Suppose that $u,v \in \mathbb R^n$ with $u,v$ not equal to $\mathbf 0$, and let $A= I + uv^\top$. a) Show that $1+v^\top u$ is an eigenvalue of $A$ and $u$ its eigenvector.
b) Define the subspace $S$ of $\mathbb  R^{n}$ to be $$S=\{x \in \mathbb R^{n}\mid v^\top x=0\}= \operatorname {null}(v^\top).$$ 
Find the dimesion of $S$. 
 A: For the first part, just note that $v^Tu \in \Bbb{R}$. Thus we have
$$
Au = (I + uv^T)\ u = u + (uv^T)u = u + u(v^Tu) = (1 + v^Tu)\ u
$$
which by definition means that $u$ is an eigenvector, corresponding to the eigenvalue $1 + v^Tu$.

For the second part, first extend $v$ to a basis $\{v,u_1,\dotsc,u_{n-1}\}$. Applying the Gram-Schmidt process we can make this into an orthonormal basis $\{v',w_1,\dotsc,w_{n-1}\}$ such that $v = \|v\| v'$. Since clearly
$$
\text{null}(v) = \text{null}(v')
$$
and since $(v')^Tw_i = 0$ for every $i \in \{1,\dotsc,n-1\}$ by construction, it follows that the dimension of $\text{null}(v)$ is $n-1$.
Alternatively let $\{e_1,\dotsc,e_n\}$ be the standard basis of $\Bbb{R}^n$ and observe that up to a change of basis we may assume that $v = e_1 = (1,0,\dotsc,0)$. Clearly $e_2,\dotsc,e_n$ are orthogonal to $e_1$, thus $\text{null}(e_1)$ must have dimension $n-1$.
A: For b) This is one more alternative:
One can think of 
$v^t$ as a linear transformation $\Bbb{R^n}$ to $\Bbb{R}$
$v$ being non zero implies rank of this transformation is$1$ so nullity is $n-1$.
