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Thanks to Higham I know that $A f(BA) = f(AB) A$ for any two matrices whose sizes are compatible.

Now I believe that $A (BA)^D = (AB)^D A$, even though the Drazin inverse is not the same function (polynomial?) for $AB$ as for $BA$.

I have validated this relationship via numerical experiments with random matrices, I just can't $prove$ it.

Can you prove (or disprove) it?

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$f$ is ... ? – Ricky Demer Jun 19 '11 at 0:58
I don't believe there are any restrictions on $f$. – Glynne Jun 20 '11 at 23:33
up vote 1 down vote accepted

These notes say that the Drazin Inverse is the matrix function corresponding to $f(z) = 1/z$, defined on the nonzero eigenvalues. Thus, by the theorem that you cite, the said equality should hold.

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Thanks. The relationship does not hold (in general) for either the Moore-Penrose inverse or the regular matrix inverse. I assume that's because of those $zero$ eigenvalues? – Glynne Jun 20 '11 at 23:36
Given the relation to $1/z$, it seems that for being viewed as a matrix function, the Drazin inverse seems to be the correct object. I did not previously know of this nice relation. – Suvrit Jun 21 '11 at 0:19

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