Inverse of a symmetric matrix consisting of submatrices is desired.

Inverse of a symmetric matrix consisting of submatrices is desired.

N(12x12) and its determinant are non-zero, B1(12x5), B1(12x4), C1(2x5) C1(2x4) are dimensional submatrices. Can someone inverse this matrix? T denotes transpose of matrix

N 0 B1 B2

0 0 C1 C2

B1T C1T 0 0

B2T C2T 0 0

[Torsten]

B1(12x5), B1(12x4), C1(2x5) C1(2x4) ?

Are you sure about B1 and B1 resp. C1 and C1 and the dimension of the C-matrices ?

Do you expect a similar structure for the inverse ?

[Walter Roberson]

The zeros need to be specified more exactly.

[ayeser]

Inverse of a symmetric matrix(23x23) consisting of submatrices is desired.

The dimensions of the submatrices are given below. (‘) denotes transpose of matrix

Can someone inverse this matrix?

Thanks

N(12x12) 0(12x2) B1(12x5) B2(12x4)

0(2x12) 0(2x2) C1(2x5) C2(2x4)

B1’(5x12) C1’(5x2) 0(5x5) 0(5x4)

B2’(4x12) C2’(4x2) 0(4x5) 0(4x4)

Note: N(12x12) submatrix can be inverted..

[Torsten]

Did you read @John D'Errico 's answer ?

And if you try to generate the entire matrix as a symbolic 23x23 array where all elements are symbolic (not numeric) then the symbolic inverse will never terminate, as it would involve on the order of factorial(23) terms.

[ayeser]

The inverse matrix(23x23) is expected to be symbolically in the input matrix structure.

[Torsten]

As you can see in @John D'Errico 's answer from the plots of the matrix structure of the inverse, this is not true.

[ayeser]

I change my question to this:

There is a symmetric linear equation system consisting of 4 equations with 4 unknowns. However, the coefficients of the system of equations consist of matrices. By making a symbolic solution to this system of equations, vectors a and b are constant vectors.

N(12x12) .x + B1(12x5).z + B2(12x4). t = a(12x1)

C1(2x5).z + C2(2x4). t = b(2x1)

B1’(5x12).x + C1’(5x2).y = 0(5x1)

B2’(4x12).x + C2’(4x2).y = 0(4x1)

Note: N(12x12) submatrix can be inverted..

The unknowns are desired to be obtained symbolically as follows.

x=......

y=......

z=......

t=........

Can someone make the solution?

thanks

[Walter Roberson]

You cannot add a 12 x 12 matrix and a 12 x 5 matrix.

syms N [12 12]
syms B1 [12 5]
syms B2 [12 4]
syms C1 [2 5]
syms C2 [2 4]
syms x y z t
syms a [12 1]
syms b [2 1]
A = [N*x + B1*z + B2 * t == a
     C1*z + C2*t == b
     (B1')*x + (C1')*y == zeros(5,1)
     (B2')*x + (C2')*y == zeros(4,1)]

Error using +
Array sizes must match.

sol = solve(A, [x, y, z, t])

[Torsten]

syms N [12 12]
syms B1 [12 5]
syms B2 [12 4]
syms C1 [2 5]
syms C2 [2 4]
syms x [12 1]
syms y [2 1]
syms z [5 1]
syms t [4 1]
syms a [12 1]
syms b [2 1]
A = [N*x + B1*z + B2*t == a
     C1*z + C2*t == b
     (B1')*x + (C1')*y == zeros(5,1)
     (B2')*x + (C2')*y == zeros(4,1)]

sol = solve(A, [x; y; z; t])

But it's samey.

[Walter Roberson]

syms N [12 12]
syms B1 [12 5]
syms B2 [12 4]
syms C1 [2 5]
syms C2 [2 4]
syms x [12 1]
syms y [2 1]
syms z [5 1]
syms t [4 1]
syms a [12 1]
syms b [2 1]
M = [N*x + B1*z + B2*t == a
     C1*z + C2*t == b
     (B1')*x + (C1')*y == zeros(5,1)
     (B2')*x + (C2')*y == zeros(4,1)]

M = 

[A, B] = equationsToMatrix(M, [x;y;z;t])

A = 

B = 

sol = A\B

[Walter Roberson]

The above will take... probably several years.

[ayeser]

Dear Walter

Can another programming language help us with this?

Best

[Torsten]

No, it's just the problem from above rephrased. And the reasons why it makes no sense trying to solve it symbolically are sufficiently explained.

[Walter Roberson]

Eventually, MATLAB crashed while trying to process the above A\B

Answer 1

[John D'Errico]

This makes little sense as you write it, AND it will be essentially impossible to perform in MATLAB using symbolic tools. syms CANNOT compute (in a finite time, measured in our lifeltimes) matrix inverses for large matrices in fully symbolic form.

