# Positive definite symmetric matrix has LDL^T decomposition with positive diagonals

• Mar 19th 2013, 06:50 AM
math2011
Positive definite symmetric matrix has LDL^T decomposition with positive diagonals
QUESTION:

Let $G$ be an $n \times n$ matrix with a factorization $G = LDL^{T}$, where $L$ is a unit lower triangular matrix and $D$ is a diagonal matrix. Show that $G$ is positive definite if and only if $D_{ii} > 0$, for $i=1,2,\ldots,n$.

ATTEMPT:

Assume $D_{ii}>0$ for $i=1,2,\ldots,n$. Then $D = D^{\frac{1}{2}} D^{\frac{1}{2}}$ is possible and
$x^T G x = x^T L D L^T x = x^T L D^{\frac{1}{2}} D^{\frac{1}{2}} L^T x = x^T (LD^{\frac{1}{2}}) (L D^{\frac{1}{2}})^T x = ( (LD^{\frac{1}{2}})^T x)^T (L D^{\frac{1}{2}})^T x = y^T y = \sum^{n}_{i=1} y_i^2 \geq 0$
where $y = (LD^{\frac{1}{2}})^T x$.
Because $LD^{\frac{1}{2}}$ has $n$ pivots it is nonsingular and hence $y = 0$ if and only if $x = 0$. Therefore $x^T G x > 0$ for all $x \ne 0$.

Assume $x^T G x = x^T LDL^T x$ is positive definite. How can I prove that $D$ has positive diagonals?
• Mar 19th 2013, 10:27 PM
chiro
Re: Positive definite symmetric matrix has LDL^T decomposition with positive diagonal
Hey math2011.

Isn't one of the definitions of a positive define matrix having positive eigenvalues?

Positive-definite matrix - Wikipedia, the free encyclopedia
• Mar 20th 2013, 02:01 AM
math2011
Re: Positive definite symmetric matrix has LDL^T decomposition with positive diagonal
Quote:

Originally Posted by chiro
Hey math2011.

Isn't one of the definitions of a positive define matrix having positive eigenvalues?

Positive-definite matrix - Wikipedia, the free encyclopedia

Yes, but how can I show that the diagonal entries of $D$ are the eigenvalues of $G$?
• Mar 20th 2013, 04:39 AM
chiro
Re: Positive definite symmetric matrix has LDL^T decomposition with positive diagonal
Is your L matrix the matrix of eigenvectors and also is that matrix orthonormal? If so then L^T = L^(-1) and you have a eigen-decomposition of your matrix G which means that D is your matrix of eigenvalues.
• Mar 20th 2013, 05:49 AM
math2011
Re: Positive definite symmetric matrix has LDL^T decomposition with positive diagonal
I don't know if L is the matrix of eigenvectors. The problem does not state that explicitly. Are you suggesting that L actually is the matrix of eigenvectors? How can I prove it?
• Mar 20th 2013, 04:04 PM
chiro
Re: Positive definite symmetric matrix has LDL^T decomposition with positive diagonal
My suggestion is to show that LDL^t is the diagonal form by showing that L is non-singular (since it is triangular with non-zero determinant) and then use this to show that L is also orthonormal.

For the ortho-normal aspect, if you can show L*L^t = I then you're done.

By showing that L^t = L^(-1) and by using the assumption that the diagonalization produces a unique decomposition, then you have shown that this is indeed the diagonalization.
• Mar 21st 2013, 05:53 AM
math2011
Re: Positive definite symmetric matrix has LDL^T decomposition with positive diagonal
Thanks for explaining but I still cannot see how to do it. L^t is a unit upper triangular matrix and L is a unit lower triangular matrix, L^t L does not necessarily equal to the identity matrix unless the L is special. The problem does not state any other information about L that directly say L is orthogonal.