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Let $A$ and $B$ be two $m\times n$-matrices over the complex numbers. I'd like to prove that $$\det\big( I_m + AB^T\big) = \det\big( I_n + B^T A\big)\,.$$

Who am I
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michelle marques
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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Luiz Cordeiro Sep 02 '19 at 20:30
  • See M.H.'s answer here – Robert Israel Sep 02 '19 at 20:53
  • Relevant: https://math.stackexchange.com/questions/844204/prove-that-detxi-m-ab-xm-n-detxi-n-ba/ – Gabriel Romon Sep 02 '19 at 20:55
  • If you know that det of a matrix is the product of its non zero eigenvalues, then you can prove this easily. – Matematleta Sep 02 '19 at 21:25
  • This was asked before (at least once), cf. https://math.stackexchange.com/questions/17831/sylvesters-determinant-identity . See also https://terrytao.wordpress.com/2013/01/13/matrix-identities-as-derivatives-of-determinant-identities/ – Hanno Sep 02 '19 at 22:06
  • Possible duplicate of Sylvester's determinant identity – nmasanta Sep 03 '19 at 01:30

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The trick is to create a block matrix with the above mentioned matrices in a particular order, and then to observe the determinant.

This link shows how to prove the answer: https://en.wikipedia.org/wiki/Sylvester%27s_determinant_theorem

You have to fill in the blanks a bit, but should be okay.

Oreomair
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