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This is part of a bigger proof that if there is a compact set, $K \subset \mathbb R^n$ such that the linear transformation $L$ maps $K$ into its interior, the eigenvalues $\lambda_i$ are all of absolute value less than 1. Clearly, if $L(K)\subset \operatorname{int}(K)$ then $L^n(K)\subset \operatorname{int}(K)$.

The easiest case seems to be: If the eigenvalue were an $n$th root of unity, then $L^n$ must have some fixed point on the boundary, a contradiction.

I can't even seem to prove it for complex eigenvalues of absolute value greater than $1$ or for eigenvalues of absolute value $1$ which are not roots of unity.

My attempts: 1) Say $|\lambda|>1$ then it seems like $L$ would take the vectors off to infinity. However, I am worried there is a case where $L$ takes some vector, lengthens it, and maps it to some other vector, and on the next iteration it undoes, or something like that.

2) Say $|\lambda|=1$. If a linear transformation on $\mathbb R^n$ has a pair of complex eigenvalues of absolute value 1 (but not roots of unity), is it true that this always corresponds to a rotation on a two dimensional subspace? If so, then it seems like this is an irrational rotation. So start with some point on the boundary of $K$ that is on the subspace of rotation, then look at the infinite union of $L^i(K)$ for $i>1$. The original point is a limit point of this union, and since $K$ was compact and since $L$ is continuous since it is a finite dimensional linear operator, $L^i(K)$ are all compact....but this isn't good because now the point is the limit point of an infinite union of compact sets which may not be compact and thus don't need to contain their limit points. But if it did contain this point this would be a contradiction.

Please and thank you for helping :D

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It's worth noting that each linear transformation must have a fixed point--namely the zero vector. –  Cameron Buie Mar 23 '13 at 16:00
    
I don't quite follow. We want to assume that $0\in int(K)$ so having this fixed point doesn't really seem to help. –  Steven-Owen Mar 23 '13 at 17:43
    
You're right, it doesn't help. I'm just pointing out that having a fixed point doesn't lead to a contradiction on its own, as you claimed. Now, you are right that having a fixed point on the boundary leads to a contradiction. –  Cameron Buie Mar 23 '13 at 17:55

3 Answers 3

up vote 1 down vote accepted

I'll try to formulate in a way that does not depend on the fact that the real $n$-dimensional space $V$ being acted upon is $\Bbb R^n$, in other words it does not depend on choosing a basis in $V$.

Let $\def\C{\Bbb C}V_\C$ be the complexification of $V$, which is a complex vector space built from the real vector space $V\oplus V$ by defining multiplication by $\def\ii{\mathbf i}\ii\in\C$ by $\ii\cdot(v,w)=(-w,v)$, and let $L_\C:V_\C\to V_\C$ be the complexification of $L$, defined by $L_\C(v,w)=(L(v),L(w))$ (it is clearly complex-linear). Now $L_\C$ as complex-linear map has the same characteristic polynomial as $L$ has as real-linear map (indeed on suitable bases they have identical matrices), so for every complex eigenvalue $\lambda=a+b\ii$ of $L$ the is a corresponding eigenvector in $(v,w)\in V_\C$. But this means that $(L(v),L(w))=(a+b\ii)(v,w)=(av-bw,bv+aw)$. In the case of interest here that $\lambda\notin\Bbb R$, that is $b\neq 0$, one easily sees that $v,w\in V$ must be $\Bbb R$-linearly independent, and so the $\Bbb R$-subspace $\langle v,w\rangle_\Bbb R\subseteq V$ is the $2$ dimensional and $L$-stable, with the restriction of $L$ to it given by the matrix $$ \begin{pmatrix}a&b\\-b&a\end{pmatrix}. $$

I guess this gives what you asked in the title.

For the "bigger proof" you need to modify the statement first to exclude taking the empty set for $K$.

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Sorry, why must they be linearly independent over the real numbers? Is it because if they weren't then $cv=w$ for real $c$ and thus multiplying them by a complex number $\lambda$ just had the effect of a real scalar since it would $(av-bcv,bv+acv)=((a-bc)v,(b+cv)v)$ where $a,b,c$ are real numbers? –  Steven-Owen Mar 25 '13 at 13:56
    
I don't know if what I said made sense, sorry. Could you just elaborate briefly on the implication from $b\neq0$ to $v,w$ being linearly independent over the reals. –  Steven-Owen Mar 25 '13 at 13:58
    
@ricky Supposing $v=\lambda x$ and $w=\mu x$ for a same vector $x$, with $\lambda,\mu\in\Bbb R^\times$, one has $\lambda(b\lambda+a\mu)x=\lambda(bv+aw)=L(\lambda w)=L(\mu v)=\mu(av-bw)=\mu(a\lambda-b\mu)x$, and taking the difference gives $b(\lambda^2+\mu^2)x=0$ in which none of the scalar and vector factors on the left can be $0$, a contradiction. –  Marc van Leeuwen Mar 25 '13 at 14:21
    
