Let $V = M_{n \times n}(R)$ with the inner product $ <A,B> = Tr(AB^t)$, and $T$ the linear operator given by $T : M_{n \times n}(R) \rightarrow M_{n \times n}(R)$ and $T(A)= A^t$ .
How can i find the minimal polynomial of T and the adjoint operator $T^t$.
I tried to compute the matrix associated of $T$. Any $A\in M_{n}$ has the form $$ A=\sum_{i,j}A_{ij}e_{ij}\in M_{n} $$ which is identified with $$ A=\sum_{i,j}A_{ij}e_{i}\otimes e_{j}\in\mathbb{R}^{n^{2}}. $$ Now, $T:\mathbb{R}^{n^{2}}\rightarrow\mathbb{R}^{n^{2}}$ by $e_{j}\otimes e_{i}\mapsto e_{i}\otimes e_{j}$, i.e., $$ T=\sum_{i,j}e_{ij}\otimes e_{ji} $$ For example, when $n=2$, \begin{eqnarray*} e_{11}\otimes e_{11} & = & \left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]\\ e_{12}\otimes e_{21} & = & \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]\\ e_{21}\otimes e_{12} & = & \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right]\\ e_{22}\otimes e_{22} & = & \left[\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{array}\right] \end{eqnarray*} But how an i get the general case? and get the minimal polynomial and characteristic of $T$ ?, and then can't find the minimal polynomial and the adjoint $T^t$, please i need help for this.
Here are some hints.
For the minimal polynomial. Notice that $\mathrm{Sym}_n(\mathbb{R})$ and $\mathrm{Skew}_n(\mathbb{R})$ (the subspaces of symmetric and skew-symmetric matrices) are eigenspaces of $T$. Moreover, we have $$M_n(\mathbb{R})=\mathrm{Sym}_n(\mathbb{R}) \oplus \mathrm{Skew}_n(\mathbb{R})$$ This tells you that the operator $T$ is diagonalizable with eigenvalues $\pm 1$. What does this say about the minimal polynomial? Alternatively, and perhaps more simply, notice that $T$ is an involution.
For the adjoint. The adjoint satisfies $$\langle T(X),\ Y\rangle = \langle X^\mathrm{T},\ Y\rangle = \langle X,\ T^\mathrm{T}(Y)\rangle$$ But we also know that $$\langle X^\mathrm{T},\ Y\rangle = \langle X, Y^\mathrm{T}\rangle$$ What does this tell you about the adjoint of $T$?