I want to calculate the SVD ($A = U\Sigma V^*$)of
$$A =
\begin{bmatrix}
0 & 2 \\
0 & 0 \\
0 & 0
\end{bmatrix}$$
but
$$A^TA =
\begin{bmatrix}
0 & 0 \\
0 & 4
\end{bmatrix}$$
which has a zero eigenvalue. The problem with this is that the columns of $U$ are given by
$$u_i = \frac{Av_i}{\sigma_i}$$
where $\sigma_i = \sqrt{\lambda_i}$.
Best Answer
No, $Av_i=\sigma_iu_i$, which is perfectly well defined even when $\sigma_i=0$. The point is $U$ can be decomposed into vectors corresponding to $\sigma_1,\cdots,\sigma_k>0$ and, when $\sigma_i=0$, you pad $U$ with vectors spanning the cokernel (i.e. whatever the range of $A$ misses) of $A$. See the example calculation here:
https://en.wikipedia.org/wiki/Singular_value_decomposition#Example