If $V$ is an $n$-dimensional vector space and $U$ is a subspace and $u_1,…,u_k$ is an orthonormal basis for $U$ and $v$ is a vector in $V$ then the orthogonal projection of $v$ in $U$ is given by
$$ \sum \langle v,u_i \rangle u_i$$
But why does $u_i$ have to be orthonormal? Why does it not hold that the projection is $ \sum \langle v,u_i \rangle u_i$ if $u_i$ is any basis? And what is $ \sum {\langle v,u_i \rangle \over \|u_i\|^2} u_i$? Is it an orthognal projection?
Best Answer
For a projection $A$ onto $U$, there has to be $uA=u$ for any $u\in U$. So try $u_j$ in your formula: $$\sum_{i=1}^k\frac{\langle u_j,u_i\rangle u_i}{\langle u_i,u_i\rangle}=u_j$$ Because $\{u_i\}$ is a basis, we have $\langle u_j,u_i\rangle=0$ when $i\neq j$. That shows $\{u_i\}$ should be orthogonal. And then $\{\frac{u_i}{\|u_i\|}\}$ is in fact orthonormal.