Eigenvalue of matrix with one dimensional column space Let $A=vw^\top$ with $v=\begin{bmatrix}1\\2\\-2\\-1\end{bmatrix}$ and $w=\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$. 
Prove that $A$ has no eigenvalue not equal to zero. (Hint: $v\perp w$).
In a previous question I proved that the column space of $A$ is one-dimensional.
 A: Suppose that $\lambda x=vw^Tx$ with $\lambda\ne0$. If you multiply by $w^T$ on the left, you have 
$$
\lambda w^Tx=w^Tvw^T=0. 
$$
So $w^Tx=0$. But then 
$$
\lambda x = vw^Tx=v0=0,
$$
so $x=0$. 
A: For $\lambda$ a scalar, let $[\lambda]$ denote the $1\times 1$ matrix whose sole entry is $\lambda$. Note that for any column vectors $a,b$, we have that $a^\top b=[a\cdot b]$ and $a[\lambda]=\lambda a$.
The matrix at hand has the form $A=vw^\top$. For any $u$, we have that $$Au=(vw^\top)u=v(w^\top u)=v[w\cdot u]=(w\cdot u)v.\tag1$$
This means that there are only two ways a nonzero vector $u$ can be an eigenvector of $A$: Either


*

*$u$ is $v$ or a multiple of $v$. And indeed, $v$ is an eigenvector of $A$ with eigenvalue $w\cdot v$, by (1).

*Or $u$ is not a multiple of $v$ and yet $(w\cdot u)v$ is a multiple of $u$. This is only possible if $w\cdot u=0$. And indeed, any nonzero vector $u$ orthogonal to $w$ is an eigenvector of $A$ with eigenvalue $0$.


Now, it just so happens in the situation at hand that $v$ itself is orthogonal to $w$, so that $w\cdot v=0$ as well. This means that the only eigenvalue of $A$ is $0$ and the eigenvectors of $A$ are precisely the nonzero vectors orthogonal to $w$. 
