Let $F$ denote $\mathbb R$ or $\mathbb C$. Let $T : V → W$, $S : V → W$ and let $R: U → V$ be linear transformations between inner product spaces $U$, $V$, $W$ over $F$. Verify the following facts:
(a) $(T + S)^∗ = T^∗ + S^∗$ (where * is the adjoint)
My attempt: We know $\langle (T+S)v, w\rangle = \langle v, (T+S)^*w\rangle$ therefore $\langle (T+S)v, w\rangle = \langle T(v),w \rangle + \langle S(v),w \rangle = \langle v, T^*(w) \rangle + \langle v, S^*(w) \rangle = \langle v, (T+S)^*(w)$
(b) For all $α ∈ F, (αT)^∗ = αT^∗$
I know that somehow the property that $\langle v, \alpha w \rangle = \overline{\alpha} \langle v,w \rangle $ will come in play.
(c) $(T^∗)^∗ = T$.
This one seems so obvious I don't know how to prove it
(d) $(TR)^∗ = R^∗T^∗.$
(e) If $T$ is invertible, then $(T^{−1})^∗ = (T^∗)^{−1}$
We haven't covered adjoints in class yet and got this assigned, so I don't know how to verify these.
I will prove e) by first proving another fact about adjoints. There may be an easier way to prove e), but this way you learn something that wasn't prescribed.
Lemma: If $id$ is the identity map on an inner product space, then $id^* = id$.
Proof: Suppose that $V$ is an inner product space. Let $v,w\in V$ and $id$ be the identity map on $V$. $$\langle w , id(v)\rangle=\langle w , v\rangle=\langle id(w) , v\rangle=\langle w , id^*(v)\rangle.$$ Therefore, $id(v)=id^*(v)$ for all $v\in V$. Hence, $id^* = id$.
Now, we shall prove e).
Proof: Suppose that $V$ and $W$ are inner product spaces. Let $T:V\to W$ be an invertible linear map and $id_V$ be the identity map on $V$.
$$id^*_V = id_V \implies (T^{-1}T)^*=id_V \implies T^*(T^{-1})^*=id_V \implies (T^{-1})^* = (T^*)^{-1}.$$