\\ 3& 6& 9\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {y}_{1}\\ {z}_{1}\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}{x}_{1}+2{y}_{1}+3{z}_{1}=0$

We can see all the eigenvector $\lambda =0$ for are spanned by vectors:

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

For $\lambda =14$, we solve:

$\left[\begin{array}{ccc}-13& 2& 3\\ 2& -10& 6\\ 3& 6& -5\end{array}\right]\left[\begin{array}{c}{x}_{3}\\ {y}_{3}\\ {z}_{3}\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}-13{x}_{3}+2{y}_{3}+3{z}_{3}=0,2{x}_{3}-10{y}_{3}+6{z}_{3}+0,3{x}_{3}+6{y}_{3}-5{z}_{3}=0$

We can choose an eigenvector:

${v}_{3}=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

Now, $\left\{{v}_{1},{v}_{2},{v}_{3}\right\}$ is an eigenbasis for ${R}^{3}$, therefore the diagonalization of A in the eigenbasis is

$\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 14\end{array}\right]$

Hence the final answer is $\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 14\end{array}\right]$. ### Want to see more solutions like these? 