Prove that $A^N v_2 = \lambda^N v_2 + N \lambda^{N-1} v_1$ for any natural number $N$

Question

$A$ is an $n \times n$ matrix with an eigenvalue $\lambda$ and a corresponding eigenvector, $v_1$.

$v_2$ is a generalized eigenvector such that $(A- \lambda I)v_2=v_1$

I'm trying to prove $A^N v_2 = \lambda^N v_2 + N \lambda^{N-1} v_1$ for any natural number $N$ by induction.

I easily proved the base case of $N=1$,

$A^{(1)}v_2=\lambda^{(1)} v_2+(1)\lambda^{0}v_1$

$A v_2=\lambda v_2+v_1$

$(A-\lambda I)v_2=v_1$

But I can't for the life of me prove $A^{k+1} v_2 = \lambda^{k+1} v_2 + (k+1) \lambda^{k} v_1$

assuming $A^k v_2 = \lambda^k v_2 + k \lambda^{k-1} v_1$ is true.

I'm fairly new to induction based proofs so I could be missing something right in front of me, but any push in the right direction would be appreciated.

Answer 1

Pre-multiplying by $A$ on both sides of the inductive hypothesis, $A^kv_2=\lambda^kv_2+k\lambda^{k-1}v_1$, we get:

$$A^{k+1}v_2=\lambda^kAv_2+k\lambda^{k-1}Av_1\stackrel{(\star)}{=}\lambda^{k}(\lambda v_2+v_1)+k\lambda^{k-1}\cdot\lambda v_1=\lambda^{k+1}v_2+(k+1)\lambda^k v_1$$

where $(\star)$ follows from the fact that: $Av_2=\lambda v_2+v_1$ and $Av_1=\lambda v_1$.