Adjoining is Linear

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Theorem

Let $H, K$ be Hilbert spaces over $\Bbb F \in \left\{{\R, \C}\right\}$.

Let $A, B \in B \left({H, K}\right)$ be bounded linear transformations.


Then the operation of adjoining $^*$ satisfies, for all $\lambda \in \Bbb F$:

$(1): \qquad \left({\lambda A}\right)^* = \overline \lambda A^*$
$(2): \qquad \left({A + B}\right)^* = A^* + B^*$


That is, $^*: B \left({H, K}\right) \to B \left({K, H}\right)$ is a linear transformation.


Proof

Proof of $(1)$

Let $\lambda \in \Bbb F$, $h \in H, k \in K$. Then:

\(\displaystyle \) \(\displaystyle \left\langle{\left({\lambda A}\right)h, k}\right\rangle_K\) \(=\) \(\displaystyle \lambda \left\langle{Ah, k}\right\rangle_K\) \(\displaystyle \)          Property $(2)$ of inner products          
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \lambda \left\langle{h, A^*k}\right\rangle_H\) \(\displaystyle \)          Definition of adjoint          
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left\langle{h, \left({\overline \lambda A^*}\right) k}\right\rangle_H\) \(\displaystyle \)          Properties $(1), (2)$ of inner products          


Thus, by Existence and Uniqueness of Adjoint, $\left({\lambda A}\right)^* = \overline \lambda A^*$.

$\Box$


Proof of $(2)$

Let $h \in H, k \in K$. Then:

\(\displaystyle \) \(\displaystyle \left\langle{\left({A + B}\right)h, k}\right\rangle_K\) \(=\) \(\displaystyle \left\langle{Ah, k}\right\rangle_K + \left\langle{Bh, k}\right\rangle_K\) \(\displaystyle \)          Property $(3)$ of inner products          
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left\langle{h, A^*k}\right\rangle_H + \left\langle{h, B^*k}\right\rangle_H\) \(\displaystyle \)          Definition of adjoint          
\(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left\langle{h, \left({A^* + B^*}\right) k}\right\rangle_H\) \(\displaystyle \)          Properties $(1), (3)$ of inner products          


Thus, by Existence and Uniqueness of Adjoint, $\left({A + B}\right)^* = A^* + B^*$.

$\blacksquare$


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