Is Orthogonal Projection independent of basis for any Basis?

Question

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.

Answer 1

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.

Answer 2

This formula for the orthogonal projection is valid iff the basis is orthogonal. Clearly it is valid if the basis is orthogonal. Assume now the formula is valid, then we get $$ w_i= P_W(w_i) = \sum_{k=1}^n \frac{\langle w_i, w_k \rangle}{\langle w_k, w_k \rangle} w_k = w_i + \sum_{k=1, k\neq i}^n \frac{\langle w_i, w_k \rangle}{\langle w_k, w_k \rangle} w_k $$ But the $w_j$'s are independent, thus we have for any $1\leq k \leq n$ with $k\neq i$ $$ \frac{\langle w_i, w_k \rangle}{\langle w_k, w_k \rangle}=0 $$ and hence $$ \langle w_i, w_k \rangle =0 $$ Therefore, our basis is orthogonal.