I am reviewing some properties of the Hermitian form and dual space using Lang's Linear Algebra book.
In one of the remarks, he mentioned that $w \mapsto L_w$ is not an isomorphism of V with the dual space if $L_w$ is given by $L_w: V \rightarrow \mathbb{C}$ such that $L_w(v) = \langle Av, w\rangle$ where $A: V \rightarrow V$ is an operator and $\langle \;,\;\rangle$ is a positive definite Hermitian form.
I understand this part as $w \mapsto T_w$ is an isomorphism from $V$ to $V^*$ if $T_w(v)$ is any non-degenerate scalar product $\langle v, w\rangle$.
However I am not able to understand Lang's reason. He gave an example stating that if $\alpha \in \mathbb{C}$, then $L_{\alpha w} = \bar{\alpha}L_w$, and it is immaterial for the existence of an element $w''$ such that $L_{\alpha w}(v) = \langle v, w''\rangle$ for all $v \in V$.
My question: If there exists a unique element $w'$ such that $L_w(v) = \langle v, w'\rangle$, then $L_{\alpha w} = \bar{\alpha}L_w = \bar{\alpha}\langle v, w'\rangle = \langle v, \alpha w'\rangle$. Why can't we just choose $w'' = \alpha w'$? What is the problem here?
To provide some context, here’s a screenshot from the textbook:
