Answer
See below
Work Step by Step
1. Find eigenvalues:
(A-$\lambda$I)$\vec{V}$=$\vec{0}$
$\begin{bmatrix}
-2-\lambda & 4\\
1 & 1-\lambda
\end{bmatrix}\begin{bmatrix}
v_1\\
v_2
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix}$
$\begin{bmatrix}
-2-\lambda & 4\\
1 & 1-\lambda
\end{bmatrix}=0$
$\left (-2- \lambda \right ) (1- \lambda)-4=0$
$(\lambda+3)(\lambda-2)=0$
$\lambda_1=-3, \lambda_2=2$
Hence, A is nondefective by Corollary 7.2.10. The eigenvectors are easily computed:
$\lambda_1=-3: v=r(-4,1) \\
\lambda_2=2: v=s(1,1)$
Set $S = \begin{bmatrix}
-4 & 1\\
1 & 1
\end{bmatrix}$
then cofactor matrix of $S$ be $M= \begin{bmatrix}
1 & -1\\
-1 & 4
\end{bmatrix}$
Hence, all eigenvectors of $A^T$ is: $v=(1,-1)\\
v=(-1,4)$
