# Prove $\det \left[\begin{smallmatrix} A&B\\\\C&D\\ \end{smallmatrix}\right] =\det(AD-BC)$

Let $$A,B,C$$ and $$D$$ be upper-triangular $$n \times n$$ complex matrices. Let

$$E=\begin{bmatrix} A&B\\C&D\\ \end{bmatrix}$$

Prove $$\det(E)=\det(AD-BC)$$.

I did this problem in the case that the matrices commute but I cannot figure out this case.

• Can you include what you did when the matrices commuted? – AHusain Jun 16 '19 at 23:36

It suffices to prove the identity when the upper triangular parts of $$A,B,C,D$$ are independent indeterminates (alternatively, prove the identity for invertible $$D$$ first, then pass $$D$$ to the limit). Try to justify the first equality below using properties of Schur complement (see also this Wikipedia entry) and the second equality below using the condition that $$A,B,C,D$$ are triangular: $$\det(E) =\det(A-BD^{-1}C)\det(D) =\det(A-BCD^{-1})\det(D) =\det(AD-BC).$$ (Edit. The following part is wrong. See darij grinberg's comment.) You may also try to prove that in Leibniz formula for determinant of $$E$$, only two generalised diagonals of $$E$$ are non-vanishing, one given by the main diagonal and the other formed by the diagonals of $$B$$ and $$D$$. But I find this argument harder to be worded clearly.
• "only two generalised diagonals of $E$ are non-vanishing": This is not true. I think $2^n$ generalized diagonals should be non-vanishing. The easiest way to get the determinant combinatorially looks like the following: Permute the rows and the columns of $E$ in such a way that the matrix becomes block-triangular with $2 \times 2$-blocks. (The permutation is $\left[1,3,5,\ldots,2n-1,2,4,6,\ldots,2n\right]$ in one-line notation, or its inverse.) Then, its determinant becomes $\prod_{i=1}^n \left(a_i d_i - b_i c_i\right)$, where $a_i, b_i, c_i, d_i$ are the diagonal entries of $A, B, C, D$. – darij grinberg Jun 17 '19 at 5:49