22
$\begingroup$

I'm trying to prove the following: Let $A$ be a $k\times k$ matrix, let $D$ have size $n\times n$, and $C$ have size $n\times k$. Then,

$$\det\left(\begin{array}{cc} A&0\\ C&D \end{array}\right) = \det(A)\det(D).$$

Can I just say that $AD - 0C = AD$, and I'm done?

  • 15
    No, you cannot just say that; it doesn't even make sense, because $AD$ cannot be computed: $A$ has size $k\times k$, and $B$ has size $n\times n$. Unless $n=k$, $AD$ doesn't make sense.2011-10-24
  • 0
    Then, what can I do?2011-10-24
  • 4
    Depends on your definition of the determinant. If it is a sum over all permutations of (in this case) $n+k$, then you should figure out which terms you know for sure are equal to $0$; the formula will drop out of that if you are careful enough. If your definition of determinant is via expansion by minors, then I suggest expanding along the first row and using induction on $k$.2011-10-24
  • 0
    Can you show how to do it both ways?2011-10-24
  • 0
    I thought of breaking the problem to 2 matricies, (A 0 in the first column, and 0 I sub n) in the second colum, and (0 I sub n in the first column, and 0 C in the second column) but how to use induction on this?2011-10-24
  • 3
    @BuddyHolly: I said that it depends on what your definition of determinant is, and sketched two possibilities. Rather than bother to even tell us what your definition of determinant is, you replied by asking me to do *two* proofs for you; why should I do a double effort when you seem unwilling to do even the small effort required to tell us what your *definition* is?2011-10-24
  • 0
    @user153012 What is missing from the top-voted answer? It looks complete to me.2015-04-12
  • 0
    @Qudit That answer does not prove the formula of the determinant of blockdiagonal matrices. It referes to the Laplace expansions, but anyway it is not a full detailed proof.2015-04-12
  • 0
    @Qudit What is wrong with the (currently top-voted) answer by joriki is that it expends all its effort in obtaining a reduction (to the case $C=0$) that is completely useless to the problem at hand. I contend that whatever method one uses to show that $\det(\begin{smallmatrix} A&0\\0&D \end{smallmatrix}) = \det A\det D$, the same also applies directly to the matrix of this question (given that the top-right block is $0$, there is just no way the bottom-left block contributes anything to the determinant). In order to make progress, one needs to split $A,D$ into _separate_ factors, see my answer2015-04-13
  • 0
    @Marc van Leeuwen Yes, I was thinking it was complete because the derivation of the formula for the determinant of block diagonal matrices using the Leibniz formula is clear to me without writing out the details. As you point out however, the same argument applies equally well to the original problem.2015-04-13
  • 0
    I just added an answer that takes Artin's approach, which is nice because its independent of explicit definitions of the determinant function. A lot of heavy hitters on this post so I feel a bit out of my league, but please check it out and leave comments!2017-02-10

6 Answers 6