The Wikipedia page about the determinant mentions the following matrix identity $$\det\begin{pmatrix}A&B\\B&A\end{pmatrix}=\det(A+B)\det(A-B),$$ valid for squared matrices $A$ and $B$ of the same size.

How is this result proved?

  • 2
    $\begingroup$ The proof is given at the wikipedia page. It follows from the Leibniz formula, for example. Several proofs are given. See also this post, which is the main step. $\endgroup$ – Dietrich Burde Aug 13 '19 at 18:10
  • $\begingroup$ A and $B$ can't be really arbitrary: they have to be square matrices of the same size. $\endgroup$ – Bernard Aug 13 '19 at 18:12
  • $\begingroup$ @DietrichBurde mmh I'm sorry where is it given? I can't see it on the wiki page $\endgroup$ – glS Aug 13 '19 at 18:12
  • $\begingroup$ @DietrichBurde I saw that post, but it's not obvious to me how to use that identity. I also saw that remark in the wiki, but I don't find that clear at all $\endgroup$ – glS Aug 13 '19 at 18:16
  • $\begingroup$ Go to this text in the wiki page:"When the blocks are square matrices of the same order further formulas hold. For example, if C and D commute (i.e., CD = DC), then the following formula comparable to the determinant of a 2 × 2 matrix holds:[13]". Then look up the reference. $\endgroup$ – Dietrich Burde Aug 13 '19 at 18:16

$$ \left( \begin{array}{cc} I&I \\ 0&I \\ \end{array} \right) \left( \begin{array}{cc} A&B \\ B&A \\ \end{array} \right) \left( \begin{array}{cc} I& -I \\ 0 &I \\ \end{array} \right) = \left( \begin{array}{cc} A+B& 0 \\ B& A-B \\ \end{array} \right) $$

  • $\begingroup$ great, thanks! Is there also a similar formula for the case with the upper-right $B$ being instead $-B$, that you know of? $\endgroup$ – glS Aug 13 '19 at 18:24

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