Orthonormal basis for the null space of almost-Householder matrix

Question

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?

Answer 1

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 $H$.

Answer 2

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,