Show that the adjoint of $A\in \operatorname{Hom}(U,V)$ is linear

Question

With $U$ and $V$ inner product spaces and $A\in \operatorname{Hom}(U,V)$ a linear map, I want to show that the adjoint of $A$ is linear.

My attempt:

* With $x\in U$ and $y_1,y_2\in V$ and $\alpha,\beta \in \mathcal{F}$: $$\begin{array}{l} \left\langle x, A^{*}\left(\alpha y_{1}+\beta y_{2}\right)\right\rangle=\left\langle A x, \alpha y_{1}+\beta y_{2}\right\rangle=\overline{\left\langle\alpha y_{1}+\beta y_{2}, A x\right\rangle} \\ =\overline{\alpha}\overline{\left\langle y_{1}, A x\right\rangle}+\bar{\beta}\overline{\left\langle y_{2}, A x\right\rangle}=\bar{\alpha}\left\langle A x, y_{1}\right\rangle+\bar{\beta}\left\langle A x, y_{2}\right\rangle \end{array}$$ and $$\begin{array}{l} \left\langle x, A^{*}\left(\alpha y_{1}\right)+A^{*}\left(\beta y_{2}\right)\right\rangle=\overline{\left\langle\alpha y_{1}+\beta y_{2}, A x\right\rangle}=\bar{\alpha}\overline{\left\langle y_{1}, A x\right\rangle}+\bar{\beta}\overline{\left\langle\bar{y}_{2}, A x\right\rangle} \\ =\bar{\alpha}\left\langle A x, y_{1}\right\rangle+\bar{\beta}\left\langle A x, y_{2}\right\rangle \end{array}$$ Since we get the same result $A^*$ is linear.

This should prove it if I understood the properties of inner products correctly. Can someone confirm?

Answer 1

Your proof is correct so far, but it is incomplete. You have shown that for all $x \in U,y_1,y_2 \in V, \alpha_1,\alpha_2 \in \Bbb F,$ it holds that $$ \langle x, A^*(\alpha_1 y_1 + \alpha_2 y_2) \rangle = \langle x, \alpha_1A^*(y_1) + \alpha_2 A^*(y_2)\rangle. $$ You must now argue that because of this, we can conclude that for all $y_1,y_2 \in V, \alpha_1,\alpha_2 \in \Bbb F,$ it holds that $$ A^*(\alpha_1 y_1 + \alpha_2 y_2) = \alpha_1A^*(y_1) + \alpha_2 A^*(y_2), $$ which is to say that $A^*$ is linear.

Answer 2

We show that $A^{*}\colon V\to U$ is linear. Let $\alpha$ be a scalar and let $x$ and $y$ be vectors in $V$. By definition of $A^{*}$ and linearity of the innerproduct in the first argument we have

\begin{align*}\langle A^{*}(\alpha x+y),z\rangle&=\langle \alpha x+y,A(z)\rangle=\alpha\langle x,A(z)\rangle+\langle y,A(z)\rangle\\ &=\alpha\langle A^{*}(x),z\rangle+\langle A^{*}(y),z\rangle=\langle\alpha A^{*}(x)+A^{*}(y),z\rangle. \end{align*} for all $z\in U$. To conclude that $A^{*}(\alpha x+y)=\alpha A^{*}(x)+A^{*}(y)$, we need the following result:

If $\langle u_{1},z\rangle=\langle u_{2},z\rangle$ for all $z\in U$, then $u_{1}=u_{2}$.

To prove this result, consider $z:=u_{1}-u_{2}$, and use the fact that $\|z\|^{2}=\langle z,z\rangle=0$ implies $z=0$.