Finding eigenvectors to eigenvalues, and diagonalization I just finished solving a problem on finding eigenvectors corresponding to eigenvalues, however, I'm not sure if it is correct. I was wondering if someone could check my work:
For the matrix $W = 
\begin{bmatrix}
    1 & 2 \\
    3 & 2\\   
\end{bmatrix}$, I must find the eigenvectors corresponding to the eigenvalues, as well as a diagonal matrix similar to W.
I was able to find that the eigenvalues were equal to $\lambda = 4, -1$. Then, I used the equation $(A - \lambda I)v = 0$ to solve for the vector.
When $\lambda = 4$, I set up the equation $\begin{bmatrix}
    1 & 2 \\
    3 & 2\\   
\end{bmatrix} - \begin{bmatrix}
    4 & 0 \\
    0 & 4\\   
\end{bmatrix}$ = $\begin{bmatrix}
    -3 & 2 \\
    3 & -2\\   
\end{bmatrix}$, which gave me the eigenvector $\begin{bmatrix}
    2\\
    3\\   
\end{bmatrix}$.
For $\lambda = -1$, I did the exact same procedure and received the eigenvector which gave me the eigenvector $\begin{bmatrix}
    1\\
    -1\\   
\end{bmatrix}$.
Did I do this part correctly? How do I find a diagonal matrix similar to $W$?
 A: I think it is worth the exercise to verify that 
$W\mathbf v = \lambda \mathbf v$ 
$W \begin {bmatrix} 2\\3 \end{bmatrix} = 4\begin {bmatrix} 2\\3 \end{bmatrix}$
and
$W \begin {bmatrix} 1\\-1 \end{bmatrix} = -\begin {bmatrix} 1\\-1 \end{bmatrix}$
which it does...in both cases.
In which case:
$W\begin{bmatrix} \mathbf v_1&\mathbf v_2 \end{bmatrix} = \begin{bmatrix} \mathbf v_1&\mathbf v_2 \end{bmatrix}\begin{bmatrix} \lambda_1\\&\lambda_2\end{bmatrix}$
Let $P = \begin{bmatrix} \mathbf v_1&\mathbf v_2 \end{bmatrix}$ and $\Lambda = \begin{bmatrix} \lambda_1\\&\lambda_2\end{bmatrix}$
$WP = P\Lambda\\
P^{-1}WP = \Lambda$
$\lambda$ is a diagonal matrix similar to W
A: we can use Row operations to obtain a diagonal matrix similar to W
W = 
\begin{bmatrix}
    1 & 2 \\
    3 & 2\\   
\end{bmatrix} 
$r_1-r_2=R_1$
gives $$W = 
\begin{bmatrix}
    -2 & 0 \\
    3 & 2\\   
\end{bmatrix}$$
then $R_2=2r_2$ gives
W = 
\begin{bmatrix}
   -2 & 0 \\
    6 & 4\\   
\end{bmatrix}
now $R_2=r_2+3r_1$ gives
$W = 
\begin{bmatrix}
    -2 & 0 \\
    0 & 4\\   
\end{bmatrix}$ 
 and $R_1=\frac{1}{2}r_1$
gives $W = 
\begin{bmatrix}
    -1 & 0 \\
    0 & 4\\   
\end{bmatrix}$
which is in diagonal form, as required, as you can see the diagonal entries are the eigenvalues you calculated 
