Cholesky decomposition: any theoretical value? Just read the Wikipedia article on Cholesky decomposition. All the applications listed there were numerical. Are there theoretical arguments to which it is important?
For instance, here there is an argument that for $M,N$ symmetric, $M>0$, we have that there exists a matrix $X$ such that $X^TMX$ and $X^TNX$ are both diagonal. They use the Cholesky decomposition, but it seems the argument works equally well with factoring $M = \left( \sqrt{M} \right)^2$. Incidentally, I'm having trouble with this argument.
 A: My personal best of the theoretical importance of the Cholesky factorisation is the ease with which you can prove its existence. Assuming $x^TMx>0$, it takes only few lines to show that there is an $N$ such that $M=N^TN$. You don't need to refer to matrix square roots and other kinds of "magic" tools which heavily rely on eigenvalue decompositions.
EDIT: I do not remember, where I saw this proof exactly. It is easy to prove it using induction. The $1\times 1$ case is trivial, assume it's true for $n$ and consider an $(n+1)\times(n+1)$ SPD matrix
$$\tag{1}
\tilde{A}=\begin{bmatrix}
\alpha & a^T \\ a & A
\end{bmatrix}.
$$
It's easy to eliminate $a$'s by
$$\tag{2}
\begin{bmatrix}
1 & 0 \\ -\alpha^{-1}a & I
\end{bmatrix}
\begin{bmatrix}
\alpha & a^T \\ a & A
\end{bmatrix}
\begin{bmatrix}
1 & -\alpha^{-1}a^T \\ 0 & I
\end{bmatrix}
=
\begin{bmatrix}
\alpha & 0 \\ 0 & A-\alpha^{-1}aa^T
\end{bmatrix}.
$$
Now it would be good if $A-\alpha^{-1}aa^T$ was positive definite, since if $A-\alpha^{-1}aa^T=LL^T$ for some lower triangular $L$, then
$$
\tilde{L}=
\begin{bmatrix}
1 & 0 \\ \alpha^{-1}a & I
\end{bmatrix}
\begin{bmatrix}
\sqrt{a} & 0 \\
0 & L
\end{bmatrix}
$$
would be the Cholesky factor of (1). That $A-\alpha^{-1}aa^T$ is SPD can be easily deduced from (2): The matrix $\tilde{A}$ is SPD by assumption, so $x^T\tilde{A}x>0$ for all nonzero $x$, so it certainly is true also for all $x$ of the form $x=[-\alpha^{-1}a^Ty,y^T]$ where $y\neq 0$, for which we have
$$
y^T(A-\alpha^{-1}aa^T)y=x^T\tilde{A}x>0.
$$
P.S.: When I wrote few lines, I thought actually "few lines of formulas" :-)
