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So I was stuck with the famous Sylvester's determinant identity and don't know how to correctly interpret the difference between the sizes of identity matrices $I_m$ and $I_n$. The line that I dont' understand is: $$\det(I_m) \det(I_n - B I_m^{-1} (-A)) = \det(I_n + BA).$$

I understand that $\det(I_m) = 1$ but what if we rewrite it in the form: $$\det(I_m) \det(I_n - B I_m^{-1} (-A)) = \det(I_mI_n - B I_m I_m^{-1} (-A)).$$

Do we have any right to multipy $I_m$ by $I_n$? Maybe there is a special property of matrix block determinants? Which part am I missing?

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user2660964
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  • $BI^{-1}(-A)=-BA$. – Angina Seng Nov 16 '18 at 19:17
  • I know that. But what about the $I_mI_n$ part? – user2660964 Nov 16 '18 at 19:20
  • You are right; that doesn't make sense. Why on earth would anyone do that? – Angina Seng Nov 16 '18 at 19:24
  • How can you then explain the line in the Sylvester's determinant identity? I know that $\det\begin{pmatrix}I_m& -A\ B& I_n\end{pmatrix} = \det(I_m I_n - B I_m I_m^{-1} (-A))$ but that doesn't make any sense for me. – user2660964 Nov 16 '18 at 19:26
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    The line you claim you "don't understand" is $$\det(I_m) \det(I_n - B I_m^{-1} (-A)) = \det(I_n + BA).$$ Since $\det I_m=1$, the left side is $\det(I_n - B I_m^{-1} (-A))$, and since $I_n - B I_m^{-1} (-A) = I_n + BA$, then that equals $\det(I_n + BA)$. – Angina Seng Nov 16 '18 at 19:30

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It looks like the question you asked was not the question you really intended to ask. Consider $$\pmatrix{I_m&-A\\B&I_n}=\pmatrix{I_m&O\\B&I_n}\pmatrix{I_m&O\\O&I_n+BA} \pmatrix{I_m&-A\\O&I_n}.$$ Take determinants. As two of three matrices on the right are triangular with $1$s on the diagonal, they have determinant $1$. Thus $$\det\pmatrix{I_m&-A\\B&I_n}=\det\pmatrix{I_m&O\\O&I_n+BA}.$$ But that's a block diagonal matrix, with determinant $\det (I_n)\det(I_n+BA) =\det(I_n+BA)$.

Angina Seng
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