Hello, Mr. Breuer,

Thank you for your answer. I think I may have not been totally precise on my question. The matrix A is nxn, and D is mxm, so that B is nxm and C is mxn.

Is your answer still valid for the case at which B and C are not square?

I undertsand that the answer does not apply for the case at which B and C are not square, otherwise your last paragraph would be incorrect. Do you still see a way to solve this?

That's exaclty what I was thinking about... the first solution would be perfect if all the block matrices were square, but that would be a very specific case to my problem.

The problem that I'm trying to address is actually about eigenvalues.

I want to obtain the eigenvalues of the block matrix M = [P Q; R S]. The matrix S is always 2x2. So, I have to calculate the determinant of [A B;C D], where A=P-sI, B = Q, C=R and D = S-sI, being s the eigenvalue. I also know that P has no null eigenvalues. Then I have some situations to evaluate:

1) For a given value of s, A is not singular and M/A is singular. Then det(M) equals det(A).det(M/A) = 0, and s is an eigenvalue. The eigenvalues can then be calculated from det(M/A)=0.

2) For a given value of s, D is not singular and M/D is singular. Then det(M) equals det(D).det(M/D) = 0, and s is an eigenvalue. The eigenvalues can then be calculated from det(M/D)=0.

3) D and A cannot be simultaneously nonsingular, otherwise M would be nonsingular (it is a sufficient condition) and s would not be an eigenvalue.

4) D and A could be simultaneoulsy singular. Then, neither M/A nor M/D could be calculated. In this case I don't know how to calculate the eigenvalues.

I want to somehow try to predict what is going to happen to my eigenvalues for some families of matrices A, B, C and D. But without knowledge of the situation 4) it is not possible to say much. I've been trying to check under which circumstances this situation is possible (because if it is not, then I shouldn't have problems), but still nothing...

I think that you can’t say nothing :

A=[1 0]

    [0 0]

B=[1 0]

    [0 1]

C=[1 0]

    [0 1]

D=[1 0]

    [0 0]

Eigenvalues(A) ={0;1}

Eigenvalues(B) ={1}

Eigenvalues(C) ={1}

Eigenvalues(D) ={0;1}

Eigenvalues(M) = {-1,0;1;2}

Now consider

A=[0 0]

    [0 1]

B=[1 0]

    [0 1]

C=[1 0]

    [0 1]

D=[1 0]

    [0 0]

Eigenvalues(A) ={0;1}

Eigenvalues(B) ={1}

Eigenvalues(C) ={1}

Eigenvalues(D) ={0;1}

Eigenvalues(M) = {(1+/-sqrt(5))/2}

suppose that you can compute the eigenvalues of M  with some formula f involving the eigenvalues of A, B , C an D , you would get simultaneously

two different results with the same input.

Hi, Marc, thank you for your answer!

Indeed, I also think it is not possible to calculate directly the eigenvalues of M using only the eigenvalues of A, B, C and D. In fact, without knowledge of the eigenvectors, one couldn't even be able to tell exaclty what the matrix is, based on the eigenvalues.

However, there is a rule for the determinant that says that if A is nonsingular, then:

det(M) = det(A).det(D-C.A-1B)          (1)

, and there is a similar rule for D.

When solving the eigenproblem, one must solve the equation det(M-sI) = 0, where s is the eigenvalue, and it would cause changes on A and D from equation (1).

Based on these, I was trying to find some relationship like the det(D-C.A-1B) . It does not have the eigenvalues of any of the four matrices isolatedly, but if one knows the four matrices, then it is possible to know some of the eigenvalues, without having to calculate all of them.

I’ve chosen this counter example because you can quickly see that there’s no equivalent formula to compute the determinant of M with the determinants of A, B, C and D (as far as you consider square matrices), and so far as you only know the values of det(A) and det(C) (which are supposed to be 0 ? right ?) .

Even in the case of square matrices, suppose that there is a function g such that det(M)= g(a,b,c,d), where a=det(A),b=det(B), c=det(C) and d=det(D);

Then, in the first case, you would get g(0,1,1,0)=0, and in the second case you would get g(0,1,1,0)=1.

The same example shows that what Peter said concerning square matrices is unfortunately wrong.

Concerning the way you try to compute the block matrix determinant,

you should consider block matrix computation like you handle the multiplication of transvection matrices in the case 2x2.

For instance , if you multiply on the right side the matrix M by the block matrix N

N=[I X]

    [0 I]

You get :

MN = [A AX+B]

          [C CX+D]

then if A is invertible, you choose X = -A^{-1}B and you can write :

MN=[A                  0    ]

        [C   -CA^{-1}B+D]

And you can see that det(M) = det(A)det(-CA^{-1}B+D)

So know, if you ask under what condition you can get such a result even if A is not invertible, the answer is that this kind of result is true as long as it is possible to find X such that AX+B = 0

So you can do this kind of computation to get the following formula  : d

et(M) = det(A)det(CX+D)

In this case, finding X such that AX=-B is equivalent to the condition "columns of B belongs to the subspace Im(A)"

It depens know of the problem you are dealing with....

It is possible  that  you should consider too the question of density of invertible matrices into the space of matrices.

If A has a 0 for eigenvalue, it is possible to find a matrix A' invertible (or a suite of matrices) very close to A for any norm in the finite dimension space of matrices.

since det is a continuous function, you can write :

det(M) = limit(  det(A')det(-CA'^{-1}B+D) ) when A' --> A

then

det(M) = limit(det(-A'CA'^{-1}B+A'D))

Finaly, if you use the relation A'^{-1} = 1/det(A')transpose(com(A')) (and if you multiply the whole relation by det(A'), it is possible to get some basic formulation involving only A, transpose(comA) B, C and D...

Hope I don't make too much mistakes in basic explanations,

Please see: Brezinski in [Br] pp.231-232, or Brezinski and Zaglia in [BZ] p. 281:

[Br] C. Brezinski, Other Manifestations of the Schur Complement, Linear Algebra and Its Applications, 111 (1988) 231-247.

[BZ] C. Brezinski, M.R. Zaglia, A Schur complement approach to a general extrapolation algorithm, Linear Algebra and Its Applications, 368 (2003) 279-301.

Brezinski in [Br] pp.231-232 (from above)

(M/A) = D - CA⁻¹B    i.e.,    det(M) = det(A)det(M/A).

We can similarly define the Schur complements

(M/B) = C - DB⁻¹A,    i.e.,    det(M) = - det(B)det(M/B),

(M/C) = B - AC⁻¹D,    i.e.,    det(M) = - det(C)det(M/C),

(M/D) = A - BD⁻¹C,    i.e.,    det(M) =   det(D)det(M/D).