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Let $A \in \mathbb{R}^{n \times p}, B \in \mathbb{R}^{p \times n}$ be two matrices. Consider the matrix-valued functions when defined:

$$z \mapsto (AB - zI_n)^{-1}, \qquad z \mapsto (BA - zI_p)^{-1} $$

These are the resolvents of matrices $AB$ and $BA$, respectively. How do I prove that for each $z$ that's neither $0$ nor an eigenvalue of $AB$, the following holds?

$$\mbox{trace} (AB - zI_p)^{-1} = \mbox{trace} (BA - zI_n)^{-1} + \frac{n - p}{z} $$

P.S. I think that for a proof, the second answer in Sylvester's determinant identity might be a way to go, using $tr(XY)=tr(YX)$, where $X,Y$ are the two block matrices that the answerer mentioned. But the differenc here is that: we're looking at $trace((AB - zI_p)^{-1})$, and $trace((BA - zI_n)^{-1})$, and not $trace(AB - zI_p), trace(BA - zI_n)$ directly.

Learning Math
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  • @RodrigodeAzevedo apparently it'll be used to prove several theorems in random matrix theory - I'm reading a lecture note and that just states this identity without a proof. * – Learning Math Oct 31 '19 at 16:06
  • @RodrigodeAzevedo I can try that, using $M^{-1}= Adj(M)/det(M)$, but care to elaborate a bit more on how that'd help to obtain a proof? – Learning Math Oct 31 '19 at 16:16
  • One determinant will be a degree-$p$ polynomial. The other a degree-$n$ polynomial. Does that help? Maybe. – Rodrigo de Azevedo Oct 31 '19 at 16:18
  • @RodrigodeAzevedo Thaks, but sorry I'm not sure how that helps. However, please note the "P.S." part in my question with the link given. In the second answer to the question here - https://math.stackexchange.com/questions/17831/sylvesters-determinant-identity, if one replaces appropriately, and use block matrix inversion formula, and that $tr(XY)=tr(YX)$, I think we'd arrive at the equation. Continued to the next comment. – Learning Math Oct 31 '19 at 19:33
  • @RodrigodeAzevedo So basically, in the link right above, in the second answer, I'd (1) replace in the first block matrix (first factor), (call $X$), the $(1,1)$ block $I$ by $zI$, and in the $(1,2)$ block by $-B$. (2) I'd also replace in the second block matrix (second factor) (call $Y$), the $(2,2)$ block $I$ by $zI$. After multiplication, we'll have in $XY$, the $(1,1)$ block $zI$, the $(2,2)$ block $zI - AB$. When we compute $YX$, we've the $(1,1)$ block $zI - BA$, and the $(2,2)$ block $zI$. Now if you invert $XY, YX$, and use the block inversion formuma for matrices, and (continued) – Learning Math Oct 31 '19 at 19:40
  • @RodrigodeAzevedo ...and that $trace(X^{-1}Y^{-1})=trace(Y^{-1}X^{-1})$, , then on the left side we'll have equality, and so we'll have equality on the riht side too. Now by block inversion formula on the right side https://en.wikipedia.org/wiki/Block_matrix#Block_matrix_inversion, we should arrive at the answer. – Learning Math Oct 31 '19 at 19:42

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