4
$\begingroup$

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 $A^T A$?

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – littleO
    Apr 18, 2016 at 3:39

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .