Subspace in Linear Algebra Find the $Proj_{\vec{w}}v$ for the five vector $v$ and subspace $W$. Let $V$ be the Eucledian space $\mathbb{R}^{4}$ and $W$ the subspace with basis $[1,1,0,1]$,$[0,1,1,0]$,$[-1,0,0,1]$
a) $v=[2,1,6,0]$
My attempt was I found out the basis perpendicular to $W$ which is $[1,-2,2,1]$ but I don't know how to proceed any further. Any help is appreciated.
 A: Are you familiar with the Gram-Schmidt process?
If so, then find an orthonormal basis for $W$.  Let us call the vectors $u_1, u_2, u_3$.
You have then that $\operatorname{proj}_W v = \langle u_1, v\rangle u_1 + \langle u_2, v\rangle u_2 + \langle u_3,v\rangle u_3$.  Keep in mind that this specific formula works only for $u_1,u_2,u_3$ that form an orthoNORMAL basis for $W$.
As to the specifics of your problem, let us run the Gram-Schmidt process on the vectors in question.
Let $r_1 = [1,1,0,1]$.
Let $r_2 = [0,1,1,0] - \operatorname{proj}_{r_1} ([0,1,1,0])= [0,1,1,0]-\frac{[1,1,0,1]}{3} = [-\frac{1}{3},\frac{2}{3},1,-\frac{1}{3}]$
Finally, $r_3 = [-1,0,0,1] - \operatorname{proj}_{r_1}([-1,0,0,1])-\operatorname{proj}_{r_2}([-1,0,0,1])=[-1,0,0,1]$
Finish by normalizing and setting $u_1 = \frac{r_1}{\|r_1\|}$, $u_2=\frac{r_2}{\|r_2\|}$, and $u_3=\frac{r_3}{\|r_3\|}$
A: Try projecting your vector $v$ onto the basis vector you found for the orthogonal complement $W^\perp$ of $W$.
$Proj_{[1,−2,2,1]}v = \frac{5}{6}[1,-2,2,1]$. This tells "how much" of $v$ lives in $W^\perp$.
Thus we have the decomposition $v=\frac{5}{6}[1,-2,2,1]+w$ where $w$ lives entirely in $W$. Solving for $w$ should give you what you want.
