Let $$A := \begin{bmatrix} -3 & 0 & 1 & 2 \\ 4 & 0 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ \end{bmatrix}$$
Compute a singular value decomposition (over $\mathbb{C}$) of $A$.
Compute the pseudo-inverse of $A$.
For (1), I have computed $$AA^T=\begin{matrix} 14 & -12 & 5 \\ -12 & 16 & 0 \\ 5 & 0 & 5 \\ \end{matrix}$$ Then to find the Eigenvalues, I computed $$\begin{matrix} 14-\lambda & -12 & 5 \\ -12 & 16-\lambda & 0 \\ 5 & 0 & 5-\lambda \\ \end{matrix}$$ and took the determinate of my new $3x3$ matrix and have now ended up with this $$=-\lambda^3+35\lambda^2-255\lambda+800$$ Now i'm stuck. I feel as though I am lost and/or have calculated wrong.
The last column of $AA^T$ should be $\left(\begin{array}{c} 5 \\ 0 \\ 5 \end{array}\right).$
Also, notice that the second column of $A$ is a zero column. the third column and the fourth column are multiple of each other. Hence the rank of the matrix is 2. Hence you are sure. One of the eigenvalue of $AA^T$ is $0$.
The characteristic equation that you should have obtained should have $0$ as one of the root. Your characteristic equation is wrong. The constant term has to be zero. Are you able to find the mistake?