Eigenvector of a square matrix whose rank is 1.

Question

Is this statement true or not?

Let $A$ be a $n\times n$ matrix such that $\mathrm{rank}(A)=1$. If $x$ is an element of $\mathrm{Col}(A)$ with $x\ne0$ then $x$ is an eigenvector of $A$.

I think this statement is true because vector $x$ is an element of $\mathrm{Col}(A)$ so the product of $A$ and vector $x$ should be a multiple of vector $x$.

Answer 1

You are right; here's a more detailed analysis.

There are two cases: $Ax=0$ or $Ax\ne0$.

If $Ax=0$, then $x$ is an eigenvector relative to $0$.

If $Ax\ne0$, then $Ax$ belongs to the column space of $A$, which is generated by $x$, as $A$ has rank $1$ and $x$, by assumption, belongs to the column space. Therefore $Ax=\lambda x$, for some $\lambda$.

Answer 2

Yes I agree with your considerations, indeed since $A\vec x$ is a linear combinations of the columns of $A$ we have that

$$A\vec x=\lambda \vec x$$

where possibly $\lambda=0$.

Answer 3

It is true that if $\mathrm{Col}(A)$ is mapped into itself since the whole space is mapped there, so any nonzero $x \in \mathrm{Col}(A)$ is an eigenvector. However you should be careful because it can happen that the corresponding $\lambda$ is still zero. Indeed this happens if $A=uv^T$ and $u$ is perpendicular to $v$.