2

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?

user483161
  • 121
  • 1
    Seems reasonable to me. As you can tell, the claim is independent of how many of the eigenvalues are one and how many are zero (see also my answer). – parsiad Mar 11 '19 at 14:59
  • For future reference, you can use \operatorname{diag}(x) for $\operatorname{diag}(x)$. – parsiad Mar 13 '19 at 06:02

1 Answers1

3

Using the fact that projection matrices are idempotent (i.e., $P^2=P$) and the binomial theorem, $$ \left(I-vP\right)^{m}=\sum_{k=0}^{m}\binom{m}{k}\left(-vP\right)^{k}=I+P\sum_{k=1}^{m}\binom{m}{k}\left(-v\right)^{k}=I-P+P\left(1-v\right)^{m}\rightarrow I-P. $$

parsiad
  • 25,154
  • One follow-up, does this also imply: $||(I-\nu P)^m f||_2^2 \to ||(I-P)f||_2^2$ where $||.||_2^2$ is the Euclidean norm and $f$ is some real-valued vector? – user483161 Mar 12 '19 at 11:42
  • 1
    If $g$ is a continuous function and $x_n \rightarrow x$, then $g(x_n) \rightarrow x$ (this fact is also known as limits and continuous functions commute). In your case, $g(x) = \Vert x f \Vert_2^2$ and $x_n = (I - v P)^n$. In short, yes, your claim is true. – parsiad Mar 13 '19 at 01:22
  • For future reference, you can use \Vert for $\Vert$ and \cdot for the $\cdot$ symbols. – parsiad Mar 13 '19 at 06:01