How can verify the result $v^Tx=0, x$ is eigenvector for A?

Question

$A=UV^T $ ,

$U = [u_1, u_2,\dots ,u_n]^T$ and $V= [v_1,v_2\dots,v_n]^T$

Let $V^TX=0 , X $ is eigenvector for A .

(1) how can I verify it and solve the eigenvalue? and

(2) if $\det(I+A)=\operatorname{Tr}(A)+1$, find the other eigenvalue, different from (1)

I cannot get it ...

Answer 1

$$A\cdot X=(U\cdot V^T)\cdot X=U\cdot(V^T\cdot X)=U\cdot 0=0 = 0X$$ Do you see what the eigenvalue is now?