Let $X_0$ be a trace-one positive definite matrix, i.e. $X_0>0$, $\mathrm{tr}(X_0)=1$. Let $A>0$ and consider the following iteration $$ X_{k+1} = X_k^{1/2}AX_k^{1/2},\quad k\geq 0,\quad (\star) $$ where $X_k^{1/2}$ denotes the (principal) square root of $X_k$.

My question:Is it true that ifthere exists$X_0$ as above such that $(\star)$ is trace-preserving starting from $X_0$, i.e. $\mathrm{tr}(X_{k+1})=\mathrm{tr}(X_{k})=1$ for all $k\geq 0$, then $A=I$?

If $X_0$ and $A$ are scalars this is clearly true. Moreover, it is easy to see that $A$ cannot be such that $A> I$ or $A< I$, but I cannot quite prove that $A$ must be the identity.

Thanks for your help.

**Addendum 1.** Note that by applying to $(\star)$ an orthogonal change of basis $T$ which diagonalizes $A$, we can rewrite $(\star)$ as
$$
\tilde{X}_{k+1}= \tilde{X}_k^{1/2} D \tilde{X}_k^{1/2} \quad (\star\star)
$$
where $\tilde{X}_k:=T^\top X_k T$ and $D:=T^\top AT>0$ is diagonal.

**Addendum 2.** A "simpler" version of this question was answered in the affirmative here.

**Addendum 3.** My attempts so far were based on working with the "simplified" dynamics $(\star\star)$. Namely, consider the partition $D=\left[\begin{smallmatrix}D_1 & 0 \\ 0 & D_2\end{smallmatrix}\right]$ with $D_1>I_{n_1}$ and $D_2<I_{n_2}$ (in the other cases, e.g. $D_1> I_{n_1}$, $D_2=I_{n_2}$, it is easy to show that $\mathrm{tr}(\tilde{X}_1)\neq 1$). Now by partitioning $\tilde X_0^{1/2}$ and $\tilde X_1$ accordingly to the block decomposition of $D$, we get
$$
\tilde{X}_1=\begin{bmatrix}(\tilde X_1)_{11} & (\tilde X_1)_{12} \\ (\tilde X_1)_{12}^\top & (\tilde X_1)_{22}\end{bmatrix} = \begin{bmatrix}(\tilde X_0^{1/2})_{11}D_1(\tilde X_0^{1/2})_{11} + (\tilde X_0^{1/2})_{12}D_2(\tilde X_0^{1/2})_{12}^\top & (\ast) \\ (\ast)^\top & (\tilde X_0^{1/2})_{12}^\top D_1(\tilde X_0^{1/2})_{12} + (\tilde X_0^{1/2})_{22}D_2(\tilde X_0^{1/2})_{22} \end{bmatrix},
$$
with the constraint $\mathrm{tr}(\tilde X_0)=1$ which now reads as
$$
\mathrm{tr}\left((\tilde X_0^{1/2})_{11}(\tilde X_0^{1/2})_{11} + (\tilde X_0^{1/2})_{12}(\tilde X_0^{1/2})_{12}^\top\right)+\mathrm{tr}\left((\tilde X_0^{1/2})_{12}^\top (\tilde X_0^{1/2})_{12} + (\tilde X_0^{1/2})_{22}(\tilde X_0^{1/2})_{22}\right)=1.
$$
Now my conjecture is that $\mathrm{tr}(\tilde X_2)\neq 1$ for all $\tilde{X_0}>0$, $\mathrm{tr}(\tilde{X_0})=1$, (indeed, it is easy to find examples for which $\mathrm{tr}(\tilde X_1)= 1$). My idea is to use the same block decomposition for $\tilde X_2$ and then exploit some "trace inequalities" applied to the diagonal blocks. However I didn't manage to conclude anything so far.

**Addendum 4.** A more general version of this question which I suspect hold true (actually, for the $2\times 2$ case, it does hold true) is the following one:

Is it true that if there exists $X_0\ge 0$, $\mathrm{tr}(X_0)=1$, such that $(\star)$ is trace-preserving starting from $X_0$, i.e. $\mathrm{tr}(X_{k+1})=\mathrm{tr}(X_{k})=1$ for all $k\geq 0$, then $X_{k+1}=X_k^{1/2}AX_k^{1/2}=X_k$ for all $k\geq 0$?

trivialbecause the trace is not less than the $n$-th root of the determinant (AM-GM), so it just blows up geometrically. $\endgroup$ – fedja Sep 20 '16 at 23:02