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Found in: Page 323

### Linear Algebra With Applications

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

# Find all ${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrices for which ${\stackrel{\mathbf{\to }}{\mathbf{e}}}_{{\mathbf{1}}}{\mathbf{=}}\left(\begin{array}{c}1\\ 0\end{array}\right)$ is an eigenvector with associated eigenvalue 5.

So, the required matrix is $\left[\begin{array}{cc}5& b\\ 0& d\end{array}\right]$.

See the step by step solution

## Step 1: Define the eigenvector

Eigenvector: An eigenvector of A is a nonzero vector v in ${{\mathbf{R}}}_{{\mathbf{n}}}$ such that ${\mathbit{A}}{\mathbit{v}}{\mathbf{=}}{\mathbit{\lambda }}{\mathbit{v}}$, for some scalar ${\mathbit{\lambda }}$.

## Step 2: Solve for A

If is an Eigen vector of A this means that:

$A\stackrel{\to }{v}={\lambda }^{\text{'}}v$

Now let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],\stackrel{\to }{v}=\left[\begin{array}{c}1\\ 0\end{array}\right]$, and $\lambda =5$.

We want to solve for A.

## Step 3: Evaluation

To do this we will replace everything into the equation $A\stackrel{\to }{v}=\lambda {v}^{\text{'}}$ and we will have:

$\stackrel{\to }{v}=\lambda \stackrel{\to }{v}\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=5\left[\begin{array}{c}1\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{c}a\\ c\end{array}\right]=\left[\begin{array}{c}5\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}a=5\phantom{\rule{0ex}{0ex}}c=0$

b,d are any real number

Thus, all matrices of the form $\left[\begin{array}{cc}5& b\\ 0& d\end{array}\right]$will satisfy the given requirements.