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I have encountered an statement several times while proving determinant of a block matrix.

$$\det\pmatrix{A&0\\0&D}\; = \det(A)det(D)$$ where $A$ is $k\times k$ and $D$ is $n\times n$ matrix. How to prove this?

Thanks in advance.

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I hope $k=d$? Or that you mean $\det(A) \det (D)$? Because if $k \neq d$ the multiplication $AD$ is not defined. – Patrick Da Silva Aug 15 '14 at 10:24
Updated the question :P – codeomnitrix Aug 15 '14 at 10:36

2 Answers 2

up vote 14 down vote accepted

This is more general result $$\det\pmatrix{A&B\\0&D} = \det A\det D$$ and to prove it notice that

$$\pmatrix{A&B\\0&D}=\pmatrix{I_k&0\\0&D}\pmatrix{A&B\\0&I_n}$$ and we develop along the $k$ first rows we find $$\det\pmatrix{I_k&0\\0&D}=\det D$$ and along the last $n$ rows we find $$\det\pmatrix{A&B\\0&I_n}=\det A$$ and the result follows.

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Together with mine that's three different proofs, I like that! +1 – Patrick Da Silva Aug 15 '14 at 10:29
Thanks Sami!! this was blowing my mind :) – codeomnitrix Aug 15 '14 at 10:34
You're welcome @codeomnitrix – user63181 Aug 15 '14 at 10:35
As a factor of Laplace expansion, the lowest-rightmost term of $\det\pmatrix{A&B\\0&I_n}=\det A$ always keeps its original sign (positive in this case). Very cool proof! Thanks! – user175861 Sep 22 '14 at 1:05

If your matrices have coefficients in an integral domain, you can pass to the field of fractions and take an algebraic closure to use the Jordan canonical form, in which case this equation becomes trivial since both sides of the equation are the product of the products of the eigenvalues of $A$ and $D$.

Otherwise you can use Laplace expansion and notice that the sum goes over all the permutations of $S_{k+n}$, in which case the only terms which "survive" are the permutations which map $\{ 1,\cdots, k\}$ to itself and $\{k+1,\cdots,k+n\}$ to itself. Then you can split the sum in two parts and obtain $\det(A) \det(D)$. This proof works over any (commutative) ring.

Hope that helps,

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Thanks Patrick :) – codeomnitrix Aug 15 '14 at 10:35

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