The adjoint of a linear operator with respect to an inner product 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} $ ?
 A: 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}
