# Trouble with adjoint of linear operators in a Hilbert space

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?

For any $x,y\in H$, we have \begin{align*} \langle x,(\alpha A+\beta B)^*y\rangle&=\langle(\alpha A+\beta B)x,y\rangle\\&=\cdots \text{(exactly what you did)}\cdots\\ &=\langle x,(\overline\alpha A^*+\overline\beta B^*)y\rangle. \end{align*} Since $x$ and $y$ were arbitrary, we have $(\alpha A+\beta B)^*=\overline\alpha A^*+\overline\beta B^*$.
This uses the following lemma: If $H$ is a complex Hilbert space and $T\in\mathcal B(H)$, then $T=0$ iff $\langle Tx,y\rangle=0$ for all $x,y\in H$.
Yes, the Riesz representation theorem is used to guarantee the existence of the adjoint. For given $y\in H$, the map $x\mapsto\langle Tx,y\rangle$ is a bounded linear functional, hence there exists a unique $T^*y\in H$ such that $\langle Tx,y\rangle=\langle x,T^*y\rangle$ for all $x\in H$. Then showing the map $y\mapsto T^*y$ is linear and bounded is a basic exercise.
Furthermore, essentially the same proof works for $T:H_1\to H_2$. You still use the Riesz representation theorem to obtain a map $T^*:H_2\to H_1$. You just have to be careful and pay attention to what space the vectors are in.