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?
Best Answer
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 $a\parallel x$.