1. ## Block Matrix Question

Let A and B be k x k matrices and let:

$\displaystyle M = \left( \begin{array}{rr} O & B \\ A & O \end{array}\right)$

Show that $\displaystyle det(M)=(-1)^kdet(A)det(B)$

How would I go about this? For example, are the rules for finding the determinants of block matrices similar to finding the determinants of normal matrices (ie multiplying diagonally and summing the result)?

Any help welcome.

2. ## Re: Block Matrix Question

You could start by trying induction on k. If A and B are 1x1 matrices then det(M) is clearly 0 - AB = (-1)1 det(A)det(B)

assume its correct for k=K and show it still holds for k=K+1

3. ## Re: Block Matrix Question

How would that look? I did my induction step for S(K+1), and ended up with $\displaystyle det(M)=0-AB=(-1)^{K+1}det(A)det(B)=(-1)^K(-1)^1det(A)det(B)$

But I'm sure this is simply wrong. I've never used induction on matrices before.

4. ## Re: Block Matrix Question

Originally Posted by CrispyPlanet
Let A and B be k x k matrices and let:

$\displaystyle M = \left( \begin{array}{rr} O & B \\ A & O \end{array}\right)$

Show that $\displaystyle det(M)=(-1)^kdet(A)det(B)$

How would I go about this? For example, are the rules for finding the determinants of block matrices similar to finding the determinants of normal matrices (ie multiplying diagonally and summing the result)?

Any help welcome.
The rules are the same. If you start with a matrix of all block matrices you can use the cofactor algorithm using those matrices as elements. Everything just becomes matrix algebra rather than real number algebra.

There's a clever way to do this. It might involve the inverse matrix of M which is easy to compute. If I come up with something I'll post it.

5. ## Re: Block Matrix Question

That would be great, thanks. If I had one example, I would understand what's involved in this kind of question.

6. ## Re: Block Matrix Question

If M is k x k, then A is m x m, and B is (k-m) x (k-m)

Let L = max(m, k-m)

if L is even it takes L/2 column swaps to go from {{0,B},{A,0}} to {{B,0},{0,A}}

if L is odd it takes (L+1)/2 swaps

each column swap adds a factor of -1 to the original determinant.

the determinant of {{B,0},{0,A}} is clearly det(A B) = det(A) det(B)

I'm just having trouble showing that (-1)k = (-1)^(L even ? : L/2, (L+1)/2))

7. ## Re: Block Matrix Question

Hi,
Previously, you essentially asked about the determinant of a "diagonal" block matrix. There you found the determinant is the product of the determinants of the diagonal matrices. Now, you know that interchanging 2 columns of a matrix changes the determinant by a factor of -1. So swap the first and (k+1)st columns, the 2nd and (k+2)nd columns, etc. for a total of k swaps. So the determinant is (-1)kdet(B)det(A).

[deleted]

9. ## Re: Block Matrix Question

Many thanks for both of your help.