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If $A,B\in M_{m \times n}$. Show that $\det(tI-BA^T)=\det(tI-B^TA)t^{m-n}$

I want to solve this question and I have an idea probably useful which is from Sylvester's determinant identity. If I can find some matrices such that $$P^{-1}\begin{bmatrix}I&\\&\ BA^T\end{bmatrix}P=\begin{bmatrix}I&\\&\ A^TB\end{bmatrix}$$, then two matrices are similar and have the same characteristic polynomial. Hope someone can give me a hint how to construct these four matrices.

Steven Lu
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1 Answers1

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Hint: Use these two facts:

$1.$ $\det(tI - BA^T) = \det(tI - AB^T)$ since for any $C$, $\det(C) = \det(C^T)$

$2.$ The algebraic multiplicity of the non-zero eigenvalues of $AB^T$ and $B^TA$ are the same (see here.)

Coriolanus
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