Minimum of the trace of a Cholesky factorizable matrix

Question

Given a matrix $P$ positive semidefinite Cholesky factorizable: $$P=SS^T$$ does this equality hold? $$\arg\min_{\theta}(\mathrm{tr}(P(\theta))=\arg\min_{\theta}(\mathrm{tr}(S(\theta)))?$$ if the diagonal elements of $P$ are greater than zero? Thanks.

Answer 1

There are many reasons to believe that the equality does not hold:

• "$\arg\min$" is a set and not necessarily a singleton;
• it can happen that minimum does not exist on either side of the equality;
• Cholesky decomposition is not unique; one can multiply $S$ on the right by a diagonal matrix whose diagonal entries are $\pm1$ to lower the trace.

At any rate, without further assumptions, the answer to your question is negative. Here is a counterexample. Let $\theta\in[0,1]$. Let $$ P(\theta) = \pmatrix{1&\theta\\ \theta&\theta^2+(1-\theta)^2}. $$ The trace of $P$ is minimised at $\theta=\frac12$. However, it is easy to show that if $P=SS^T$ is a Cholesky decomposition, $S$ must be of the form $$ \pmatrix{u&0\\ \theta u&(1-\theta)v}, $$ where $u,v=\pm1$. Therefore, depending on the sign of $v$, the trace of $S$ is minimised at either $\theta=0$ or $\theta=1$, but not at $\theta=\frac12$.