x,y are nx1 vectors of real numbers. Matrix A is a product of x and y'.
$$A=xy^T$$
Prove that for any x and y there exists $\lambda \in \Bbb R$ that meets equation $ A^{k}=\lambda^{k-1}A$, $k \in \Bbb N$.
My idea was to solve it through math induction.
For k=1 we have that $A=\lambda^0A$, which definitely holds.
The if we suppose that $A^n=\lambda^{n-1}A$ holds, then $$AA^n=A\lambda^{n-1}A$$
How do we prove that $A\lambda^{n-1}=\lambda^{n}$?
You can proceed as follows: $$A^2=(xy^t)(xy^t)=x\underbrace{(y^tx)}_{\text{scalar, call it }c}y^t=cxy^t=cA.$$ Now you can get $$A^3=cA^2=c(cA)=c^2A,$$ and so on.