Calculate M=AC+CA ?

[loveislonely]

Hello,

 

I am facing a problem now. Any one can help?

 

A and C are two real n*n matrices, while A is a symmetrical matrix and C is not. Now I need to calculate a matrix M, which satisfies M=AC+CA, on computer. Since C is not symmetrical, so far I have to use the same matrix multiplication routine twice to get the results of AC and CA, respectively. I am wondering whether there is a way that I can use the matrix multiplication routine only once and get the matrix M.

 

Thank you a lot.

[Bignose]

I don't really see how since with matrix multiplication in general [math] \mathbf{AC} \neq \mathbf{CA}[/math].

 

I looked at something like [math] ({\mathbf{CA}}^T)^T = (\mathbf{A}^T \mathbf{C}^T)^T = (\mathbf{A} \mathbf{C}^T)^T [/math] but you still have to compute the [math] (\mathbf{A} \mathbf{C}^T) [/math] which doesn't appear to save anything.

 

For that matter, even if both A & B are symmetric, in general [math] \mathbf{AB} \neq \mathbf{BA}[/math]. [math] \mathbf{AB} = (\mathbf{BA})^T[/math]

Edited June 27, 2008 by Bignose

[loveislonely]

Hi, since using a matrix multiplication routine takes a long time especially when the matrices are very big, and in my program I have to use the matrix multiplication routine twice per loop, then after all the loops, it might take one day to get the result. That is why I wanted to find a way to reduce the number of matrix multiplication. I was wondering if there is a way that we can derive CA from AC (or AC from CA)? Thank you.

 

I don't really see how since with matrix multiplication in general [math] \mathbf{AC} \neq \mathbf{CA}[/math].

 

I looked at something like [math] ({\mathbf{CA}}^T)^T = (\mathbf{A}^T \mathbf{C}^T)^T = (\mathbf{A} \mathbf{C}^T)^T [/math] but you still have to compute the [math] (\mathbf{A} \mathbf{C}^T) [/math] which doesn't appear to save anything.

 

For that matter, even if both A & B are symmetric, in general [math] \mathbf{AB} \neq \mathbf{BA}[/math]. [math] \mathbf{AB} = (\mathbf{BA})^T[/math]

 

Yeah, you are right, if A and B are both symmetrical, I can easily derive BA from AB (BA=(AB)^T). The problem now is A is symmetrical but B is not. So I am wondering whether there is a way to reduce the number of matrix multiplication routine usage. Many thanks.

 

Since M=AC+CA, I am thinking whether there is a way to make it like M=A(C+C^T)? Thank you.

Edited June 27, 2008 by loveislonely

[Bignose]

I just don't see how. I'd look into some of the matrix multiplication routines that MATLAB uses. MATLAB is by and far the fastest matrix multiplication I've ever seen.