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I'd like a hint rather than a full solution.

The problem I am considering is the following:

$X$ is an $n\times m$ matrix $Y$ is $m\times n$

Show that

$(I - XY)^{-1}\cdot X = X\cdot(I - YX)^{-1}$

The first $I$ is of dimension $n$ and the second $I$ is of dimension $m$.

alan213123
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1 Answers1

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Hint

Verify that $$(I-XY)X=X(I-YX)$$

  • Okay....this is easy enough to verify. Am I now trying to work backwards from here or am I going about this the wrong way. – alan213123 Sep 19 '14 at 21:33
  • +1 great hint. A quick question: is the invertibility of $I-XY$ equivalent to that of $I-YX$? – Kim Jong Un Sep 19 '14 at 21:34
  • I wonder if it would be a better hint to flip the sides of the equation. – augurar Sep 19 '14 at 21:46
  • I'm still having trouble making the leap from verifying the above equation to the one in the OP – alan213123 Sep 19 '14 at 21:54
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    @KimJongUn Invertibility of $I-XY$ is equivalent to $1$ not being an eigenvalue of $XY$; but $XY$ and $YX$ have the same non zero eigenvalues. – egreg Sep 19 '14 at 22:11
  • @KimJongUn See also https://math.stackexchange.com/questions/17831/how-can-we-prove-sylvesters-determinant-identit – Anne Bauval Feb 08 '24 at 19:18