# 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 $T^* = A T A^{-1}$ ?

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}