Adjoint of Sum = Sum of Adjoints?

Question

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?

Answer 1

$\langle (A + B)^\dagger x, y \rangle = \langle x, (A + B)y \rangle = \langle x, Ay + By \rangle$ $= \langle x, Ay \rangle + \langle x, By \rangle = \langle A^\dagger x, y \rangle + \langle B^\dagger x, y \rangle; \tag 1$

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

$(A + B)^\dagger = A^\dagger + B^\dagger . \tag 5$

Answer 2

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)