Find the orthogonal projection of a vector v onto U

Question

I am trying to solve this question:

Let $u$ be a unit vector in $R^n$ and let $U$ be the subspace spanned by $u$. Show that the orthogonal projection of a vector $v$ onto $U$ is given by

$\operatorname{proj}_{U}v = (uu^T)v$,

and thus that the matrix of this projection is $uu^T$. What is the rank of $uu^T$?

Where $u^T$ is the transpose of $u$. Any help is appreciate! I have no idea how to begin this other than knowing

$\operatorname{proj}_{U}v = \frac{(v, u)u}{(u, u)}$

Thanks!

Answer 1

The projection formula you have is the solution. $\left(u,u\right)=1$, since $u$ is a unit vector. So \begin{align*} \text{Proj}_{U}v & =\left(v,u\right)u=\left(u^{T}v\right)u=\left(uu^{T}\right)v. \end{align*}

Answer 2

The general formula in any inner product space, whether $u$ be a unit vector or not is

$$\operatorname{Proj}_U(v)=\frac{\langle\, v,u\,\rangle}{\langle\, u,u\,\rangle}u=\frac{\langle\, v,u\,\rangle}{\lVert u\rVert^2}u$$ Here $\lVert u\rVert=1$, hence the formula comes down to $$\operatorname{Proj}_U(v)=\langle\, v,u\,\rangle u.$$ Now in $\mathbf R^n$ equipped with the Euclidean inner product, $\,\langle\, v,u\,\rangle={}^{\mathrm t\mkern-1mu}v\mkern2mu u$, when $u$ and $v$ are represented by column vectors.