# Long-term behaviour of a linear transformation (Is the domain eventually mapped onto the dominant eigenspace?)

As far as its coordinate representation is concerned, the domain of a linear transformation will eventually (i.e. after infinitely many iterations of the transformation) be mapped onto the dominant eigenspace.

True or False?

p.s. Revised hypothesis: Let T be a linear transformation that has a square matrix representation with no complex eigenvalues. A nonzero vector transformed iteratively by T tends towards the dominant eigenspace (oscillatingly if the dominant eigenvalue has a negative sign). In coordinate space, T will, after infinitely many iterations, evetually map the original domain onto the dominant eigenspace.

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Try not to give orders please. Also, where did this statement come from (is it homework) and what have you tried already? Maybe it helps too if you formulate the question more precisely with a bit of mathematical notation than just words. What do you mean by coordinate representation? – Dario Aug 4 '12 at 18:44
So, what do you think of Dario's answer? – Gerry Myerson Aug 5 '12 at 5:39
It's not clear what "after infinitely many iterations" means. Say your linear transformation is $T(x,y)=(2x,3y)$. Where are you after infinitely many iterations? – Gerry Myerson Aug 5 '12 at 9:14
You have to replace all your woolly statements about "infinite iterations" and "eventually" with precise statements about limits. I'll make a precise statement in an answer. – Gerry Myerson Aug 5 '12 at 11:03
@GerryMyerson Your given transformation T has a standard matrix with dominant eigenvalue 3 and corresponding eigenvector $(0,1)$. Thus, after n itertaions, an arbitrary vector $(a,b)$ is transformed to $(2^na,3^nb)$. So after infinitely many iterations, it will become parallel to $(0,1)$, the dominant eigenvector. – Ryan Aug 6 '12 at 15:48

Call the linear transformation $T$, and the vector space on which it acts, $V$. Assume $V$ is finite-dimensional.

In the special case where the $V$ has a basis $B$ consisting of eigenvectors of $T$, and there is a dominant eigenvalue $\lambda$, the following is true: if $v$ is any element of $V$, and if the expression for $v$ with respect to the basis $B$ has a nonzero component in the eigenspace $W$ of $\lambda$, then $\lim_{n\to\infty}\lambda^{-n}T^nv$ exists and is nonzero and is in $W$.

The easiest way to see this is to write $v$ as a linear combination of vectors in $B$ and then compute $\lim_{n\to\infty}\lambda^{-n}T^nv$ (this will also give you some idea what to do when the conditions are not met). That is, let $B=\{{x_1,x_2,\dots,x_r\}}$ and let $$v=a_1x_1+a_2x_2+\cdots+a_rx_r$$ where $Tx_i=\lambda_ix_i$. Then $$T^nv=a_1\lambda_1^nx_1+a_2\lambda_2^nx_2+\cdots+a_r\lambda_r^nx_r$$ Now suppose $\lambda_1$ is dominant, that is, $|\lambda_1|\gt|\lambda_i|$ for $i=2,3,\dots,r$. Then $$\lambda_1^{-n}T^nv=a_1x_1+a_2(\lambda_2/\lambda_1)^nx_2+\cdots+a_r(\lambda_r/\lambda_1)^nx_r$$ and when you take the limit as $n\to\infty$ you get $a_1x_1$.

Things get more complicated if there isn't a single dominant eigenvalue but you can still use this approach and get pretty far with it.

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I'm completely lost. What's v? Which vector space is B a basis of (the domain?)? Why is there a negative power? I don't know where to start understanding the answer. Sorry! – Ryan Aug 5 '12 at 11:35
The domain and codomain have to be the same vector space, otherwise there's no such thing as iterating the transformation. But I'll edit a bit. – Gerry Myerson Aug 5 '12 at 12:10
I've read your answer and also a bit on eigens and discrete dynamical systems, and it seems that my conjecture/generalisation was missing quite a few qualifications, although the heart was in the right place haha. There cannot be complex eigenvalues, the component of the initial vector in the direction(s) of the dominant eigenspace must be nonzero, and apparently the dominant eigenvalue must be the only one that is at least 1 (this last qualification puzzles me though).... Thanks for your help. – Ryan Aug 9 '12 at 8:50

What about a 2D-rotation by an irrational angle? Will this ever (after iterating) have an eigenspace in $\mathbb R^2$.

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The place for a revised hypothesis is in the question, not in a comment on an answer. Either that, or in a new question. – Gerry Myerson Aug 5 '12 at 9:15
Oh thanks! But the angle doesn't need to be irrational, does it? – Ryan Aug 5 '12 at 10:25
Right, for some reason I just wanted to be sure that the transformation isn't the identity after every $k \cdot n$ iterations. However it should work in either case. – Dario Aug 5 '12 at 19:35
Ok I have read up more and just learnt that all matrices with complex eigenvalues have rotational information encoded within. Fascinating! – Ryan Aug 8 '12 at 16:53