how to find a representing matrix and to show that $(T_P)^*=T_{P^*}$

Question

i can't solve this question correctly. could you help me with it please?

a) let $V=M_{n \times n}^C$ be the standard inner product space, and $P \in V$ be an invertible matrix. we'll define a linear transformation(mapping) as following $T_p:M_{n \times n}^C \to M_{n \times n}^C$ defined by $T_pX=P^{-1}XP$ for every $X \in V$prove that $(T_P)^*=T_{P^*}$

b)let $T_p:M_{2 \times 2}^C \to M_{2 \times 2}^C$ defined by $T_pX=P^{-1}XP$ when $P=\begin{pmatrix}i&1\\ -1&-i\end{pmatrix}$. find the matrix that defines $(T_P)^*$ in the standard base of $M_{2 \times 2}^C$

what i did:

a)since P follows $P*P^{-1}=P^{-1}P=I$, i put it in $T_pX=P^{-1}XP$, but i don't get that $(T_P)^*=T_{P^*}$

b)i tried to calculate P using the given mapping(transformation),in order to find the matrix that represents $(T_P)^*$ under the basic standart of $M^C_{2 \times 2}$(since as far as i know $T_P(X)=AX$ (where A is the desired matrix), but still couldn't find it.

please help me show $(T_P)^*=T_{P^*}$ and find the matrix that defines $(T_P)^*$

thank you very much for helping

Answer 1

$\newcommand\tr{\operatorname{tr}}$ I believe the standard inner product on complex matrices is $$ \left<X,Y\right> = \tr(XY^*) $$ where $Y^*$ is the conjugate transpose of $Y$. Using the identity $\tr(ABA^{-1}) = \tr(B)$, we have \begin{align*} \left<T_P X,Y\right> &= \tr(P^{-1} X P Y^*) = \tr(PP^{-1} X P Y^*P^{-1}) = \tr(X P Y^*P^{-1}) \\&= \tr(X ((P^{-1})^*Y P^*)^*) = \tr(X ((P^*)^{-1}Y P^*)^*) \\&= \left<X,T_{P^*} Y\right> \end{align*} This establishes that $(T_P)^* = T_{P^*}$.

For (b), I believe the standard basis of $V$ is \begin{align*} E_{11} &= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} &E_{12} &= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} &E_{21} &= \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} &E_{22} &= \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \end{align*} For each matrix $E$, you need to compute $T_{P^*}E$ and decompose it in the basis $\mathcal{B} = \left\{E_{11},E_{12},E_{21},E_{22}\right\}$. Using a crude Sage worksheet, I computed $$ \left[T_{P^*}\right]_{\mathcal B} = \frac{1}{2}\begin{bmatrix} 1 & -i & i & 1 \\ i & -1 & -1 & -i \\ -i & -1 & -1 & i \\ 1 & i & -i & 1 \end{bmatrix} $$

Answer 2

The adjoint operator $T_P^*$ is defined by $\langle T_PX,Y\rangle=\langle X,T_P^*Y\rangle$. Obtain that $\langle T_PX,Y\rangle=P^{-1}XP:Y=\sum_{ijkl}P^{-1}_{ij}X_{jk}P_{kl}Y_{il}=\sum_{ijkl}(P^{-1})^*_{ji}X_{jk}P^*_{lk}Y_{il}=\sum_{ijkl}X_{jk}(P^{-1})^*_{ji}Y_{il}P^*_{lk}=\langle X,(P^{-1})^*YP^*\rangle=\langle X,(P^{*})^{-1}YP^*\rangle=\langle X,T_{P^*}Y\rangle$.

Therefore we deduce that $(T_P)^*=T_{P^*}$. Now for (b) you can directly use the result from (a).