Answer
$\vec{b_1}=\begin{bmatrix}
1\\
1
\end{bmatrix}$ and $\vec{b_2}=\begin{bmatrix}
1\\
3
\end{bmatrix}$
Work Step by Step
$A=P^{-1}DP$
$det\begin{bmatrix}
-\lambda&1\\
-3&4-\lambda
\end{bmatrix}=(-\lambda)(4-\lambda)+3=0$
$\lambda=1, \lambda=3$
When $\lambda=1$,
$A-\lambda I=\begin{bmatrix}
-1&1\\
-3&3
\end{bmatrix}$ and $Nul(A-\lambda I)=\begin{bmatrix}
1\\
1
\end{bmatrix}$
When $\lambda=3$,
$A-\lambda I=\begin{bmatrix}
-3&1\\
-3&1
\end{bmatrix}$ and $Nul(A-\lambda I)=\begin{bmatrix}
1\\
3
\end{bmatrix}$
$P=\begin{bmatrix}
1&1\\
1&3
\end{bmatrix}$ and $D=\begin{bmatrix}
1&0\\
0&3
\end{bmatrix}$
By Theorem 8, the basis for $R^2$ is the columns of P, $\begin{bmatrix}
1\\
1
\end{bmatrix}$ and $\begin{bmatrix}
1\\
3
\end{bmatrix}$
