# Uniqueness of Cholesky factorization

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.

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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

## 1 Answer

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 $1$.

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Thanks Sivaram!!! – newbietolinalg Mar 14 '12 at 4:25