Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am given $A$ a symmetric positive definite matrix, and $U$ which the Cholesky factor of $A$. I am also told that if $V$ is an upper triangular matrix such that $A$ = $V^TV$. I have to show that there exists a diagonal matrix $D$ whose entries on the main diagonal are either $−1$ or $1$ such that $V = DU$.

My thoughts are as follows. Based on the first statement, $A$=$U^T U$ as we are given that $U$ is a cholesky factor of $A$. Also based on this I can equate $A = V^T V = U^T U$, but I got stuck after this. How should I introduce $D$ in this and show that diagonal of $D$ should be $1$ or $-1$ to get $V = DU$?

Any help would be appreciated.

share|improve this question
    
I changed the title (apart from formatting your text) since the title was too long. Hope it is fine. –  user17762 Mar 14 '12 at 3:47
add comment

1 Answer 1

Note that $V^TV = U^TU$ implies $\left(VU^{-1} \right)^{T} \left(VU^{-1} \right) = I$. Now note that $U^{-1}$ is also upper triangular since inverse of upper triangular is also upper triangular and $VU^{-1}$ is a product of two upper triangular matrices which is again upper triangular. Hence, you have $I = R^TR$ where $R$ is an upper-triangular matrix. This implies $R$ has to be diagonal with entries $\pm 1$ since the identity matrix has only diagonal entries and each equals $1$.

share|improve this answer
    
Thanks Sivaram!!! –  newbietolinalg Mar 14 '12 at 4:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.