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Let us consider the $2n \times 2n$ block matrix $$\mathbf{X} = \pmatrix{\mathbf{P} & \mathbf{Q}\\ \mathbf{R} & \mathbf{S} }$$ where $\mathbf{P} , \mathbf{Q} , \mathbf{R} , \mathbf{S}$ are square matrices of order $n$ which commute pairwise. Show that $$\det \mathbf{X} = \det (\mathbf{PS} - \mathbf{QR}).$$

The problem in this case I am having is when none of the matrices $ \mathbf{P} , \mathbf{Q} , \mathbf{R} , \mathbf{S} $ are invertible. Any help will be greatly appreciated.

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Maybe this helps: Determinant of Block matrices and the linked pdf with a proof: blocks.pdf –  draks ... Aug 31 '12 at 19:10
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1 Answer 1

up vote 5 down vote accepted

It's enough for $P$ and $R$ to commute. When $P$ is invertible,

$\det(X) = \det(PS - PRP^{-1}Q)$. If $P$ and $R$ commute and $P$ is not invertible, consider

$$f(t) = \det \pmatrix{P+tI & Q\cr R & S\cr}$$

For all but finitely many $t$, $P+tI$ is invertible, and it commutes with $R$, so $$f(t) = \det((P+tI)S - (P+tI) R (P+tI)^{-1} Q) = \det((P+tI)S - RQ)$$ Now both $f(t)$ and $\det((P+tI)S - RQ)$ are polynomials in $t$, so if they are equal for all but finitely many $t$ they are equal for all $t$.

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How does one know that "For all but finitely many $t$, $P+tI$ is invertible"? –  nbubis Aug 31 '12 at 19:32
    
How many eigenvalues can $P$ have? –  copper.hat Aug 31 '12 at 19:33
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