An identity involving the Drazin inverse - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:21:23Z http://mathoverflow.net/feeds/question/68176 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68176/an-identity-involving-the-drazin-inverse An identity involving the Drazin inverse Glynne 2011-06-19T00:31:25Z 2011-06-19T01:35:02Z <p>Thanks to Higham I know that \$A f(BA) = f(AB) A\$ for any two matrices whose sizes are compatible.</p> <p>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\$. </p> <p>I have validated this relationship via numerical experiments with random matrices, I just can't \$prove\$ it. </p> <p>Can you prove (or disprove) it?</p> http://mathoverflow.net/questions/68176/an-identity-involving-the-drazin-inverse/68183#68183 Answer by S. Sra for An identity involving the Drazin inverse S. Sra 2011-06-19T01:35:02Z 2011-06-19T01:35:02Z <p><a href="http://benisrael.net/GI-LECTURE-7.pdf" rel="nofollow">These notes</a> say that the Drazin Inverse is the matrix function corresponding to \$f(z) = 1/z\$, defined on the <em>nonzero</em> eigenvalues. Thus, by the theorem that you cite, the said equality should hold.</p>