linear algebra - 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 $W$?

Answer

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