Solving for Eigenvectors of an Already Diagonalized Matrix

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I'm a bit stuck in solving for the eigenvectors of the following equation. \begin{align} \begin{bmatrix} 2m+M-2k\omega & 0 \\ 0 & \dfrac{1}{12}Ml^2 + \dfrac{1}{2}ml^2 - \dfrac{k}{2}l^2\omega \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} =0 \end{align}

I found the eigenvalues using Mathematica, but finding the Eigenvectors was problematic, so I attempted it by hand. The eigenvalues are: \begin{align} \omega_1=\dfrac{3m}{2k} \ \ , \ \ \omega_2 = \dfrac{6m+M}{6k} \end{align}

In solving for the eigenvector corresponding to $\omega_1$, I get to the following equation.

\begin{align} \begin{bmatrix} M-m & 0 \\ 0 & \dfrac{1}{12}Ml^2 - \dfrac{1}{6}ml^2 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} =0 \end{align}

I'm totally unsure on how to solve this for the eigenvector. Essentially, my confusion lies in both $a_1$ and $a_2$ equal to zero here, since there are no off-diagonal terms.

Any help would be appreciated, thank you.