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# A question on matrix calculation

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Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

7 replies to this topic

### #1ronm  Members

Posted 05 December 2012 - 01:13 PM

Hello all, let say I have a PD symmetric matrix Sigma (order n) and two matrices A and B as order rXn. Now consider the expression:

(A * Sigma * A') - (B * Sigma * B')

Can I write this expression as (C * Sigma * C')?

### #2Álvaro  Members

Posted 05 December 2012 - 01:37 PM

No, you can't in general do that. If n < r, (A * Sigma * A') - (B * Sigma * B') will generally have rank 2*n, but (C * Sigma * C') can only have rank n.

### #3ronm  Members

Posted 05 December 2012 - 02:03 PM

okay, let say r = 1 and n > 1. In this restricted scenario what will be the case?

Thanks,

### #4Brother Bob  Moderators

Posted 05 December 2012 - 02:41 PM

If by PD you mean positive definite (I searched, but no other relevant terms showed up, just disregard if you mean something else), then no, you cannot find such a relation. You have two quadratic forms with A and B that are always positive since Sigma is positive definite. The difference between them may be negative if the quadratic form of B is greater than the one of A. Since sigma is positive definite, you cannot find a C that results in a negative value.

### #5Álvaro  Members

Posted 05 December 2012 - 03:40 PM

If r=1, (A * Sigma * A') - (B * Sigma * B') is a 1x1 matrix. If its element happens to be positive, then yes, you can find a vector C (many, actually) such that (C * Sigma * C') has the same value.

How about you tell us what you are trying to do, instead of going back and forth with descriptions of the problem that don't quite make sense?

### #6ronm  Members

Posted 05 December 2012 - 09:18 PM

Thanks for your comment. Basically I want to have a Statistical test as H0: A * Sigma * A' = B * Sigma * B'

Here the only population parameter is Sigma, which can be estimated from sample observations. Therefore I know the distribution of Sigma_Sample. And from this I can derive the distribution of C * Sigma * C'.

Therefore if can somehow write (A * Sigma * A') - (B * Sigma * B') = C * Sigma * C' then I can construct a Test Statistic for my testing.

It will also be very helpful if you can propose some better alternative of my testing problem.

Thanks and regards,

### #7ronm  Members

Posted 07 December 2012 - 11:25 AM

any suggestion pls?

Thanks,

### #8Inferiarum  Members

Posted 07 December 2012 - 12:43 PM

Why can you not derive the distribution of (A * Sigma * A') - (B * Sigma * B')?

Old topic!

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