Suggested languages for you:

Americas

Europe

Q3E

Expert-verified
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 eigensspace .use the geometric multiplicities of the eigenvalues to determine whether a matrix is diagonalizable. For each of the matrices A in Exercise 1 through 20 ,find all (real) eigenvalues. Then find a basis of each eigenspaces ,and diagonalize A, if you can. Do not use technology.$\left[\begin{array}{cc}6& 3\\ 2& 7\end{array}\right]$

Given eigenvalue, find a basis of the associated eigensspace.

$\mathrm{det}\left(\mathrm{A}-\mathrm{\lambda l}\right)=0$

Now, $\left\{{v}_{1}.{v}_{2}\right\}$ is an Eigen basis for ${R}^{2}$ , so the diagonalization of in this eigenbasis is

$\left[\begin{array}{cc}4& 0\\ 0& 9\end{array}\right]$

See the step by step solution

## Step 1: definition of matrices

A function is defined as a relationship between a set of inputs that each have one output.

Given,

$det\left(A-\lambda l\right)=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}6-\lambda & 3\\ 2& 7-\lambda \end{array}\right]=0\phantom{\rule{0ex}{0ex}}\left(6-\lambda \right)\left(7-\lambda \right)-6=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\lambda }^{2}-13\lambda +36=0\phantom{\rule{0ex}{0ex}}\left(\lambda -4\right)\left(\lambda -9\right)=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\lambda }_{1}=4,{\lambda }_{2}=9$

We solved,

$\lambda =4\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}det\left(A-4l\right)x=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}2& 3\\ 2& 3\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}2{x}_{1}+3{x}_{2}=0$

Basic of this eigenspace is

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

## Step 2: multiply the matrices

Similarly $\lambda =9\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(A-9l\right)x=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}-3& 3\\ 2& -2\end{array}\right]\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}-3{x}_{1}+3{x}_{2}=0,2{x}_{1}-{x}_{2}=0\phantom{\rule{0ex}{0ex}}{x}_{1}-{x}_{2}=0$

Basic of Eigen space is

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

Hence,

Now, $\left\{{v}_{1},{v}_{2}\right\}$ is an Eigen basis for ${R}^{2}$ , so the diagonalization of in this eigenbasis is

$\left[\begin{array}{cc}4& 0\\ 0& 9\end{array}\right]$