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 if *there 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.




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**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.




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**Addendum 2.** A "simpler" version of this question was answered in the affirmative [here](https://mathoverflow.net/questions/249684/trace-of-a-nonlinear-matrix-equation/249738).




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**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.




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**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 $k\geq 0$?
> 
> 
> 



Trace of a nonlinear matrix equation (cont'd)
Actually, you have completely solved it yourself, just didn't dare to acknowledge it. In my notation, you have $(X\circ X^T)v(A)=(Y\circ Y^T)v(I)$ when $Y^2=XAX$. Similarly, $(Z\circ Z^T)v(I)=(Y\circ Y^T)v(A)$ when $Z^2=YAY$. Taking the trace, we must have 
$$
1=\langle(X\circ X^T)v(I),v(I)\rangle=\langle(Y\circ Y^T)v(I),v(I)\rangle=\langle(Z\circ Z^T)v(I),v(I)\rangle\,.
$$
However,
$$
\langle(Y\circ Y^T)v(I),v(I)\rangle=\langle(X\circ X^T)v(A),v(I)\rangle
$$
and
$$
\langle(Z\circ Z^T)v(I),v(I)\rangle=\langle(Y\circ Y^T)v(A),v(I)\rangle
\\
=
\langle(Y\circ Y^T)v(I),v(A)\rangle=\langle(X\circ X^T)v(A),v(A)\rangle
$$
so 
$$
\langle(X\circ X^T)(v(A)-v(I)),(v(A)-v(I))\rangle=0
$$
whence $X\circ X^T$ and, thereby, $X$ must be degenerate unless $v(A)=v(I)$, i.e., $A=I$.


This story definitely has a few morals but I'll abstain from spelling them out :-).


