How can we apply this simple eigenvector expression 'repeatedly'? Let $A,B$ be linear operators on a complex vector space $V$ and suppose
$$ABu = (\alpha + 2)Bu$$
where $u \in V$ is an eigenvector of $A$ with eigenvalue $\alpha$ and $\alpha \in \mathbb{C}$. We can interpret this as either $Bu = 0$ or $Bu$ is an eigenvector of $A$ with eigenvalue $\alpha+2$. I'm reading a proof which proves this equation above about the operators $A,B$ as a lemma and then claims that 'by applying the lemma repeatedly' we get the formula
$$AB^ku = (\alpha+2k)B^ku$$
I have tried many things but cannot see how to apply it repeatedly. The lemma seems to a statement about $ABu$, sticking many more $B$'s seems like it would prevent us from using it.
The context: $\pi$ is a representation of the Lie algebra $sl(2,\mathbb{C})$ acting on $V$. Above I was letting $A = \pi(H)$ and $B = \pi(X)$ where
$$H = \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} \quad \quad X = \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}$$
Calculation shows $[X,H] = HX - XH = 2X$, and because $\pi$ is a Lie algebra homomorphism, we have $\pi([H,X]) = [\pi(H),\pi(X)] = 2\pi(X)$. Playing with this equation is what proves the lemma above. Maybe playing with this repeatedly is what is needed.
 A: Assume that $ABu=(\alpha+2)Bu$
$[\pi(H),\pi(X)] = 2\pi(X) \Rightarrow AB-BA=2B$
$AB-BA=2B \Rightarrow AB^2-BAB=2B^2$
$AB^2-BAB=2B^2 \Rightarrow AB^2u-BABu=-(\alpha+2)B^2u+AB^2u=2B^2u$
$-(\alpha+2)B^2u+AB^2u=2B^2u \Rightarrow AB^2u=(\alpha+4) B^2u$
$BA-AB=2B \Rightarrow AB^3-BAB^2=2B^3$
$AB^3-BAB^2=2B^3 \Rightarrow AB^3u-BAB^2u=AB^3u-(\alpha+4)B^3u=2B^3u$
$AB^3u-(\alpha+4)B^3u=2B^3u \Rightarrow AB^3=(\alpha+6)B^3u$ if we repeatedly apply the lemma we will get
$AB^ku=(\alpha +2k)B^ku$ let's prove with induction. Assume $AB^ku=(\alpha +2k)B^ku$
$AB-BA=2B \Rightarrow AB^{k+1}-BAB^{k}=2B^{k+1}$
$AB^{k+1}-BAB^{k}=2B^{k+1} \Rightarrow AB^{k+1}u-BAB^{k}u=AB^{k+1}u-(\alpha+2k)B^{k+1}u=2B^{k+1}u$
$AB^{k+1}u-(\alpha+2k)B^{k+1}u=2B^{k+1}u \Rightarrow AB^{k+1}u=(\alpha+2(k+1))B^{k+1}u$
A: First, thank you for your hard work Nevzat!
Here is the simple answer. Interpret the lemma as
$$u \text{ is an eigenvector of } A \text{ eigenvalue } \alpha \Longrightarrow Bu \text{ is an eigenvector of } A \text{ eigenvalue } \alpha + 2$$
But then applying this implication with the eigenvector $Bu$ gives $B^2u$ an eigenvector of $A$, eigenvalue $(\alpha +2) + 2 = \alpha + 4$. This is the repeated application, and gives the formula.
