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Q8E

Expert-verifiedFound in: Page 345

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**For a given eigenvalue, find a basis of the associated eigenspace. Use the geometric multiplicities of the eigenvalues to determine whether a matrix is diagonalizable. For each of the matrices A in Exercises 1 through 20, find all (real) eigenvalues. Then find a basis of each eigenspace, and diagonalize A, if you can. Do not use technology**

**${\mathbf{8}}{\mathbf{(}}{\left(\begin{array}{ccc}1& 1& 0\\ 0& 2& 2\\ 0& 0& 3\end{array}\right)}$ **

The eigenbasis for the ${\mathrm{\mathbb{R}}}^{3}$, so the diagonalization of A in the eigenbasis is $\left(\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right)$

**Algebraic versus geometric multiplicity If λ is an eigenvalue of a square matrix A, **

**then gemu(λ) ≤ almu(λ).**

**This matrix is upper triangle, so its eigenvalues are the entries units main diagonal, which are**

**$\begin{array}{l}{\lambda}_{1}=1\\ {\lambda}_{2}=2\\ {\lambda}_{3}=3\end{array}$**

**For $\lambda =1$, we solve**

**role="math" localid="1659588503956" $\left(\mathrm{A}-\mathrm{l}\right)\mathrm{x}=0\phantom{\rule{0ex}{0ex}}\left(\begin{array}{ccc}0& 1& 0\\ 0& 1& 2\\ 0& 0& 2\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\phantom{\rule{0ex}{0ex}}{x}_{2}=0,{x}_{2}+2{x}_{3}=0,2{x}_{3}=0\phantom{\rule{0ex}{0ex}}{x}_{2}=0,{x}_{3}=0$**

**The basic of this eigenspace is**

**$\left\{\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\right\}=\left\{{v}_{1}\right\}$ **

First, $\lambda =2$,

role="math" localid="1659588665728" $\left(A-2\lambda \right)x=0\phantom{\rule{0ex}{0ex}}\left(\begin{array}{ccc}-1& 1& 0\\ 0& 0& 2\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\phantom{\rule{0ex}{0ex}}-{x}_{1}={x}_{2}=0,2{x}_{3}=0,{x}_{3}=0\phantom{\rule{0ex}{0ex}}{x}_{1}-{x}_{2}=0,{x}_{3}=0$

The basic for this eigenspace is $\left\{\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]\right\}=\left\{{v}_{2}\right\}$

Now, $\lambda =3$ , we solve

$\left(A-3l\right)x=0\phantom{\rule{0ex}{0ex}}\left(\begin{array}{ccc}-2& 1& 0\\ 0& -1& 2\\ 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\phantom{\rule{0ex}{0ex}}-2{x}_{1}+{x}_{2}=0,-{x}_{2}+2{x}_{3}=0$

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

Therefore, $\left\{{V}_{1},{V}_{2},{V}_{3}\right\}$ is an eigenbasis for ${\mathrm{\mathbb{R}}}^{3}$ , so the diagonalization of A in the eigenbasis is $\left(\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 3\end{array}\right)$

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