Find $T^*$ where $T(A)=(\begin{smallmatrix}1 & 2\\ \:3 & 4\end{smallmatrix})^{-1}A(\begin{smallmatrix}1 & 2\\\:3 & 4 \end{smallmatrix})$

$A,B,C \in M_2(\mathbb{R})$

$T:V\to V $ such that $T(A)=\begin{pmatrix}1 & 2\\ \:3 & 4 \end{pmatrix}^{-1}A\begin{pmatrix}1 & 2\\ \:3 & 4 \end{pmatrix}$

Find $T^*$

I know how to solve this by finding $T(e_1),T(e_2),T(e_3),T(e_4)$ such that $E:=(e_1,e_2,e_3,e_4)$ is the standard basis m and then find $[T^*]_E$

I try to solve it by using the inner product.

$tr(C^t\begin{pmatrix}1 & 2\\ \:3 & 4 \end{pmatrix}^{-1}A\begin{pmatrix}1 & 2\\ \:3 & 4 \end{pmatrix})=\langle TA, C \rangle = \langle A,T^*C \rangle=tr((T^*C)^tA)$

I get stuck , help please ?

Thanks !

Solution 1:

You can use the property that the trace is invariant under cyclic permutations. For clarity, let $P_1 = \begin{pmatrix} 1 & 2 \\ 3 & 4\end{pmatrix}^{-1}$ and $P_2 = P_1^{-1}.$ Then your map $T$ is just $T(A) = P_1 A P_2.$ Now consider

\begin{align*} \langle T(A),C\rangle &= \text{tr}(C^\top T(A)) \\ &= \text{tr}(C^\top P_1 AP_2)\\ &= \text{tr}(P_2 C^\top P_1 A) \\ &= \text{tr}((P_1^\top C P_2^\top)^\top A)\\ &= \langle A, P_1^\top C P_2^\top\rangle \end{align*}

and therefore, $T^\ast(C) = P_1^\top C P_2^\top.$