Trace of a nonlinear matrix equation (cont'd)

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.

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 $k\geq 0$?

• @FedorPetrov I can show that cycles of any length are impossible if $A\ne I$. Alas, this doesn't help much with the original problem unless $X_j$ are uniformly non-degenerate (which is never true in the interesting case $\det A<1$), so we are still nowhere... – fedja Sep 20 '16 at 4:47
• Certainly. Assume $A=diag(a_1,\dots,a_n)$. For a self-adjoint matrix $Z$, let $v(Z)$ be the vector whose coordinates are the diagonal entries of $Z$. Assume $Y^2=XAX$. Then $v(Y^2)=(X\circ X^T)v(A)$ while $v(X^2)=(X\circ X^T)v(I)$ where $Z\circ T$ is the Schur (element-wise) product of $Z$ and $T$. Since $X\circ X^T$ is positive-definite, we get $\langle v(Y^2)-v(X^2),v(A)-v(I)\rangle>0$ if $v(A)\ne v(I)$, so we drift in one direction every time with no chance to return. – fedja Sep 20 '16 at 21:17
• @NawafBou-Rabee Yes, we can show that the entire sequence converges to a degenerate operator (degenerate fixed points are plentiful). What's next? – fedja Sep 20 '16 at 21:45
• @NawafBou-Rabee As I said from the very beginning, $\det A<1$ is the only interesting case. Note that any power of a positive number is still positive, so no degenerate situations occur in that case. – fedja Sep 20 '16 at 22:41
• @NawafBou-Rabee Yeah, but that case is trivial because the trace is not less than the $n$-th root of the determinant (AM-GM), so it just blows up geometrically. – fedja Sep 20 '16 at 23:02

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 :-).

First, let me modify the given problem for the sake of simplicity.

Given an $n \times n$ matrix $X_0>0$ with unit trace, i.e., $$X_0 = \sum_{1 \le i \le n} \lambda_i v_i v_i^T \;, \quad \sum_{1 \le i \le n} \lambda_i = 1$$ where $\{ \lambda_i \}$ and $\{ v_i \}$ are the positive eigenvalues and orthonormal eigenvectors of $X_0$, respectively. Set $A>0$ to be $$A = \sum_{1 \le i \le n} \alpha_i v_i v_i^T \tag{\diamond}$$ where $\{ \alpha_i \}$ are positive eigenvalues of $A$, and we stress that $A$ has the same eigenvectors as $X_0$. If the solution to the matrix recurrence relation $$X_{k+1} = X_k^{1/2} A X_k^{1/2}$$ with initial data $X_0$ satisfies $\operatorname{trace}(X_{k})=1$ for all natural numbers $k \ge 0$, then $A=I_n$, i.e., $\alpha_i = 1$ for all $1 \le i \le n$.

Remark. The difference between this formulation and the OP's formulation is that $A$ has the same eigenvectors as the given seed $X_0$. Admittedly, this form of $A$ is restrictive, but it is nicer to work with because, as we will see, it simplifies the subsequent calculations.

For the sake of contradiction, suppose that $A \ne I_n$, i.e., not all $\alpha_i = 1$ for $1 \le i \le n$. Since $$X_0^{1/2} = \sum_{1 \le i \le n} \sqrt{\lambda_i} v_i v_i^T$$ after one step of the recurrence relation we have that: \begin{align*} X_1 &= \sum_{1 \le i,j \le n} \sqrt{\lambda_i} \sqrt{\lambda_j} (v_i^T A v_j) v_i v_j^T \\ &= \sum_{1 \le i,j,k \le n}\sqrt{\lambda_i} \sqrt{\lambda_j} \alpha_k (v_i^T v_k) (v_k^T v_j) v_i v_j^T \\ &= \sum_{1 \le i,j \le n}\sqrt{\lambda_i} \sqrt{\lambda_j} \alpha_i \delta_{ij} v_i v_j^T \qquad \text{($\delta_{ij}$ is the Kronecker delta)}\\ &= \sum_{i=1}^n \lambda_i \alpha_i v_i v_i^T \end{align*} and the unit trace requirement implies that $$\sum_{i=1}^n \lambda_i \alpha_i = 1 \;.$$ Iterating the above calculation $k$ times, we see that the unit trace requirement implies that $\sum_{i=1}^n \lambda_i \alpha_i^k = 1$ which, according to our hypotheses, must hold true for any natural number $k \ge 0$. A more transparent way to write this requirement is as: $$\mathbf{V}^T \boldsymbol{\lambda} = \mathbf{1} \tag{\star}$$ where we have introduced an infinite Vandermonde-like matrix and two vectors: $$\mathbf{V} = \begin{bmatrix} 1 & \alpha_1 & \cdots & \alpha_1^k & \cdots \\ 1 & \alpha_2 & \cdots & \alpha_2^k & \cdots \\ \vdots & \vdots & \ddots & \vdots & \cdots \\ 1 & \alpha_n & \cdots & \alpha_n^k & \cdots \end{bmatrix} \;, \quad \boldsymbol{\lambda} = \begin{bmatrix} \lambda_1 \\ \vdots \\ \lambda_n \end{bmatrix} \;, \quad \mathbf{1} = \begin{bmatrix} 1 \\ 1 \\ \vdots \end{bmatrix}$$ If $\alpha_1 = \cdots = \alpha_n = 1$, then the solution set of ($\star$) basically contains all unit trace matrices that are positive definite. (This is the trivial case.) However, if even one of the eigenvalues of $A$ is not equal to one (the case at hand), then the solution set to this infinite (overdetermined) system of equations is empty, and there exists no $X_0$ that satisfies the hypotheses given above, which is a contradiction that is resolved only if $\alpha_1 = \cdots = \alpha_n = 1$ or $A=I_n$.

