# Operators that are not represented as matrices , operating on matrices.

I am currently going through "Log-gases and random matrices" by PJ Forrester. I'm coming from a totally different academic background, and I cannot understand a point of his notation. More precisely, he defines an antinunitary time reversal operator $T=\mathbf{Z}_{2N}K$ where $\mathbf{Z}_{2N}$ is the tensor product of the $N\times N$ identity matrix and of the $2\times2$ matrix $\begin{bmatrix}0 &-1\\ 1 & 0 \end{bmatrix}$, and where $K$ is the complex conjugate operator. So if I understand correctly, $T$ acts on a matrix $\mathbf{A}$ like so $$T\mathbf{A} =\mathbf{Z}_{2N}\bar{\mathbf{A}},$$ right? Where I stop following is how one should deal with $K$ in "mixed operations", i.e. since it is not presented as a martix, how does one invert $K$, apply it "from the right", etc.

The property i'm trying to understand is the following: commutation relations between Hermitian matrices $\mathbf{A}$ and $T$ lead to $$\mathbf{A} = T \mathbf{A} T^{-1} = \mathbf{Z}_{2N}K\mathbf{A} K^{-1} \mathbf{Z}_{2N}^{-1} = \mathbf{Z}_{2N}K\mathbf{A} K \mathbf{Z}_{2N}^{-1} = \mathbf{Z}_{2N}\bar{\mathbf{A}}\mathbf{Z}_{2N}^{-1}$$ From what I gather, $T$ is more or less treated as a matrix (at least with regard to inversion operation), and $K$ is it's own inverse (that, i understard). The puzzling point for me is the last equality. Does The second $K$ disappears because it is applied (with no effect) to the real matrix $\mathbf{Z}_{2N}^{-1}$? I cannot seem to find anything on the subject easily... I just need a confirmation of my understanding, and possibly further reading material!

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For your first question, if $y=Kx$, then $y=\bar{x}$ and hence $x=\bar{y}=Ky$. Therefore the inverse operator of $K$ is $K$ itself and hence $T^{-1}=\left(\mathbf{Z}_{2N}K\right)^{-1}=K^{-1}\mathbf{Z}_{2N}^{-1}=K\mathbf{Z}_{2N}^{-1}$.
For your second question, the answer is no. What specific matrix $K$ applies to is irrelevant. For any vector $x$, we have $KAKx=KA\bar{x}=\overline{A\bar{x}}=\bar{A}x$. Therefore $KAK=\bar{A}$.