Basis of two projectors that act on a two-dimensional subspace

Question

This comes up in the proof of Jordan's lemma: https://cims.nyu.edu/~regev/teaching/quantum_fall_2005/ln/qma.pdf. In particular, I am wondering about the statement on page 4: "Finally, it follows easily from (1) that inside $S$, $\Pi1$ and $\Pi2$ are rank-one projectors."

To summarize the claim, if we have two projectors $\Pi1$ and $\Pi2$ and a two dimensional subspace spanned by vectors $v$ and $v^{\perp}$ such that $(\Pi1 + \Pi2)v = \lambda v$ and $(\Pi1 + \Pi2) v^{\perp} = \lambda_2 v^{\perp}$, $\Pi1$ and $\Pi2$ are rank-1 projectors in this basis.

What is an easy way to see this?

Answer 1

Let $S$ denote $S = \operatorname{span}(\{v,v^\perp\})$. Let $\Pi_1|_S$ denote the restriction of $\Pi_1$ to the (invariant) subspace $S$. Note that for this portion of the proof, we have already assumed that $\Pi_1 v$ is not a multiple of $v$, and can similarly deduce that $\Pi_2v$ is not a multiple of $v$. On the other hand, for each $j = 1,2$:

1. Because $\Pi_j$ is self-adjoint/Hermitian with invariant subspace $S$, $\Pi_j|_S$ is self-adjoint (its matrix relative to the orthonormal basis $\{v,v^\perp\}$ will be Hermitian),
2. $(\Pi_j|_S)^2 = \Pi_j^2|_S = \Pi_j|_S$,
3. $\Pi_j|_S$ is neither equal to the identity (the only rank-2 projector) nor the zero operator (the only rank-0 projector), since $v$ is not an eigenvector of $\Pi_j$.

From observations 1,2, we see that each $\Pi_j|_S$ is a projector. From this, the fact that $\Pi_j|_S$ is an operator over a 2-dimensional space, and observation 3, we can deduce that $\Pi_j|_S$ must have rank $1$.