Blockwise Symmetric Matrix Determinant This question arises from another one of mine, but separate enough that I feel it deserves its own thread. Wikipedia says that $$det\begin{bmatrix}A&B\\\B &A \end{bmatrix} = det(A+B)det(A-B)$$ _Regardless_ of whether or not A and B commute. Using the general formulation $$det\begin{bmatrix}A&B\\\C &D \end{bmatrix} = det(A)det(D - CA^{-1}B)$$ We see that this becomes $$det(AD- ACA^{-1}B)$$ Or for my original matrix, $$det(AA - ABA^{-1}B)$$ I see how this becomes det((A-B)(A+B)) if A and B commute, but how is it valid if they _don't_ commute? Can anyone prove this please?

I don't think whoever wrote that section of the Wikipedia article meant that the first formula in your question follows from the second one. The author probably only meant to say that when $A$ and $B$ do not commute, we can still evaluate the determainant of the block matrix by $\det(A+B)\det(A-B)$. At any rate, the two formulae hold because \begin{align} &\pmatrix{I&-I\\\ 0&I}\pmatrix{A&B\\\ B&A}\pmatrix{I&I\\\ 0&I}=\pmatrix{A-B&0\\\ B&A+B},\\\ &\pmatrix{A&B\\\ B&A}=\pmatrix{A&0\\\ B&I}\pmatrix{I&A^{-1}B\\\ 0&A-BA^{-1}B}. \end{align} So, if you really want to relate $\det(A-B)\det(A+B)$ to $\det(A)\det(A-BA^{-1}B)$, you may combine the above two equalities: $$ \pmatrix{I&-I\\\ 0&I}\pmatrix{A&0\\\ B&I}\pmatrix{I&A^{-1}B\\\ 0&A-BA^{-1}B}\pmatrix{I&I\\\ 0&I}=\pmatrix{A-B&0\\\ B&A+B}, $$ but I don't think doing so is very meaningful.