Prove range, span, and nullity using orthogonal Matrix. I am not sure where to find information on this proof. I'm don't know where to begin. I think it is in the SVD chapter in the book I am learning from, but I cannot find anything similar. Any help would be appreciated. 
Let $u \in \mathbb{R}^n$ and suppose that $\| u \|_2 = 1$. If $P=uu^\top$, prove that  
(i)
$\mathcal{R}(P) = \text{span}\{u\}$.
(ii)
$\mathcal{N}(P) = \mathcal{H}(u) := \{ v \in \mathbb{R}^n \mid \langle u,v \rangle = 0 \}$.
(iii)
$\dim(\mathcal{H}(u)) = n-1$.
 A: Since you mention about SVD, I will use properties about SVD to address this question. 
For a matrix $A$, let its SVD be $$A=U\Sigma V^T,$$
where $U$ and $V$ are orthogonal matrices and $\Sigma$ is a diagonal matrix with nonnegative entries.
The following properties are useful:


*

*The columns of $U$ corresponding to the non-zero singular values of $A$ are an orthonormal set of basis vector for the range of $A$.

*The columns of $V$ which correspond to zero singular values form an orthonomal basis for the null space of $A$.


Now, consider our matrix $P$.
$$P=uu^T=u(1)u^T=\begin{bmatrix}u & U_2\end{bmatrix}\begin{bmatrix}1 & \\ & \textbf{0}\end{bmatrix}\begin{bmatrix}u & U_2\end{bmatrix}^T$$
is a singular value decomposition of $P$.
Using the first result that I just quoted, we immediately see that $\{u\}$ is a basis for the range of $P$.
Using the second result, we immediately see that columns of $U_2$ forms a basis for the null space of $A$, and they satisfy the condition they are orthogonal to $u$.
A: No need to go with SVD.
An element in $v\in\mathcal{R}(P)$ is of the form $v=Px$, for some $x\in\mathbb{R}^n$. Then
$$
v=uu^{\top}x=\langle u,x\rangle u
$$
as it can be readily verified. Therefore $\mathcal{R}(P)\subseteq\operatorname{span}\{u\}$. On the other hand, $u=uu^{\top}u=\langle u,u\rangle u=u$, so $u\in\mathcal{R}(P)$. In particular, the rank of $P$ is $1$.
Next, $v\in\mathcal{N}(P)$ if and only if
$$
0=uu^{\perp}v=\langle u,v\rangle u
$$
and, since $u\ne0$, this happens if and only if $\langle u,v\rangle=0$.
The rank-nullity theorem now yields
$$
\dim\mathcal{H}(u)=\dim\mathcal{N}(P)=n-\operatorname{rank}(P)=n-1
$$
A: No deep formal proof, but some pointers:
For (i), just look at what $uu^{\top}$ means, the matrix is a linear combination of $u$, specifically $P=[u^1 u \dots u^n u]$. Together with $\mathcal{R}(P) = \{Px \; | \; x\in \mathbb{R}^n \}$ which means the set containing all the linear combinations of the columns of $P$, which happens to be a linear combination of $u$.  
For (ii), $\mathcal{N}(P) = \{x\in \mathbb{R}^n \; | \; Px=0\}$ which means we seek $x$ such that $uu^{\top}x=0$. From the problem: $\|u\|_2=1 \implies uu^{\top}=1$ thus $\langle u,x\rangle=0$ defines the Nullspace. 
Finally for (iii), this follows from the whole concept of Range and Nullspace really, but a little more formally, from your definition of $\mathcal{H}(u)$ you see that the space is spanned by the hyperplane defined by $\langle u,v\rangle=0$, hyperplanes are subspaces of codimension $1$, so there you have it: $n-1$. 
