Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
1-\lambda & 4\\
2 & 3-\lambda \\
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
1-\lambda & 4\\
2 & 3-\lambda \\
\end{bmatrix}=0$
$\left (1- \lambda \right ) (3- \lambda)=0$
$\lambda^2-4\lambda -5=0=0$
$\lambda_1=5, \lambda_2=-1$
2. Find eigenvectors:
For $\lambda_1=5$
let $B=A-\lambda_1I$
$B=\begin{bmatrix}
1-\lambda & 4\\
2 & 3-\lambda \\
\end{bmatrix}=\begin{bmatrix}
-4 & 4\\
2 & -2 \\
\end{bmatrix}$
Then,
$B\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[B|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1 & -1 & 0 \\
0 & 0& 0
\end{array}\right)
\]
let $r$ is a free variable.
Then $x_1= x_2=r \\
$
$\vec{V}=(1,1) \\
\rightarrow dim(E_1)=1$
For $\lambda_1=-1$
let $C=A-\lambda_1I$
$C=\begin{bmatrix}
1-\lambda & 4\\
2 & 3-\lambda \\
\end{bmatrix}=\begin{bmatrix}
2 & 4\\
2 & 4 \\
\end{bmatrix}$
Then,
$C\vec{V}$=$\vec{0}$
Use reduced row echelon form
$[C|\vec{0}]$=
\[
\left(\begin{array}{@{}cc|c@{}}
1 & 2 & 0 \\
0 & 0& 0
\end{array}\right)
\]
let $s$ is a free variable.
Then $x_1= -2x_2=-2S $
$\vec{V}=(-2,1) \\
\rightarrow dim(E_2)=1$
Hence, the matrix $A$ is nondefective.
