So, I have the information that $z\in \mathbb{R}^n$ is said to be orthogonal to the subspace $W \subset \mathbb{R}^n$. The orthogonal projections of a vector $v \in \mathbb{R}^n$ onto the subspace W is the vector $\tilde{w} \in W$ that makes the difference $z = v - \tilde{w}$ orthogonal to W. Now if $\textbf{u}$ forms a basis for the subpace $\textbf{W}$, show that the orthogonal projection of vector $v \in \mathbb{R}^n$ on the subspace W is $\tilde{w} = \frac{v^Tu}{||u||_2^2}u$. I have been lost on this proof for about an hour and have no idea what to do, if I coould get some help that would be great. Thanks.
I know that $z^T \cdot W = 0$, but other than that I have been out of luck trying different things
Let $u_1,\ldots,u_n$ be the basis of $W$ and let $U=[u_1,\ldots,u_n]$. We want $z=v-w$ such that $w\in W$ and $z\perp W$. Since $w\in W$, there is a $\hat{w}$ such that $w=W\hat{w}$. The condition $z\perp W$ is equivalent to $U^Tz=0$. Putting these two together gives $$ 0=U^Tz=U^T(v-U\hat{w})\quad\Rightarrow\quad w=U\hat{w}=U(U^TU)^{-1}U^Tv. $$
If $n=1$ so that $W=\mathrm{span}\{u\}$, this gives $$ w=u(u^Tu)^{-1}u^Tv=\frac{u^Tv}{\|u\|_2^2}u. $$