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The question is from an exercise in Gilbert Strang's Linear Algebra and its Applications.

The powers $A^k$ approach zero if all $|\lambda_i|<1$, and they blow up if any $|\lambda_i|>1$. Peter Lax gives four striking examples in his book Linear Algebra. $$A = \left( \begin{array}{cc} 3& 2 \\ 1& 4 \\ \end{array} \right)\qquad B = \left( \begin{array}{cc} 3 & 2 \\ -5 & -3 \\ \end{array} \right)\qquad C = \left( \begin{array}{cc} 5& 7 \\ -3& -4 \\ \end{array} \right)\qquad D = \left( \begin{array}{cc} 5& 6.9 \\ -3& -4 \\ \end{array} \right)$$ $$\|A^{1024}\|>10^{700}\qquad B^{1024}=I\qquad C^{1024}=-C\qquad \|D^{1024}\|<10^{-78}$$ Find the eigenvalues $\lambda=e^{i\theta}$ of $B$ and $C$ to show that $B^4=I$ and $C^3=-I$.

Here is my question:

Why are these examples so special? Is it because that all of them contain the number "1024"? Or such examples are hard to construct?

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This sounds pretty subjective. I don't see anything too striking about them at all, other than maybe the fact that all components are obviously $>1$ in magnitude yet the powers of the middle two are kept (relatively) stable while the last shrinks. – anon Aug 10 '11 at 3:16
As I see, one gets $D$ by a "small" perturbation on $C$. – Jack Aug 10 '11 at 3:21
up vote 16 down vote accepted

There is nothing particularly special about $1024$; it just happens to be convenient to compute matrices to $2^n$ powers because you can repeatedly square.

What's supposed to be striking (of course this is subjective) is that $A, B, C, D$ all have small, superficially similar-looking entries, but after sufficient iteration have very different qualitative behavior; moreover, if you didn't know a lot of linear algebra, you would be hard-pressed to guess which matrices would admit which behavior just by looking at them.

Examples $A, D$ are not hard to construct in the sense that a random matrix you choose will do one of those two things. Examples $B, C$ are more fine-tuned, but not hard to construct by hand once you understand how to construct matrices with a given characteristic polynomial.

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How to choose the random matrix? – Jack Aug 10 '11 at 17:36
Any way you want. Just write down four small integers. (If you're asking how to make a precise statement, note that periodic behavior is impossible unless the determinant is a root of unity, and the set of matrices with this property has measure zero in the space of all matrices.) – Qiaochu Yuan Aug 10 '11 at 17:44

The examples display three different behaviors:

    1. $\|A^k\|\to\infty$
    2. $B^k$ and $C^k$ cycle with periods of $4$ and $6$ respectively, always with norm $1$

$1024$ is just a large number that displays these behaviors strikingly. Since $1024=0\pmod{4}$ and $1024=4\pmod{6}$, so $B^{1024}=I$ and $C^{1024}=-C$.

Peter Lax said the examples were "striking", not "special".

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