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.

  • 3
    $\begingroup$ Can you include what you did when the matrices commuted? $\endgroup$ – 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.

  • 1
    $\begingroup$ "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 $A, B, C, D$. $\endgroup$ – darij grinberg Jun 17 '19 at 5:49
  • $\begingroup$ @darijgrinberg You are right. I stand corrected. Perhaps you may elaborate your comment into an answer. $\endgroup$ – user1551 Jun 17 '19 at 6:41

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