$A = U\Sigma V^T$
When given a 2x2 matrix, I know how to get as far as to find the eigenvalues and square root them to find the values that go along the diagonal in $\Sigma$. However, let's say
$A = \begin{bmatrix} 0 & 0 \\ -2 & 0 \end{bmatrix}$
So after doing $A^T A$, you'd arrive at $\begin{bmatrix} 4 & 0 \\ 0 & 0 \end{bmatrix}$ with 4 and 0 being eigenvalues, 2 and 0 being the singular values.
...Normally to find eigenvalues, you do $(A-\lambda I)x = 0$ and then with that new matrix, you RREF it and then convert it to parametric vector form to get your corresponding eigenvector.
In the aforementioned example, I'd be inclined to use $A^T A$ as my "A" in $(A-\lambda I)x= 0$. However,
$\begin{bmatrix} 4 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & -4 \end{bmatrix}$ parametric vector forming that becomes...
$\begin{bmatrix} 0 \\ 0 \end{bmatrix}$
but it should be
$\begin{bmatrix} 1 \\ 0 \end{bmatrix}$.
Help identifying what I am messing up would be much appreciated. Here's work I have that's a bit clearer: enter image description here
