Linearity of Adjoint operator I was looking through my quantum mechanics textbook and found the following property of adjoint operators:
$$(\hat A+\hat B)^\dagger = \hat A^\dagger +\hat B^\dagger,$$
where $\hat{A}$ and $\hat{B}$ are linear operators.
How would you prove this?
 A: You can prove it by equality of matrix element:
$$(A^\dagger)_{ij}=\langle i|A^\dagger|j\rangle=\langle Ai|j\rangle=\langle j|A i\rangle^*=\langle j|A|i\rangle^*$$
Now look at the matrix element of $C=A+B$
$$(C^\dagger)_{ij}=\langle i|C^\dagger|j\rangle=\langle Ci|j\rangle=\langle j|C i\rangle^*=\langle j|C|i\rangle^*$$
$$(C^\dagger)_{ij}=\langle j|(A+B)|i\rangle^*=\langle j|A|i\rangle^*+\langle j|B|i\rangle^*$$
$$(C^\dagger)_{ij}=A^*_{ji}+B^*_{ji}=(A^\dagger)_{ij}+(B^\dagger)_{ij}$$
$$(A+B)^\dagger=A^\dagger+B^\dagger$$
QED
A: Below is a proof of the identity without going to any basis. I have dropped all the hats from the operators.
$$\bbox[4px,border:1px solid black]{\textbf{Spoilers Ahead}}$$
Theorem: Consider a inner product space $\left(\mathbb{V}, \langle \cdot, \cdot \rangle\right)$ over a field $\mathcal{F}$. Then,
$\left(A + B \right)^{\dagger} = A^{\dagger} + B^{\dagger}$ for any $A, B \in \mathcal{L}(\mathbb{V},\mathbb{V})$; where
$$\mathcal{L}(\mathbb{V},\mathbb{V}) := \{T \, | \, T \text{ is a linear operator }T:\mathbb{V} \longmapsto \mathbb{V} \}$$
Proof:
By definition, $R^{\dagger}$ is the operator in $\mathcal{L}(\mathbb{V},\mathbb{V})$ such that $\langle x, R^{\dagger} y \rangle := \langle Rx, y\rangle$ for any $x,y \in \mathbb{V}$. Let $A,B \in \mathcal{L}(\mathbb{V},\mathbb{V})$; we have: $\langle x, A^{\dagger} y \rangle := \langle Ax, y \rangle^{\color{red}{1}}$ and $\langle x, B^{\dagger} y \rangle := \langle Bx, y \rangle^{\color{red}{2}}$. For $O := A+B$, one obtains:
$$\langle x, O^{\dagger} y \rangle :=\langle Ox, y\rangle = \langle Ax + Bx, y\rangle = \langle Ax, y \rangle+ \langle Bx,y \rangle \stackrel{(\color{red}{1},\color{red}{2})}{=} \langle x, A^{\dagger} y\rangle + \langle x, B^{\dagger} y \rangle = \langle x, \left(A^{\dagger} + B^{\dagger}\right)y \rangle$$
Denoting $P := A^{\dagger} + B^{\dagger}$, one obtains:
$$\langle x, O^{\dagger} y \rangle = \langle x, Py \rangle \Longrightarrow \langle x, (O^{\dagger} - P)y \rangle = 0_{\mathcal{F}} \ \text{for all } x,y \in \mathbb{V}$$
Setting $x = (O^{\dagger} - P)y \in \mathbb{V}$, we get:
$$\langle x, x \rangle =0_{\mathcal{F}} \Longrightarrow x = 0_{\mathbb{V}} \Longrightarrow (O^{\dagger} - P)y = 0_{\mathbb{V}} \Longrightarrow O^{\dagger} y = Py \text{ for all } y \in \mathbb{V}$$
Consequently, by the very definition of equality of two operators, we have
$$O^{\dagger} y = Py \ \forall y \in \mathbb{V} \Longleftrightarrow O^{\dagger} = P \Longleftrightarrow \bbox[4px,border:1px solid black]{(A+B)^{\dagger} = A^{\dagger} + B^{\dagger}}$$