If the above seems harder than it should be, morally if $L(v)\in\langle v\rangle_\Bbb R$, then $v$ has a real eigenvalue, which would also be the eigenvalue of $(v,w)$ in $V_\Bbb C$, contradicting that the latter has eigenvalue $a+b\mathbf i$. –  Marc van Leeuwen Mar 25 '13 at 14:27
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Yes, in my comment $\lambda$ was just a real scalar, no relation to an eigenvalue. What you do amounts to taking $x=w$, $\lambda=c$, $\mu=1$ in what I said, so I am saying $c(bc+a)w=c(bv+aw)=L(cw)=L(v)=av-bw=(ac-b)w$, and taking the difference between the two ends gives $b(c^2+1)w=0$ contradiction. Seems like you're saying essentially the same thing. –  Marc van Leeuwen Mar 25 '13 at 14:41
  1. If $\lambda$ is an eigenvalue of a real matrix $A$, then it has a nonzero (complex) eigenvector $v$ corresponding to it.
  2. If $v$ is real, then so is $\lambda$, and we're done, because if we had $\lvert \lambda\rvert\geq 1$, there would be either a point in $K$ which explodes into infinity with $A^n$, or a fixed point of $A^2$ on $K$'s border (this is because $K$ must contain a ball around $0$).
  3. Otherwise, it's easy to see that then $\overline v$ corresponds to $\overline \lambda$ (because $\overline A=A$ and $\overline{Av}=\overline{\lambda v}$).
  4. Furthermore, we can assume that $v+\overline v$ is a nonzero real vector (equal to twice the real part of $v$). Otherwise $v$ would be purely imaginary, $iv$ would be a real eigenvector corresponding to $\lambda$, so we can use the same argument as above.
  5. From now on, we will only consider restrictions of $A$ and $K$ to the plane spanned by real and imaginary parts of $v$. We can assume that $v=(1,i)^T$, other coordinates zero (by changing coordinates if necessary).
  6. Then we have that $A^n(v+\overline v)=\lambda^nv+\overline\lambda^n\overline v$, likewise for $-i(v-\overline v)$; choose $r,\theta$ so that $\lambda=re^{i\theta}$. You can see by inspection that $A^n$ is a dilation by $r^n$ followed by a rotation by $n\theta$. This means that if $r\geq 1$, then either it blows up some points in $K$ into infinity, or it keeps the points of maximal distance from $0$ on the circle they belong to, both yielding a contradiction.
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Thank you for the help. A couple clarification questions: 1) What do you mean in 5. by real and imaginary parts of $v$? Do you mean writing $v$ as $x+iy$ where $x,y$ are real vectors and looking at the span of ${x,y}$ with the scalar field being $C$? 2) In 6. when you say likewise for $-i(v-\bar v)$ are you just saying $A^n(-i(v-\bar v))= -iA^n(v-\bar v)=-i(\lambda^nv-\bar\lambda^n\bar v)$. So what? 3) In 6. again, what is this point in $K$? It seems like we are only talking about complex vectors. –  Steven-Owen Mar 23 '13 at 19:12
    
@ricky: 1) no, i mean the real span, 2) yes, that's what I meant; $v+\overline v$ and $i(v-\overline v)$ are (up to a real scalar) the real and complex parts of $v$, so you can determine what $A$ is by looking at their images; 3) this point in $K$ is just any nonzero point in $K$ (restricted to the plane where $A$ is the dilation rotation) –  tomasz Mar 23 '13 at 22:13

Here is some rough ideas. Let $x_0=\arg\min_{x\in K}\|x\|_2$. By considering the sequence $x_n=L^nx$, we see that $\{x_n\}$ has a convergent subsequence and in turn $L$ has a fixed point $u$ in the interior of $K$. This $u$ must be the origin. Otherwise, as $L$ is linear, the farthest boundary point of $K$ on the span of $u$ is also a fixed point of $L$, which is a contradiction to the problem's assumption.

Now, define $K'=\{x\circ z:\, z=(z_1,\ldots,z_n)^T\in\mathbb{C}^n,\, |z_j|=1\ \forall j\,\}$, where $x\circ z$ denotes the Hadamard (i.e. entrywise) product of $x$ and $z$. Then $K'$ is a compact set in $\mathbb{C}^n$ and the natural extension of $L$ to $\mathbb{C}^n$ maps $K'$ into its interior. If $(\lambda,v)$ is an eigenpair of $L$ over $\mathbb{C}$, by considering the farthest boundary point of $K'$ on the line spanned by $v$, we see that $|\lambda|<1$.

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