# Why do eigenspaces attract a sequence?

I was reading about eigen-stuff and I came across a very interesting visualization.

We know that, a matrix when multiplied with a vector in a $$2D$$ space simply maps that vector to someplace else, while that 'someplace else' is guaranteed to be the vector's span if the vector is an eigenvector of the matrix. Or conversely, any vector which stays on it's span after transformation is an eigenvector. This is true for all the vectors along that particular direction. So collectively, we can call it an eigenspace. If we apply the same linear transformation to any vector in the eigenspace, it just scales (positively or negatively) that vector without changing it's direction. The matrix multiplication can be done as many times as possible and all the resulting vectors will line up in that same direction. Now it depends on the eigenvalue of course whether the vectors will spread apart on the line or densely populate at $$0$$. This is a very non-technical view obviously.

Now what interested me is the fact that if we keep applying the matrix transformation to any vector that is not an eigen vector, the resulting vectors slowly start to polarize themselves towards the span line until they end up following the direction of the eigen vector of the matrix with every multiplication. These 'non-eigenvector' multiplications are generally kept from being repeated for their complexity, but if one does it then s/he will find that this is indeed the case!

This is where I found the idea I'm talking about https://setosa.io/ev/eigenvectors-and-eigenvalues/ There is also a mind-blowing fibonacci sequence example following the above observation..

My question is simple - why? Why does the eigenspace attract the sequence of multiplying the transformation matrix with any non-eigen vector? I don't see the intuition...

First this is not necessarily true, for example recursively applying $$A=\begin{pmatrix} 1 & \\ & -1\end{pmatrix}$$ to $$\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ will not bring it closer to any of the two eigenspaces.
This is true under the assumption that $$A$$ can be diagonalized and has a unique eigenvalue with maximal absolute value. Then vector $$v=\sum_i e_i$$ where $$e_i$$ is an eigenvector with eigenvalue $$s_i$$, and $$A^n v= \sum s_i^n e_i$$. Note that if $$|e_1|>|e_2|\ge |e_3|\ge\cdots$$, then the compoent $$s_1^n$$ grows exponentially faster than $$s_i^n$$ for $$i\ge 2$$, so eventually and pretty quickly $$s_1^ne_1$$ will become the dominant component of the vector $$v$$.
As "Just a user" pointed out, this is not always the case; consider the matrix that rotates a vector in $$\mathbb{R}^2$$ by an angle $$\theta$$, or one that reflects over an axis, etc. However, consider a matrix the describes how the populations of two cities change over the course of a year, assuming that the total population doesn't change; people only move from one city to the other. Naturally, if the cities begin in the state where many are clustered in one city, than many will leave to the emptier city, but still a few will choose to move to the clustered; the rates at which the populations switch are probabilistic. Systems like these are called Markov Chains. So, one can guess that an e.vector of a Markov matrix will represent an equilibrium state (e.g. 50% and 50%), because if we begin in this state, we expect to end in this state as well (note that one can change the rules of the transformation to allow for multiple equilibria). If we begin in a state that is not an equilibium state, then we expect every iteration of the matrix to bring it closer to equilibrium.