Assuming the usual inner product $\langle x, y\rangle = \bar x^\mathsf{T} y$ on a complex vector space $V$, and defining the projection $\mathbf P_W\colon V\to W\leqslant V$ as
$$\mathbf P_W(v) = \sum_{k=1}^n \frac{\langle v, w_1\rangle}{\langle w_1,w_1\rangle} w_1,$$
where $\{w_1,\dots,w_n\}$ is an arbitrary basis (not assuming orthogonal nor unit), is it true that $\mathbf P_W(v)$ is always well defined?
And if this is true, how do I prove it?
My attempted strategy was to show that $\mathbf P_W(v)$ as the unique vector $w$ such that $\langle v-w,x\rangle = 0$ for all $x\in W$. Uniqueness would then imply independence of basis.
Indeed, it would suffice if I can prove this for all the basis vectors $w_1,\dots,w_n$, then for all $x$ would follow. So for $1\leqslant j\leqslant n$, we have
\begin{align*}
\langle v – \mathbf P_W( v), w_j\rangle &= \bigg\langle v – \sum_{i=1}^n\frac{\langle v, w_i\rangle}{\| w_i\|^2}\, { w_i}, w_j\bigg\rangle\\
&= \langle v, w_j\rangle – \sum_{i=1}^n \frac{\langle v, w_i\rangle}{\| w_i\|^2}\langle w_i, w_j\rangle\\
&= \langle v, w_j\rangle – \langle v,w_j\rangle – \underbrace{\sum_{i\neq j} \frac{\langle v, w_i\rangle}{\| w_i\|^2}\langle w_i, w_j\rangle}_{=0?},
\end{align*}
and if we insisted that the basis were orthogonal, we would be done. But I don't know if the remaining terms are still zero.
Best Answer
Consider the following counterexample: let $V = \mathbb{R}^3$ and $W = span\{(1, 0,0), (0, 1, 0) \}$. Take $w_1 = \frac{1}{\sqrt{2}} (1, 1, 0)$, $w_2 = (1, 0, 0)$ and $v = (1,1,1)$. Then \begin{equation} v - P_W(v) = (1, 1, 1) - (1, 1, 0) - (1, 0, 0) = (-1, 0, 1) \end{equation} and we see that it is not perpendicular to $W$. The right definition for $P_W$ would be the one that first applies the Gram-Schmidt procedure to the basis for W and then plugs that new, orthonormal basis into your formula.