Computing right null vector with smallest eigenvalue

Question

I know that the null space can be calculated from the equation $Ax=0$ by computing $svd(A).$ If we compute svd, $svd(A)=USV,$ $S$ are the singular values, $V$ are the eigenvectors of $A^TA$ and U are the eignevectors of $AA^T$. The vector I need is the last column of $V.$ i.e. the vector corresponding to the smallest eigenvalue What I want to know is how can I compute this vector without finding svd of $A$?

Answer 1

Any null vector corresponds to an eigenvalue (or singular value) of zero, thus it doesn't really make sense to associate one with the "smallest" eigenvalue in this case (because they are all zero). Another way of saying this is that, in the SVD, among singular vectors/eigenvectors corresponding to $0$, there is no canonical ordering of them (there's no fixed notion of "the last column" if multiple columns correspond to $0$).

But you can use $QR$ decomposition to do this, it is faster than SVD.

https://stackoverflow.com/questions/2181418/computing-the-null-space-of-a-matrix-as-fast-as-possible