4
$\begingroup$

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.

$\endgroup$
1
  • 3
    $\begingroup$ Can you include what you did when the matrices commuted? $\endgroup$ – AHusain Jun 16 '19 at 23:36
1
$\begingroup$

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.

$\endgroup$
2
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.