1
$\begingroup$

I noticed that for any linear transformation -- with at least one eigenvector with a non-zero eigenvalue -- repeatedly applying the transformation to a non-zero vector approaches an eigenvector. That is to say for matrix $A$ with eigenvectors $v_1,...,v_k$ (kind of hand-wavy limit here) $$\lim_{n\to\infty} A^n \vec{x}=a\vec{v_i}, a\in \mathbb{R}$$ This is trivial when $\vec{x}$ is in the span of the eigenvectors since $$\lim_{n\to\infty} A^n \vec{x}=A^n(a_1\vec{v_1}+...+a_k\vec{v_k})=a_1\lambda_1^n\vec{v_1}+...+a_k\lambda_k^n\vec{v_k}$$ and the limit will get closer to the span of the eigenvector with the largest $\lambda$

What confuses me is that choosing an $\vec{x}$ outside the span of the eigenvectors -- even one perpendicular to the space of eigenvectors -- will also approach one of the eigenvectors. Can a formal reason be given for this?

Two transformations that exhibit this behavior apparently are $$\begin{bmatrix} 1 & -1 \\ 1 & 3 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 & \frac{1}{2} \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \end{bmatrix}$$ with 1-dimensional and 2-dimensional eigenvector space respectively. (maybe errors here)

1 Answers 1

1

As currently written, your statement is incorrect. What you should say instead is that $$ v = \lim_{n \to \infty} \frac{A^nx}{\|A^nx\|} $$ will generally be an eigenvector, assuming that this sequence converges. Note that this sequence will not necessarily converge! For a simple example, take $A = -1$.

That being said: it will work "most of the time". In fact, this is precisely what the power iteration method does.