$A\in M_n(R)$ be invertible, ler $x,y\in R^n$ with $x\neq 0, y^TA^{-1}x\neq 0$. Define $B=xy^TA^{-1}$

Question

$A\in M_n(R)$ be invertible, let $x,y\in R^n$ with $x\neq 0, y^TA^{-1}x\neq 0$. Define also $B=xy^TA^{-1}$. Show that:

1. $y^T A^{-1}x$ is eigenvalue of $B$ with multiplicity $1$
2. $0$ is eigenvalue of $B$ of multiplicity $n-1$

I am getting stuck on this one. I am unable to find the vector that I need to multiply to $B$ to show that it is an eigenvalue

Answer 1

Hint: $B$ is rank-1 and all of its columns are in the direction of $x$.

Answer 2

First, notice that $$ \text{rank}(xy^T A^{-1}) \le \min \big(\text{rank}(xy^T), \text{rank}(A^{-1})\big) = 1, $$ since matrix $xy^T$ is rank-$1$ matrix. Hence, matrix $B$ has only $1$ non-zero eigenvalue and the rest are $0$.

To find the only non-zero eigenvalue take the eigenvector $v = x$ yielding $$ xy^T A^{-1} \cdot x = \lambda \cdot x \Leftrightarrow x \cdot (y^T A^{-1} x) = (y^T A^{-1} x) \cdot x. $$ Therefore $\lambda = y^TA^{-1} x$.