# Solving $Ax = 0$ with Singular Value Decomposition

In my computer vision class, we have seen the problem of solving $Ax = 0$ numerous times. Initially, we solved the problem by computing the smallest eigenvectors corresponding to the smallest eigenvalues of $A$.

However, we are now using singular value decomposition and extracting the the rightmost column vector of $V$ from $[U, D, V]$.

I am confused, because my understanding is that the columns in $V$ correspond to eigenvectors of $A^T A$. Therefore, how can the solution in one case be an eigenvector of $A$ and in another an eigenvector of $A^T A$?

• Keep in mind that $A$ and $A^T A$ have the same null space. Also,if you want to find a unit vector $v$ such that $\|Av\|$ is as small as possible, the SVD method does exactly that. Apr 18, 2016 at 3:39

Because a vector that corresponds to the zero eigenvalue of $A$ also corresponds to the zero eigenvalue of $A^TA$. This is not the case for nonzero eigenvalues, but since you're looking for the null space, that is what is of interest to you.