If $A \in M_n$ and $rankA=1$ then at least one column of A can be eigenvector of the matrix A?

Question

$RankA=1$, that means that one column of matrix A is linearly independent,and all other columns can be presented through that one column.

I proved that one of the columns can be eigenvector on the following example

$A \in M_n$ \begin{bmatrix} a_1 & a_2 & \cdots & a_n \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 \\ \end{bmatrix}

because A is upper triangular matrix,eigenvalues are the elements on the diagonal, and that's $x_1=a_1$ and $x_2=0$.For $x_2=0$ we can see that first column is eigenvector of A.

But with that one example,I can't prove that is true for every $A\in M_n$.I was wondering how to prove that for every $A$?

Answer 1

If $rank(A)=1$, then wlog, we can write $A$ as $$ A=[a_1,a_2,...,a_n]=[a_1,\beta_2a_1,...,\beta_n a_1]=a_1[1,\beta_2,...\beta_n] $$ where $a_1\neq \bf 0$ and $\beta_2,...,\beta_n$ some real number.

plug it in the equation $$ Ax=\lambda x $$ we get $$ a_1[1,\beta_1,...\beta_n]x=\lambda x $$ let $\lambda=[1,\beta_1,...\beta_n]a_1$ then $a_1=x$.