I am a little stuck with how to proceed with this question.
Let W be a subspace of a finite dimensional inner product space V. Prove that $proj_w(v)$ is independent of basis.
I think that I have to show that if I have two orthogonal basis such as $\{v_1,….,v_k\}$ and $\{w_1,….w_k\}$ then I need to show that $\frac{<v,v_1>}{|v_1|^2}v_1+….+\frac{<v,v_k>}{|v_k|^2}v_k=\frac{<v,w_1>}{|w_1|^2}w_1+….+\frac{<v,w_k>}{|w_k|^2}w_k$
However I am not really sure how to start show this or even if I am on the correct path to begin with
Best Answer
A general projection onto a subspace $W$ is not uniquely defined. If you complete a basis of $W$ to a basis of $X$, then you end up with $X=W\oplus V$ where $V$ is the span of the added basis elements. The projection onto $W$ along V is the typical language.
Orthogonal projection is unique for a given inner product, but there are many inner products you can put on a space. Given an inner product, the orthogonal projection of $v$ onto a subspace $W$ is the unique $w \in W$ such that $$ (v-w) \perp W. $$ This is the same orthogonal projection you learned in Calculus. And it's the same as the closest-point projection of $v$ onto $w$. You can see that $w$ is unique because \begin{align} &(v-w_1)\perp W,\;\; (v-w_2)\perp W \\ &\implies ((v-w_2)-(v-w_1))\perp W \\ &\implies (w_2-w_1) \perp W \\ &\implies (w_2-w_1)\perp (w_2-w_1)\;\;\ \mbox{(because $w_2-w_1\in W$)}\\ &\implies w_2-w_1 = 0. \;\;\; \mbox{(because $\|w_2-w_1\|^2=(w_2-w_1,w_2-w_1)=0$)} \end{align} If you have an orthonormal basis $\{ e_1,e_2,\cdots,e_n \}$ of $W$, then the unique orthogonal projection is $$ P_{W}x = (x,e_1)e_1 + (x,e_2)e_2 + \cdots + (x,e_n)e_n $$ because $$ (x-\sum_{j=1}^{n}(x,e_j)e_j) \perp \{ e_1,e_2,\cdots,e_n \}. $$