Find the eigenvalues and eigenvectors of this matrix $$ \begin{bmatrix} -3 & 3 \\ -6 & 3 \end{bmatrix} $$
I found the eigenvalues $$λ = ±3i$$ However, I am having trouble with finding the eigenvectors. I tried checking steps with online calculators but I couldn't quite follow what was being done.
The matrix for $λ=3i$: $$ \begin{bmatrix} -3-3i & 3 \\ -6 & 3-3i \end{bmatrix} $$
I began by dividing $R_1$ with $(-3-3i)$ to get the first element to be $1$.
$$ \begin{bmatrix} 1 & \frac{1}{(-1-i)} \\ -6 & 3-3i \end{bmatrix} $$
Now I would multiply $R_1$ with 6 and subtract that from $R_2$, but that doesn't give me the correct eigenvector. How should I reduce this matrix?
The answer should be: $$v_1= \begin{bmatrix} 1 \\ 1 +i \end{bmatrix} $$
$$v_2= \begin{bmatrix} 1 \\ 1 -i \end{bmatrix} $$
Thank you!
Note that$$\frac1{-1-i}=-\frac12+\frac i2$$and that therefore the second line of the matrix $\left[\begin{smallmatrix}1&\frac1{-1-i}\\-6&3-3i\end{smallmatrix}\right]$ is the first line times $-6$. So, solving the equation $x+\frac y{-1-i}=0$ is the same thing as solving the equation $-6x+(3-3i)y=0$. One solution will be $(1,1+i)$, since $1+\frac{1+i}{-1-i}=1-1=0$.