You say this:

N(12x12) and its determinant are non-zero, B1(12x5), B1(12x4), C1(2x5) C1(2x4)

First, we are not told what the B2 and C2 sizes are. I'll assume the second arrays you gave as B1 and C1 are actually B2 and C2.

I'll make that assumption then, and as a test case, I want to point out the result will be a full matrix in general. Just to give you a feeling, consider this example:

N = randn(12); N = N+N';
B1 = randn(12,5);C1 = randn(2,5);
B2 = randn(12,4); C2 = randn(2,4);

A = [N,zeros(12,2),B1,B2; ...
     zeros(2,12),zeros(2,2),C1,C2; ...
     B1',C1',zeros(5,5),zeros(5,4); ...
     B2',C2',zeros(4,5),zeros(4,4)];

You can see the non-zeros in the A matrix here:

spy(A)

And see that inv(A) will be a full matrix, completely filled in.

spy(inv(A))

Can you compute the matrix inverse, using these block matrices? I suppose there is no reason why you could not just perform a simple Gaussian Elimination on those blocks, as long as the necessary elements will be non-singular. However, MATLAB does not have that capability, at least not today.

It might seem you could try. That is, just create a 4x4 symmetric matrix, with variables representing those blocks, then compute the symbolic inverse. The problem is, MATLAB would need to know that the matrix products in that result do NOT commute. And this is why the above scheme will fail. The symbolic toolbox will assume commutativity.

And if you try to generate the entire matrix as a symbolic 23x23 array where all elements are symbolic (not numeric) then the symbolic inverse will never terminate, as it would involve on the order of factorial(23) terms.

As such, your question cannot be solved using MATLAB. You would be best off just doing the linear algebra yourself. I expect it will get a bit nasty looking, but you could in theory survive. Just not using MATLAB.

Answer 2

[Walter Roberson]

N = symmatrix('N', [12 12]);
B1 = symmatrix('B1', [12,5]); C1 = symmatrix('C1', [2,5]);
B2 = symmatrix('B2', [12,4]); C2 = symmatrix('C2', [2,4]);
A = [N,zeros(12,2),B1,B2; ...
     zeros(2,12),zeros(2,2),C1,C2; ...
     B1',C1',zeros(5,5),zeros(5,4); ...
     B2',C2',zeros(4,5),zeros(4,4)]

A = 

invA = inv(A)

invA = 

[John D'Errico]

Lol.

[ayeser]

I couldn't see the open version of the inverse matrix.

[Torsten]

MATLAB can't see it either.

It just writes that inv(A) = A^(-1) which doesn't help much.

[Torsten]

You could split your matrix into a matrix of the form [A B;C D] where A is 12x12 (invertible), B is 12x11, C is 11x12 and D is 11x11 and apply

http://msvlab.hre.ntou.edu.tw/grades/now/inte/Inverse%20%26%20Border/border-LuTT.pdf

I don't know if this leads to anything useful.

Answer 3

[Paul]

I wasn't going to post this as it didn't seem like a useful answer, but, since @Torsten raised the idea of using the block matrix inversion formula ....

A = symmatrix('A',[12 12]);
B = symmatrix('B',[12 11]);
C = symmatrix('C',[11 12]);
D = symmatrix('D',[11 11]);

% M = [A B;C D]

invM =  [ ...
      inv(A) + inv(A)*B*inv(D-C*inv(A)*B)*C*inv(A) , -inv(A)*B*inv(D-C*inv(A)*B);
    -(inv(D-C*inv(A)*B))*C*inv(A)                  ,  inv(D-C*inv(A)*B)  ]

invM = 

invM = subs(invM,C,B.')

invM = 

N = symmatrix('N',[12 12]);
Z = symmatrix('Z',[12 2]);
B1 = symmatrix('B1', [12,5]); C1 = symmatrix('C1', [2,5]);
B2 = symmatrix('B2', [12,4]); C2 = symmatrix('C2', [2,4]);

invM = subs(invM,inv(A),inv(N))

invM = 

Hmm, not all of the inv(A) terms are subbed with inv(N). Don't know what that's all about.

On my local installation of R2021b, the substitution worked the way I intended and thought it would.

I guess the workwaround would be to start the problem with N instead of A for the upper left block. Or just subs A for N (instead of their inverses).

invM = subs(invM,A,N)

invM = 

invM = subs(invM,B,[Z B1 B2])

invM = 

invM = subs(invM,D,[zeros(2) , C1 , C2;[C1';C2'],zeros(9)])

invM = 

invM = subs(invM,Z,sym(zeros(12,2)))

invM = 

At this point, I couldn't get much further. Coudn't even figure out how to subs in values for N, B1, B2, C1, and C2.

[ayeser]

thanks