Eigenvalues of a matrix that is a product of a vector and transpose vector

Question

Find eigenvalues, eigenvectors and rank of matrix $A$.

$$\textbf{a}=\begin{bmatrix}a_1 \\ a_2 \\ a_3 \\ \vdots \\ a_n \end{bmatrix}, \quad \textbf{b} \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ \vdots \\ b_n \end{bmatrix}$$

$$A = \textbf{a} \cdot \textbf{b}^T$$

I tried finding the rank through $\dim V=\operatorname{rank} A+\dim(\ker A)$ and

$v \in \ker A $

$Av=0$

but haven't produced result.

I was also thinking about $a(b^Tx)=\lambda x$

Any ideas?

Answer 1

Hint: (Rank part) Consider the formula about rank of product of matrices $$\text{rank}(AB)\leq\min\{\text{rank}(A),\text{rank}(B)\}$$ (Eigenvalue part) Note that $(ab^T)x=a(b^Tx)=(b^Tx)a=\lambda x$ implies $a\parallel x$.

Answer 2

The set of $0$-eigenvectors is the null-space of the $1\times n$ matrix $b^T$