question about fast symmetric matrix multiplication

[joegao1]

can some one give me a hint?

I want to program the code for matrix multiplication with as less arithmetical / multiplication operations as possible.

my task is to calculate the matrix multiplication A*B*A' , where
- A' refers to the transpose of matrix A
- matrix B is symmetric matrix.
- (optional) it will be much better if the result of (A*B) or (A*B)' can be stored as temperal matrix, as
this value is required thereafter.

I have been thinking about splitting matrix B into two parts,
such that : B = (L + L') , where L is the lower triangle matrix of B .

any hint or link or paper is welcomed !!!

danke
Joe

[curiously enough]

You can multiply two matrices using three for loops, the first two are for the elements of the product matrix, and the third is for multiplying a row from the first matrix by a column from the second matrix(which will give you one of the elements of the product matrix).
It is like this:

for(int i=0;i<rows;i++)//rows is the number of rows in the first matrix
  for(int j=0;j<cols;j++)//cols is the number of columns in the second matrix
    for(int k=0;k<dim;k++)/*dim is the number of columns in the first matrix as well as the number of rows in the second matrix(since the number of columns of the first matrix is obigatory equal to the number of rows of the second matrix)*/
         P[i][j]+=A[i][k]*B[k][j];//finds element P[i][j] of the product matrix P

As for multiplying three matrices, you can conclude what to do from the above.

[arnaudk]

Get the right tool for the job: use a linear algebra package. Have a look at GSL, or boost or blitz++.
The fact that B is symmetric does not lead to any simplifications as far as I can tell since A is arbitrary.

[JosAH]

I have been thinking about splitting matrix B into two parts,
such that : B = (L + L') , where L is the lower triangle matrix of B .

Is diag(B) == 0?

kind regards,

Jos

[joegao1]

thanks for all the responses !!!

You can multiply two matrices using three for loops, the first two are for the elements of the product matrix, and the third is for multiplying a row from the first matrix by a column from the second matrix(which will give you one of the elements of the product matrix).
It is like this:

for(int i=0;i<rows;i++)//rows is the number of rows in the first matrix
  for(int j=0;j<cols;j++)//cols is the number of columns in the second matrix
    for(int k=0;k<dim;k++)/*dim is the number of columns in the first matrix as well as the number of rows in the second matrix(since the number of columns of the first matrix is obigatory equal to the number of rows of the second matrix)*/
         P[i][j]+=A[i][k]*B[k][j];//finds element P[i][j] of the product matrix P

As for multiplying three matrices, you can conclude what to do from the above.

yes, this is what I already have now. But it is kind of expensive as I am programming on an embedded system. I am looking for a compact version of that.

Get the right tool for the job: use a linear algebra package. Have a look at GSL, or boost or blitz++.
The fact that B is symmetric does not lead to any simplifications as far as I can tell since A is arbitrary.

thanks for the links. I will check them out. and it is true that matrix A is arbitrary.

Is diag(B) == 0?

kind regards,

Jos

sorry, I make a mistake . it should be like : B= L+L' + D , where D is the diagonal matrix.

[JosAH]

sorry, I make a mistake . it should be like : B= L+L' + D , where D is the diagonal matrix.

I don't think that'll buy you much except that A*B*A == A*(L+L'+D)*A' ==
A*L*A'+A*L'*A'+A*D*A'; only A*L will be a common (transposed) factor; so
X = A*L then X*A'+A*X'+A*D*A'; now Y=X*A' will be the common (transposed)
factor so: Y+Y'+A*D*A'.

kind regards,

Jos

[joegao1]

I don't think that'll buy you much except that A*B*A == A*(L+L'+D)*A' ==
A*L*A'+A*L'*A'+A*D*A'; only A*L will be a common (transposed) factor; so
X = A*L then X*A'+A*X'+A*D*A'; now Y=X*A' will be the common (transposed)
factor so: Y+Y'+A*D*A'.

kind regards,

Jos

thanks Jos,

I think this is the best way to implement the multiplication.

[hami009]

@arnaudk
Hi,
I am trying to use GSL library with my C++ program. Can you please tell what steps I need to take to link the library with bcc32 compiler. I have already installed GSL on my machine also I used IMPLIB.exe to link the gsl library to my prgram but it didn't work I used following command

Implib /Ic:/WinGnu32/lib libgsl.a libgsl.dll

Can you please guide me with step by step information how can I link GSL library to C++ program.

Cheers,

[sustik]

Hi, I wonder what solution did you end up using? Could you exploit that B is symmetric? Can you email me at [e-mail]? I would really like to know more about the applicaion this came from. Thanks!
Matyas