I need a check on the following problem:
Let us consider the following algorithm:
- $B_0$ given, SPD matrix.
- Compute its Cholesky factorization $B_k = L_k L_k^T $, for $k=0,\ldots$
- Define $B_{k+1} = L_k^T L_k$.
Show that $B_k$ is similar to $B$, i.e. $B_k=M^{-1}BM$
First of all, I note that each $B_k$ is SPD. Indeed, given $x \in \mathbb{R}^n$:
$$x^T B_k x = x^T L_k^T L_k x = ( L_k x)^T L_kx \geq 0$$ and equality holds iff $x=0$
- Now, consider $k=1$. If $(\lambda, v)$ is an eigenpair of $B$ I note that $(\lambda, L_0^T v)$ is an eigenpair of $B_1 = L_0^T L_0$: $$L_0^TL_0 (L_0^T v) = L_0^T B v = \lambda L_0^Tv $$
Hence, $B_1$ and $B_0$ have the same eigenvalues. Let's call $D$ the diagonal matrix with those eigenvalues.
- Now, since $B$ is symmetric, there exists $P$ s.t. $$D = P^{-1}BP$$
Also $B_1$ is symmetric, hence there exists $M$ s.t. $$B_1= M D M^{-1} = M P^{-1} B P M^{-1}$$
Let's call $\tilde{M} = PM^{-1}$, hence I have shown that:
$$B_1 = \tilde{M}^{-1} B \tilde{M}$$
which is the thesis for $k=1$.
If it's okay, then I can just iterate and there should be no problem in the generalization
I assume that SPD means symmetric and positive definite.
Yes, your proof is fine. It is important to note that $L_0^Tv$ is a non-zero eigenvector because of the invertibility of $L_0$. We could make the remainder of your proof shorter by observing that similarity is an equivalence relation: because $B_0$ is similar to $D$ and $B_1$ is similar to $D$, it must hold that $B_0$ is similar to $B_1$.
That said, the entire proof can be shortened into the following line: because $L_0$ is invertible, we have $$ \begin{align} L_0^{-1}B_0 L_0 &= L_0^{-1}(L_0L_0^T)L_0 = (L_0^{-1}L_0)(L_0^TL_0) = B_1 \end{align} $$
$$ %\begin{align} %B_1 &= L_0^TL_0 = L_0^{-1}L_0 (L_0^TL_0)L_0^{-1}L_0 %\\ & = %L_0^{-1}(L_0 L_0^T)(L_0L_0^{-1})L_0 %= L_0^{-1}(L_0L_0^T)L_0 = L_0^{-1}B_0 L_0. %\end{align} $$