I'm having trouble understanding the adjoint of a linear operator in a Hilbert space. I have the following problem but I'm unsure of my answers:
Let A and B be bounded linear operators mapping a complex Hilbert space H into itself. Prove that:
a) $(\alpha A + \beta B)^*=\bar\alpha A^* + \bar\beta B^*$
b) $(AB)^* = B^*A^*$
c) $(A^*)^* = A$
d) $I^* = I$, where $I$ is the identity operator.
So I believe I just use the inner product, using the Riesz representation theorem to find a specific vector $y$ such that $\langle Ax, y\rangle = \langle x, A^*y\rangle$. So my solution for the first one (the rest are similar) is:
\begin{align*}\langle(\alpha A+\beta B)x, y\rangle &= \alpha\langle Ax, y\rangle + \beta \langle Bx, y\rangle\\ &=\alpha\langle x, A^*y\rangle + \beta\langle x, B^*y\rangle\\ &=\langle x, \bar\alpha A^*y\rangle + \langle x, \bar\beta B^*y\rangle\\ &=\langle x, (\bar\alpha A^* + \bar\beta B^*)y\rangle, \end{align*} therefore $\bar\alpha A^* + \bar\beta B^* = (\alpha A + \beta B)^*$.
Is this the right idea? We use the inner product for the adjoint of an operator on Hilbert spaces into themselves because the Riesz representation theorem ensures we can always find a special $y$ to have the equality $\langle Ax, y\rangle = \langle x, A^*y \rangle$? We would not use the inner product for the adjoint if we were going between two different Hilbert spaces, correct?