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

QUESTION:

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

ATTEMPT:

Assume $\displaystyle D_{ii}>0$ for $\displaystyle i=1,2,\ldots,n$. Then $\displaystyle D = D^{\frac{1}{2}} D^{\frac{1}{2}}$ is possible and

$\displaystyle x^T G x = x^T L D L^T x = x^T L D^{\frac{1}{2}} D^{\frac{1}{2}} L^T x = x^T (LD^{\frac{1}{2}}) (L D^{\frac{1}{2}})^T x = ( (LD^{\frac{1}{2}})^T x)^T (L D^{\frac{1}{2}})^T x = y^T y = \sum^{n}_{i=1} y_i^2 \geq 0$

where $\displaystyle y = (LD^{\frac{1}{2}})^T x$.

Because $\displaystyle LD^{\frac{1}{2}}$ has $\displaystyle n$ pivots it is nonsingular and hence $\displaystyle y = 0$ if and only if $\displaystyle x = 0$. Therefore $\displaystyle x^T G x > 0$ for all $\displaystyle x \ne 0$.

Assume $\displaystyle x^T G x = x^T LDL^T x$ is positive definite. How can I prove that $\displaystyle D$ has positive diagonals?

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

Re: Positive definite symmetric matrix has LDL^T decomposition with positive diagonal

Quote:

Originally Posted by

**chiro**

Yes, but how can I show that the diagonal entries of $\displaystyle D$ are the eigenvalues of $\displaystyle G$?

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.

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?

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.

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.