Orthonormal basis for the null space of almost-Householder matrix A matrix $H$ is defined as:
$$H = I - vv^T$$
where $v$ is a unit vector.


*

*What is the rank of $H$?  

*What would be an orthonormal basis for the null space of $H$?

*How do we find the number of zero eigenvalues and their
associated eigenvector?

 A: The characteristic polynomial of $H$ is
$$
p(\lambda)=|\lambda I-H|=(\lambda-1)^n+Tr(vv^T)(\lambda-1)^{n-1}
$$
for $vv^T$ is a Rank-$1$ matrix and all principal minors above $2$ are $0$.
Let $v=\pmatrix{v_1\\\vdots\\v_n}$. Then 
$$
vv^T=\pmatrix{v_1v_1\cdots v_1v_n\\ \vdots \hspace{15 mm} \vdots \\ v_nv_1 \cdots v_nv_n}
$$
So 
$$Tr(vv^T)=\sum\limits_{k=1}^nv_k^2=1$$
And we have
$$
p(\lambda)=\lambda(\lambda-1)^{n-1}
$$
So there are $n-1$ eigenvalue of $1$ and $1$ eigenvalue of $0$. And 
$$\text{Rank}(H)=n-1$$
Since $Hv=v-vv^Tv=v-v=0$, $v$ is the only eigenvector for $0$, and the only orthonormal basis vector for the null space of $H$. 
A: Pick any vector $w \in V$ where $V$ is an inner product space of dimension $n$ ; assume the basis has been chosen so that $v^{\top} w = \langle w,v \rangle$. Then 
$$
Hw = w - v v^{\top} w = w - \langle w,v \rangle v. 
$$
It follows that if $v$ has norm $1$, $H$ is the orthogonal projection operator to the $(n-1)$-dimensional subspace of $V$ orthogonal to $v$ ; call it $W$. Its eigenvalues are therefore $0$ and $1$, where $1$ has (geometric) multiplicity $n-1$ and $0$ has multiplicity $1$. An orthogonal basis for $W$ together with $v$ gives an orthogonal basis of eigenvectors for $H$. In particular, the rank of $H$ is $n-1$. 
Hope that helps,
