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Given, in $\mathbb {R^2}$, an inner product $(u,v) = u^t \cdot \begin{pmatrix} 3 & 2 \\ 2 & 3 \end{pmatrix} \cdot v$ and an operator $S \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ 0 \end{pmatrix} $. Find $S^*$ (the adjoint operator).

So, here's what I did:

I took the normal basis for $\mathbb R^2$, did the Gram Schmiddt algorithm, and got an orthogonal basis. Then I displayed $S$ in the basis and got the matrix that represents it, then I transposed and conjugated it to get what is $S^*$, but then I got this: (I don't know if it is correct): $[S^*]_b = \begin{pmatrix} \frac{1}{\sqrt{3}} & 0 \\ \frac{-2}{\sqrt{15}} & 0 \end{pmatrix}$. How do I find what $S^*$ does here?

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Here is another way:

All representations are with respect to the standard basis.

Remember that the adjoint is computed with respect to an inner product. So, the adjoint with respect to $(\cdot, \cdot)$ will be different that the adjoint with respect to $\langle \cdot, \cdot \rangle$.

You can choose a basis so that $(\cdot, \cdot)$ becomes the standard inner product with respect to this basis, compute $S$ in this basis, transpose and then find the representation of the transpose in the original (standard) basis. Alternatively, you can just use properties of matrix multiplication and transposition to obtain the same result, which is what I do below. It amounts to the same thing, of course, but avoids computing another basis in the interim calculations.

To avoid confusion, let $S^\square$ by the adjoint of $S$ with respect to $(\cdot, \cdot)$. Let $(x,y) = x^T A y$, where $A$ is the matrix used to define the inner product above.

The defining characteristic of $S^\square$ is $(S^\square x, y) = (x, Sy)$, or, expanding the definition, $(S^\square x)^T A y = x^T A Sy$. Continuing, we have

$(S^\square x)^T A y = x^T (S^\square)^T A y = x^T A Sy$. Since this is true for all $x,y$, we have $(S^\square)^T A = AS$, from which we get $(S^\square)^T = A S A^{-1}$, and finally $S^\square = A^{-T} S^T A^T$.

Noting that $A$ is symmetric, we have $A=\begin{bmatrix} 3 & 2 \\ 2 & 3 \end{bmatrix}$, $A^{-1} = \frac{1}{5} \begin{bmatrix} 3 & -2 \\ -2 & 3 \end{bmatrix}$ and $S= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, which gives $S^\square = \frac{1}{5} \begin{bmatrix} 9 & 6 \\ -6& -4 \end{bmatrix}$.

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