Recall the Cholesky Theorem on $LL^{T}$-Factorization:
Theorem: If $A$ is a real, symmetric, and positive definite matrix, then it has a unique factorization, $A=LL^{T}$, in which $L$ is lower triangular with a positive digonal.
I need to use the Cholesky Theorem to show that the following two propertie sof a symmetric matrix $A$ (i.e., a matrix where $A = A^{T}$) are equivalent:
- $A$ is positive definite (i.e., $x^{T}Ax > 0$ for every nonzero vector $x$).
- There exists a linearly independent set of vectors $x^{(1)},x^{(2)},\cdots,x^{(n)}$ in $\mathbb{R}^{n}$ such that $A_{ij} = \left( x^{(i)}\right)^{T}\left( x^{(j)} \right)$.
I kind of see where the second part comes from, since if $A$ were a matrix we got as a result of multiplying together two other matrices $B$ and $C$, then each entry $A_{ij}$ would be the product of a row vector and a column vector. But, I'm really kind of lost as to how to prove this.
$1 \implies 2$: If $A$ is positive definite, let $LL^\top$ be as in the Cholesky theorem. Then let $x^{(i)}$ be the $i$th row of $L^\top$.
$2 \implies 1$: If $X$ is the matrix with columns $x^{(i)}$, then $A=X^\top X$. This is positive definite because $x^\top A x = x^\top XX^\top x = \|X^\top x\|^2 \ge 0$, with equality if and only if $x=0$ since $X$ is full rank.