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If I am to find a non-zero vector in the kernel of a singular matrix $A$, that is, an eigenvector to the eigenvalue 0, how do I get the power iteration with shift to converge to a such vector?

Obviously I cannot choose shift 0, as the $A$ will be numerically singular.

Thanks in advance

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    If you shift to $A + \epsilon I$ for $\epsilon > \|A\|$, then $A$ will no longer be singular. I don't know if you necessarily need to avoid a singular $A$, though; it suffices to have eigenvalue associated with the eigenvalue $0$ to be non-zero (at least, it is if you use something like the $QR$ method). – 2017-02-17
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    However: if you're looking for the eigenvector associated with $0$, you're really just solving $Ax = 0$. Why not apply the usual "row-reduction" technique? – 2017-02-17

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