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If $A$ and $B$ are Positive semidefinite matrices, then show that $\operatorname{Col}(A B) \cap \text{Null}(A B)=$ $\{0\} $.

From this, I have been able to show that eigenvalues of $AB$ are non-negative. I tried to show that the Null space of $AB$ and $(AB)^2$ are equal for proving the required statement, but I could not show so. How can we approach this problem?

Cherryblossoms
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Balasubramannyan S
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  • I flagged this as a duplicate. I assume that these are hermitian (or real symmetric) PSD matrices and that you know $\text{rank}\big(C\big)=\text{rank}\big(C^2\big)\iff \operatorname{col}(C) \cap \text{null}(C)= \big{0\big} \iff $ eigenvalue $\lambda = 0$ is semi-simple. This is a pre-req for diagonlizability and $C=AB$ here is diagonalizable. That said you may just want to skip to the end and just see my rank based argument for "semisimplicity of =0" in the link. – user8675309 Mar 18 '21 at 18:00

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