If $\sum_j \overline{E_j AE_j^\dagger}=\sum_j E_j AE_j^\dagger$ for all $A$, does $ \overline{E_j AE_j^\dagger}= E_j AE_j^\dagger$

Question

I'm currently optimizing a statistical algorithm and want to make sure that an assumption I'm making is correct. Let $E_j$ be a finite sequence of $n \times n$ complex matrices that satisfies: $\sum_j E_j E_j^\dagger=I.$ If $$\sum_j \overline{E_j AE_j^\dagger}=\sum_j E_j AE_j^\dagger$$ for all symmetric real matrices $A$, can we conclude that $ \overline{E_j AE_j^\dagger}= E_j AE_j^\dagger$ for all $j$? Here the bar denotes the complex conjugate and the dagger is the adjoint. I believe this is true as we can rewrite the equality as $$\sum_j \overline{E_j} AE_j^T-E_j AE_j^\dagger=0$$ for all $A$ where $E_j^T$ denotes the transpose of $E_j$ and I'm not sure how else we could construct a sequence of matrices that would hold for any given $A$.

Answer 1

Let $U=[u_{i,j}]\in M_n(\mathbb{C})$ and let $(E_{i,j})$ be the canonical basis of $M_n(\mathbb{R})$. The first thing to do is to see what is the meaning of

for every real symmetric matrix $A\in M_n(\mathbb{R})$, $\overline{U}AU^T=UAU^*$.

Note that $(E_{i,j}+E_{j,i})$ is a basis over $\mathbb{R}$ of the vector-space of symmetric real matrices.

for every $k,l,i,j$, one has $\overline{u_{k,i}}u_{l,j}+\overline{u_{k,j}}u_{l,i}\in\mathbb{R}$. In particular, when $i=j$, $\overline{u_{k,i}}u_{l,i}\in\mathbb{R}$.

It's not difficult to find (up to eventual errors) that $U$ has one of the following two forms:

Case 1. $U=e^{i\theta}V$ where $\theta\in \mathbb{R},V\in M_n(\mathbb{R})$.

Case 2. $U=[a_1,\cdots,a_n]^T[b_1e^{i\theta_1},\cdots,b_ne^{i\theta_n}]$ where the $a_i's,b_i's,\theta_i's\in\mathbb{R}$.

Answer 2

You are essentially asking whether the conditions that $\sum_jE_jAE_j^\dagger$ is real for every real symmetric $A$ and that $\sum_jE_jE_j^\dagger=I$ would imply that $E_jAE_j^\dagger$ is also real for every $j$ and for every real symmetric $A$. The answer is negative. Here is a counterexample. Let $$ E_1=\frac1{\sqrt{3}}\pmatrix{1-i\\ &i},\quad E_2=\frac1{\sqrt{3}}\pmatrix{i\\ &1-i}, \quad A=\pmatrix{a&b\\ c&d}. $$ Then $$ \begin{align*} E_1AE_1^\dagger &=\frac13\pmatrix{1-i\\ &i}\pmatrix{a&b\\ c&d}\pmatrix{1+i\\ &-i} =\frac13\pmatrix{2a&(-1-i)b\\ (-1+i)c&d}\tag{1}\\ E_2AE_2^\dagger &=\frac13\pmatrix{i\\ &1-i}\pmatrix{a&b\\ c&d}\pmatrix{-i\\ &1+i} =\frac13\pmatrix{a&(-1+i)b\\ (-1-i)c&2d}\tag{2}. \end{align*} $$ Hence $$ E_1AE_1^\dagger+E_2AE_2^\dagger =\frac13\pmatrix{3a&-2b\\ -2c&3d}, $$ which is real for every real $A$ (not just for the symmetric ones), but from $(1)$, it is obvious that $E_1AE_1^\dagger$ is not real whenever $A$ is real but not diagonal.