Suppose $P$ is a $n \times n$ projection matrix with two eigenvalues equal to one. Is it true that and how can I show that $$ (I-\nu P)^m \to (I-P), \quad for \ m \to \infty, $$ with $0 < \nu < 1$.
My attempt: Let $P=ODO^T$ be an eigendecomposition with $OO^T=I$ and $D=diag(d_1, \dots, d_n)$ where $d_1=d_2=1$ and $d_k=0 \ (k=3, \dots, n)$. Then $$ (I-\nu P)^m = O(I-\nu D)^mO^T=Odiag((1-\nu d_k)^m)O^T $$ For $k=1, 2$, $(1-\nu d_k)^m \to 0$ and for $k=3, \dots, n$, $(1-\nu d_k)^m = 1$. Since those are the eigenvalues of the orthogonal complement $I-P$, the claim follows.
Is my reasoning correct?
Using the fact that projection matrices are idempotent (i.e., $P^2=P$) and the binomial theorem, $$ \left(I-vP\right)^{m}=\sum_{k=0}^{m}\binom{m}{k}\left(-vP\right)^{k}=I+P\sum_{k=1}^{m}\binom{m}{k}\left(-v\right)^{k}=I-P+P\left(1-v\right)^{m}\rightarrow I-P. $$