• Thanks! The approach is interesting, although restricted to a particular case. – Ludwig Sep 18 '16 at 13:56
• Does the restriction that $A$ has the same eigenvectors as $X_0$ reduce the practicality of this result? – Nawaf Bou-Rabee Sep 18 '16 at 15:18
• Unfortunately, yes – Ludwig Sep 18 '16 at 15:19
• Could you please elaborate? What is the motivation for this question? – Nawaf Bou-Rabee Sep 18 '16 at 15:20
• It's a long story, but basically I would like to show that the iteration $(\star)$ has no invariant trajectories in the set of positive definite trace one matrices if $A\neq I$ – Ludwig Sep 18 '16 at 15:29

I sketch here a solution for the $2\times 2$ case. My hope is that the outlined approach can be used to find a general solution.

As noticed in Addendum 1, we can restrict wlog to diagonal positive definite $A$'s $$A := \begin{bmatrix}d_1 & 0 \\ 0 & d_2\end{bmatrix}.$$ I consider the case $d_1>1$ and $0<d_2<1$, the other cases being trivial. Define $X_0^{1/2}$ as $$X_0^{1/2} := \begin{bmatrix}a & b \\ b & c\end{bmatrix}.$$ Since $X_0>0$ and hence $X_0^{1/2}>0$ it holds $a>0$, $b>0$, and $ac-b^2>0$. Moreover since $\mathrm{tr}(X_0)=1$, we have that $$a^2+2b^2+c^2=1. \quad (1)$$ Now after one iteration step we obtain \begin{align} X_1 &= X_0^{1/2}AX_0^{1/2}= \begin{bmatrix} a^2d_1+b^2d_2 & \ast \\ \ast & b^2d_1+c^2d_2\end{bmatrix}. \end{align} Due to the fact that $\mathrm{tr}(X_1)=1$, it follows that $$a^2d_1+b^2(d_1+d_2)+c^2d_2=1. \quad (2)$$ Let us define $$X_1^{1/2} := \begin{bmatrix} a_1 & b_1 \\ b_1 & c_1\end{bmatrix}$$ Since $X_1=X_1^{1/2}X_1^{1/2}$, we get \begin{align} a_1^2+b_1^2=a^2d_1+b^2d_2 \ \ \text{ and }\ \ b_1^2+c_1^2=b^2d_1+c^2d_2. \quad(\#) \end{align} Now consider the second iteration step $$X_2 = X_1^{1/2}AX_1^{1/2}= \begin{bmatrix} a_1^2d_1+b_1^2d_2 & \ast \\ \ast & b_1^2d_1+c_1^2d_2\end{bmatrix}.$$ We have \begin{align} \mathrm{tr}(X_2) &= d_1(a_1^2+b_1^2)+d_2(b_1^2+c_1^2)\\ &\overset{(\#)}{=}a^2d_1+2b^2d_1d_2+c^2d_2^2=1,\quad (3) \end{align} by virtue of $(\#)$ and of the trace constraint $\mathrm{tr}(X_2)=1$. By collecting $(1)$, $(2)$, and $(3)$ we arrive at the following linear system $$\begin{bmatrix} 1 & 2 & 1 \\ d_1 & d_1+d_2 & d_2 \\ d_1^2 & d_1d_2 & d_2^2\end{bmatrix}\begin{bmatrix}a^2 \\ b^2\\ c^2\end{bmatrix}=\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix}.$$ The solution $(\hat{a}^2, \hat{b}^2, \hat{c}^2)$ of the previous system — you can evaluate it manually or using some symbolic toolbox, as I did — is such that $$\sqrt{\hat a^2 \hat c^2}-\hat b^2 = \frac{|(d_1-1)(d_2-1)|+(d_1-1)(d_2-1)}{(d_1-d_2)^2}=0$$ since $d_1>1$ and $0<d_2<1$, by assumption. But, in this case, we get that $X_0^{1/2}$ is singular, which is a contradiction since $X_0>0$.

Update. It is possible to generalize the above procedure to the case $$A = \begin{bmatrix}d_1I_{n_1} & 0\\ 0&d_2I_{n_2}\end{bmatrix}$$ with $d_1>1$ and $0<d_2<1$ scalars. Indeed, by considering the following block decomposition of $X_0^{1/2}$ $$X_0^{1/2} := \begin{bmatrix}A & B \\ B^\top & C\end{bmatrix},$$ by replacing $a$, $c$, and $b$ with $\mathrm{tr}(A^2)$, $\mathrm{tr}(C^2)$, and $\mathrm{tr}(B^\top B)$, and by following almost verbatim the above solution we get the desired result.