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Found in: Page 345

### Linear Algebra With Applications

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

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

See the step by step solution

## Step 1: Algebraic Versus.

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

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

## Step 2: To find the values.

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 $\left\{\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]\right\}=\left\{{v}_{2}\right\}$

## Step 3: To find the values.

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