Adjoint of Sum = Sum of Adjoints? Didnt find this anywhere, just verifying. 
We know that: $$ (A+B)^T= A^T+B^T $$.
Does it follow that:$$ (A+B)^† = A^† +B
^† $$ for all A and B matrices (the dagger here representing the adjoint)?
If so what is the proof?
 A: $\langle (A + B)^\dagger x, y \rangle = \langle x,  (A + B)y \rangle = \langle x, Ay + By \rangle$
$= \langle x, Ay \rangle + \langle x, By \rangle = \langle A^\dagger x, y \rangle + \langle B^\dagger x, y \rangle; \tag 1$
from this we have
$\langle (A + B)^\dagger x - A^\dagger x - B^\dagger x, y \rangle = 0; \tag 2$
since this binds for all $y$, we have
$(A + B)^\dagger x - A^\dagger x - B^\dagger x = 0, 
 \tag 3$
or
$(A + B)^\dagger x = A^\dagger x + B^\dagger x, 
 \tag 4$
holding for all $x$; we thus conclude that
$(A + B)^\dagger = A^\dagger + B^\dagger . 
 \tag 5$
A: This statement is not true in general.
Wikipedia tells us that for 3 by 3 matrices the first element of the adjoint obtains as,
$$
a_{1,1}^\dagger = a_{2,2}a_{3,3} - a_{2,3}a_{3,2} \quad \text{ and } \quad b_{1,1}^\dagger = b_{2,2}b_{3,3} - b_{2,3}b_{3,2},
$$
but
$$
\begin{align}
(a_{1,1} + b_{1,1})^\dagger &= (a_{2,2} + b_{2,2})(a_{3,3} + b_{3,3}) - (a_{2,3} + b_{2,3})(a_{3,2} + b_{3,2}) \\ &= a_{2,2}a_{3,3} + b_{2,2}b_{3,3} - a_{2,3}a_{3,2} - b_{2,3}b_{3,2} + a_{2,2}b_{3,3} + b_{2,2}a_{3,3} - a_{2,3}b_{3,2} - b_{2,3}a_{3,2} \\ &=
a_{1,1}^\dagger + b_{1,1}^\dagger + a_{2,2}b_{3,3} + b_{2,2}a_{3,3} - a_{2,3}b_{3,2} - b_{2,3}a_{3,2} \\ &\neq a_{1,1}^\dagger + b_{1,1}^\dagger 
\end{align}
$$
You may also check numerically with Matlab:
A = randn(3);
B = randn(3);
adjoint(A+B)
adjoint(A)+adjoint(B)

