I have an operator $T (x_1, x_2) := (x_2, x_1)$ which is self-adjoint with respect to standard inner product, now I have a new inner product on $\Bbb R^2$
$$\langle x,y\rangle = x_1y_1 + \frac{1}{2} (x_1y_2 + x_2y_1) + \frac{1}{3} x_2y_2.$$
I find the matrix of this inner product to be $$A = \begin{pmatrix} 1 &\frac{1}{2}\\ \frac{1}{2} & \frac{1}{3}\end{pmatrix}.$$
How can I find the adjoint of $T$ – $(T^*)$ with respect to this new inner product? Is $T^* = A T A^{-1} $ ?
Best Answer
By definition, the adjoint operator of $T$ corresponding to the inner $\langle \cdot, \cdot\rangle_A$ is given by \begin{align} \langle v, Tv\rangle_A = \langle T^\dagger v, v\rangle_A. \end{align} where $T^\dagger$ will be the symbol of the adjoint with respect to $A$. However, since \begin{align} \langle v, Tv\rangle_A = \langle v, ATv\rangle = \langle v, ATA^{-1}Av\rangle = \langle (ATA^{-1})^\ast v, Av\rangle= \langle (A^\ast)^{-1}T^\ast A^\ast v, v\rangle_A \end{align} then it follows \begin{align} T^\dagger = (A^\ast)^{-1}T^\ast A^\ast. \end{align}
Edit: In your case, we have that \begin{align} T= \begin{bmatrix} 0& 1\\ 1 & 0 \end{bmatrix} \ \ \ \Rightarrow \ \ \ T^\ast = T \end{align} and \begin{align} A^\ast = \begin{bmatrix} 1 &\frac{1}{2}\\ \frac{1}{2} & \frac{1}{3}\end{bmatrix} \ \ \ \ (A^{\ast})^{-1} = \frac{1}{\frac{1}{3}-\frac{1}{4}}\begin{bmatrix} \frac{1}{3} &-\frac{1}{2}\\ -\frac{1}{2} & 1\end{bmatrix} = \begin{bmatrix} 4 & -6\\ -6 & 12 \end{bmatrix} \end{align} which means \begin{align} T^\dagger = \begin{bmatrix} 4 & -6\\ -6 & 12 \end{bmatrix} \begin{bmatrix} 0& 1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 &\frac{1}{2}\\ \frac{1}{2} & \frac{1}{3}\end{bmatrix} = \begin{bmatrix} 4 & -\frac{5}{3}\\ 9 & 4 \end{bmatrix}. \end{align}