The Question:
Let $V$ be a finite-dimensional complex inner product space.
Let $W$ be a subspace of $V$ with basis $\{ e_1,\dots,e_k\}$.
Let $T:V \rightarrow W$ be a linear transformation defined by
$$T(v) = \langle v,e_1 \rangle e_1+\langle v,e_2 \rangle e_2+ \cdots + \langle v,e_k \rangle e_k$$
Show that the Image of $T$ is $W$, $\;$ i.e. $\text{Im}(T) = W$.
My Attempt:
Showing one direction of the inclusion is easy:
\begin{align} w \in \text{Im}(T) & \implies \exists \; v \in V \,|\, T(v) = w \\ & \implies w = T(v) = \langle v,e_1 \rangle e_1+\langle v,e_2 \rangle e_2+ \cdots + \langle v,e_k \rangle e_k \\ & \implies w \in \text{span} \{e_1,\dots,e_k\} \\ & \implies w \in W \end{align}
But I don't know how to show the converse.
I suppose that it suffices to show that $\exists \; v_i \in V \,|\, T(v_i)=e_i$ for each $i=1,\dots,n$, but when I tried to construct such $v_i$ it became really messy.
I have also considered Gram-Schimdt but it doesn't seem to help.
Any hints?
Hint: Because $V$ is finite dimensional, it suffices to show that $\ker T|_{W} = \{0\}$.
With that in mind: if $T(v) = 0$, then we can deduce that $\langle v,e_k \rangle = 0$ for $k = 1,\dots,n$. Using the fact that $\{e_1,\dots,e_n\}$ is a basis and the linearity of our inner product, we can show that this implies that $\langle v,v \rangle = 0$, which is to say that $v = 0$.