Proving the image of inner product map is whole subspace

Question

I'm doing a specimen exam question and they often have typos and missed pieces of necessary information. I think the question I'm doing might be one such example, but am not sure:

We're given that $W$ is a subspace of a finite dimensional complex inner-product space $V$.

We're asked to prove that if $\left \{e_{1}, e_{2},..., e_{k} \right \}$ is a basis for $W$ and $\phi: V \rightarrow W$ is a linear transformation given by

$$\phi(v)=<v,e_{1}>e_{1} + <v,e_{2}>e_{2} + ... + <v,e_{k}>e_{k}$$

where $v \in V$. Then $im(\phi)=W$ and $ker(\phi)=W^{\perp}$.

I can't seem to do this without assuming that the basis is orthonormal. Do we need his be true?

Answer 1

Clearly $im(\phi) \subseteq W$. We now extend $\left \{ e_{1},...,e_{k} \right \}$ to a basis $\left \{ e_{1},...,e_{n} \right \}$ for $V$ with $k<n. \\$ Consider $v_{j}=\sum_{i=1}^{n}a_{i}e_{i}$ such that $<v_{j},e_{i}>=\delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta symbol.

This gives us $k$ equations and we have $n$ unknowns $(a_{1},...,a_{n})$ so there exists a nonzero solution $v_{j}$ for each $e_{j} \in \left \{ e_{1},...,e_{k} \right \}$.

If we now set $v'_{j}=\frac{1}{<v_{j},e_{j}}v_{j}$ then $\phi(v'_{j})=e_{j}$.

Therefore, $e_{j} \in im(\phi)$ for $j=1,2,...,k$.

So, any $(W \ni) \hspace{3mm} w=\sum_{i=1}^{k}b_{i}e_{i} \in im(\phi)$.

Which gives $im(\phi)=W$ as required.

Answer 2

Clearly, the image of $\phi$ is a subset of $W$ and $W^\perp$ is a subset of $\mathrm{Ker}\,\phi$. It remains to prove equality. If a vector $w\notin W^\perp$, then $\langle w,e_i\rangle\neq 0$ for some $e_i$. The independence of $\{e_i\}$ shows that $w$ is not in the kernel. So, $\mathrm{Ker}\, \phi=W^\perp$ and the dimension argument $$\dim \mathrm{Ker}\,\phi+\dim \mathrm{Im}\,\phi =\dim\,W+\dim\,W^\perp$$ together with $\mathrm{Im}(\phi)\subseteq W$ implies that $\mathrm{Im}(\phi)=W$.