Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
2-\lambda & 1\\
-1 & 4-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
2-\lambda & 1\\
-1 & 4-\lambda
\end{bmatrix}=0$
$(2- \lambda)(4-\lambda)+1=0$
$(\lambda^2-3)^2=0$
$\lambda_1=\lambda_2=3$
Hence, A is nondefective by Corollary 7.2.10. The only eigenvector are easily computed:
$\lambda=3: v=r(1,1)$
Hence, $A$ is detective.
We obtain matrix $J_3= \begin{bmatrix}
3 & 1\\
0 & 3
\end{bmatrix}$
There exists a matrix $S=\begin{bmatrix}
v_1,v_2
\end{bmatrix}$ such that $S^{-1}AS =J_3$
Now we assume $v_1=(1,1)\\
v_2=(a,b)$
From exercise 35, we have $(A-\lambda 1)v_2=v_1$
then $\begin{bmatrix}
-1 & 1\\
-1 & 1
\end{bmatrix}\begin{bmatrix}
a\\
b
\end{bmatrix}=\begin{bmatrix}
1\\
1
\end{bmatrix} \\
\rightarrow -a+b=1$
Hence, $S=\begin{bmatrix}
1 & b-1\\
1 & b
\end{bmatrix}$
