Is the set of diagonal matrices $D^{-1/2}$ such that $I +D^{-1/2}AD^{-1/2}$ is positive semi-definite convex?

Question

Let $A$ be an arbitrary real symmetric matrix. I was able to convince myself that the set of diagonal matrices $D$ such that $D+A \succeq 0$ is convex. Here $M \succeq 0$ means $M$ is positive semi-definite.

I thought of this in two ways:

1. Directly showing that $D_{1} + A \succeq 0$ and $D_{2} + A \succeq 0$ then $tD_{1}+(1-t)D_{2} + A \succeq 0$.

2. Another way, I thought of showing this is write $\{ D: D+A \succeq 0 \} = \cap_{v \in \mathbb{R}^{n}} \{ D : v^{T}(D+A)v \geq 0 \}$ and think of $\{ D : v^{T}(D+A)v \geq 0 \}$ as closed half-spaces.

I noticed that $D+A \succeq 0$ iff $I+D^{-1/2}AD^{-1/2} \succeq 0$. I tried the first direct method above by considering $tD^{-1/2}_{1} + (1-t)D^{-1/2}_{2}$ but this doesn't work out as cleanly. So my question is

Is the set of diagonal matrices $D^{-1/2}$ such that $I +D^{-1/2}AD^{-1/2} \succeq 0$ convex?

I guess I could try this using a half space argument, but I don't have a precise definition of what a half-space is

Answer 1

We have for $$ A= \pmatrix{ 0 & 1\\ 1 & 0} $$ that $$ I + \pmatrix{a & 0 \\ 0 & b} A \pmatrix{a & 0 \\ 0 & b} = \pmatrix{ 1 & ab\\ ab& 1}. $$ This matrix is positive semidefinite iff $a^2b^2\le 1$. The set of such points is not convex, for instance $d_1=(2,1/2)$ and $d_2=(1/2,2)$ are viable choices, but the point $\frac12(d_1+d_2)=(5/4,5/4)$ is not.

Hence the matrices $$ D_1^{-1} = \pmatrix{2 & 0 \\ 0 & 1/2}, D_2^{-1} = \pmatrix{1/2 & 0 \\ 0 & 2} $$ satisfy the condition, while $\frac12(D_1,D_2)$ does not.

Answer 2

Your reformulation is confusing. In the original setting, you are asking whether the set $\mathscr{D}$ of diagonal matrices $D$ that makes $D+A$ positive semidefinite is convex. Note that members of $D$ are not necessarily positive definite.

In your reformulation, however, you are asking whether $\mathscr S=\{D^{-1/2}:D\succ0,\ D\in \mathscr{D}\}$ is convex. Here, members of $\mathscr S$ are necessarily positive definite and they are in general not members of $\mathscr D$, but inverse square roots of those positive definite members of $\mathscr D$.

As shown in the other answer, $\mathscr S$ is not convex. The set $\mathscr D$ in your original setting is indeed convex, however, because $\left[tD_1+(1-t)D_2\right]+A=t(D_1+A)+(1-t)(D_2+A)$.