Let be $V$ and $W$ vector spaces of finit dimension over $\mathbb{C}$ with inner product $\left \langle , \right \rangle_{v}$ and $\left \langle , \right \rangle_{w}$ respectively. Consider $T:V \rightarrow W$ a surjective linear transformation such that exists $T^{*}:W \rightarrow V$ and $\left \langle Tv,w \right \rangle_{w}=\left \langle v,T^{*}w \right \rangle_{v}$, $\forall v\in V$ and $w \in W$
Prove that $T^{*}$ is injective
- My attempt to prove if $T^{*}$ is injective:
Let be $w_1, w_2 \in W$ such that $T^{*}w_1=T^{*}w_2$, that implies that
\begin{align} \left \langle v, T^{*}w_1\right \rangle_{v}&=\left \langle v, T^{*}w_2\right \rangle_{v} \ \ \ \ \ \ \ \ \ \forall v \in V\\ \iff \left \langle Tv, w_1\right \rangle_{w}&=\left \langle Tv, w_2\right \rangle_{w}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ w_1&=w_2\\ \therefore T^{*}& \text{ is injective} \end{align}
Am I correct? I would really appreciate your help!
Update:
Let be $w_1, w_2 \in W$ such that $T^{*}w_1=T^{*}w_2$.
$\forall v \in V$, we have that: \begin{align} 0&=\left \langle v, 0\right \rangle_{v}\\&=\left \langle v, T^{*}w_1-T^{*}w_2\right \rangle_{v}\\&=\left \langle v, T^{*}(w_1-w_2)\right \rangle_{v}\\&=\left \langle Tv, w_1-w_2\right \rangle_{w} \end{align}
We have, $\left \langle Tv, w_1-w_2\right \rangle_{v}=0$, but $T$ is surjective, and that means that $\forall w\in W, \exists v \in V$, such that $w=Tv$.
That implies that there exists $v^{'} \in V$ such that $Tv^{'}=(w_1-w_2)\in W$. And
\begin{align} \left \langle Tv^{'}, w_1-w_2\right \rangle_{w}&=\left \langle w_1-w_2, w_1-w_2\right \rangle_{w}=0\\ \iff w_1-w_2&=0 \iff w_1 = w_2 \end{align} Would it be okay like this?
Why do folks insist on not taking advantage of the fact that injectivity of linear transformations can be checked at $\mathbf{0}$?
Say $T^*\mathbf{w}=\mathbf{0}$. We just need to show that this implies that $\mathbf{w}=\mathbf{0}$.
Let $\mathbf{v}$ be such that $T\mathbf{v}=\mathbf{w}$ (using surjectivity of $T$). Then $$0 = \langle \mathbf{v},\mathbf{0}\rangle = \langle \mathbf{v},T^*\mathbf{w}\rangle = \langle T\mathbf{v},\mathbf{w}\rangle = \langle\mathbf{w},\mathbf{w}\rangle.$$ Therefore, $\mathbf{w}=\mathbf{0}$, so $T^*$ is injective.