Let $A$ be positive definite, show that there is $S\in \operatorname{GL}(n, \mathbb{R})$ such that $A=SS^T$

Question

Let $A$ be positive definite, show that there is $S\in \operatorname{GL}(n, \mathbb{R})$ such that $A=SS^T$

I know that this hints at the Cholesky decomposition. But I don't have to prove that $A$ can be decomposed into a lowever triangular matrix and its conjugate but rather that a decomposition exists.

I know that $A$ is positive definite iff $x^TAx>0$. This is not much to work with. Another way of characterizing A positive definite Matrix is by looking at its Eigenvalues which all have to be greater than zero. This is a dead end for me too. Can anyone give me a hint?

Answer 1

If you write down the eigendecomposition $A=UDU^\top$ where $U$ is orthogonal and $D$ is diagonal, you can define $D^{1/2}$ to be the result of taking the square root of each entry of $D$ (this is possible because the eigenvalues are positive). Then check that $S=UD^{1/2} U^\top$ works.