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Topic: Matrix multiplication by its transpose
Replies: 11   Last Post: Jun 13, 2010 11:04 AM

 Messages: [ Previous | Next ]
 luca Posts: 11 Registered: 5/19/10
Matrix multiplication by its transpose
Posted: Jun 9, 2010 7:40 AM

Hi,

i have fear of asking this question, because it could be stupid, but i
will try it nevertheless:

suppose i have a matrix A N x 8, where N is of order 10^3-10^5 and a
vector v of size Nx1 (all are real elements).

To compute the element (i, j) of A i need to do some computation
(derivative of image pixels, some multiplications and so on). After
computing A, i need to compute its pseudo-inverse: pinvA = (A^T *
A)^-1 * A^T (where A^T is the transpose of A and (..)^-1 is the
inversione of the quantity between parenthesis) and finally i need to
compute the multiplication of the pesudoinverse by v:

pInvA * v

The result will be a vector x0 of size 8 x 1.

Here are the sequence of steps i do:

1] compute A
2] multiply A^T * A, without forming A^T (this step require 64*N
double floating point multiplications)
3] compute the inverse of A^T * A with the LU decomposition (A^T * A
is a 8x8 matrix, so its inversion
is not so slow to compute). let B = (A^T * A)^-1
4] compute v1 = A^T * v
5] compute x0 = B * v1

The step 2] require a lot of work. Is there a way to form A^T * A
without doing so much multiplications?!??!

I have another question. A^T * A is a ill-conditioned matrix (its
condition number is of order O(10^11)).
Surprising enough, the algorithm is working reasonably well but
sometimes fails. I am trying to understand
the source of the fails...What i am asking is: O(10^11) is really a
But if so, why it works at all?!??? Should i seriously address this
"problem" or not?
How can i lower the condition number of a large matrix?

Thank you,
Luca

Date Subject Author
6/9/10 luca
6/9/10 borchers@nmt.edu
6/9/10 luca
6/9/10 Dave Dodson
6/9/10 aruzinsky
6/10/10 aruzinsky
6/10/10 luca
6/10/10 aruzinsky
6/11/10 Peter Spellucci
6/13/10 Eli Osherovich
6/13/10 aruzinsky
6/13/10 aruzinsky