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What's the easiest and simplest way to show that there isn't $A \in M_2(\mathbb R)$ such that $A^{2004}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}$?

The nicest solution will award its writer with 500 points.

Thank you

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11  
For some reason, I have the feeling that this is a seven-year old problem... –  Georges Elencwajg Aug 19 '11 at 10:01
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@Georges: the problem is for "seven year olds", or the problem is "seven years old"? (I kid, I know what you're saying... ;) ) –  J. M. Aug 19 '11 at 10:04
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@Nir, I guess somebody has to mention this to you: current year 2011 minus seven years invoked by Georges equals 2004 the power of the matrix in your question. Where is the Cynicism or the math-snobbery or whatever? –  Did Aug 19 '11 at 10:49
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@Nir: as Didier explained, my comment meant that since 2011=2004+7, the problem must date from 2004, seven years ago. It is indeed not unusual for professors to sneak the current year into a problem, as a mild, harmless joke. I hadn't even realized that my sentence could be interpreted as having something to do with a seven-year old person, as is perfectly clear from my answer to J.M.'s comment. Since this is all very clearly explained in Didier's comment, I wonder why you persist in your insulting interpretation: "Read carefully and you'll find it [cynicism, math-snob]" –  Georges Elencwajg Aug 19 '11 at 11:22
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@Nir: some patience and good cheer would be helpful, you know... it doesn't help you and other people to be onion-skinned on the Internet. –  J. M. Aug 19 '11 at 11:23

3 Answers 3

up vote 20 down vote accepted

Suppose $A^{2004}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}.$

Then $B=A^{1002}$ is a real 2x2 matrix such that $B^{2}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}.$ We show that this is not possible.

Let $\quad B=\begin{pmatrix} a &b \\ c&d \end{pmatrix}.$

Then $B^{2}=\begin{pmatrix} a^{2}+bc &ab+bd \\ ac+cd&d^{2}+bc \end{pmatrix}=\begin{pmatrix} -1 &0 \\ 0&-2 \end{pmatrix}\quad \quad \quad (1)$

Considering the (1,2) entry in equation (1) yields $$ab+bd=0.$$If $b \neq 0 $ then we must have $a=-d$ but then the diagonal entries of $B^{2}$ would be equal contradicting (1). If $b=0$ then by considering the (1,1) entry in equation (1) we see that $a^{2}=-1$, which is not possible for $a$ real.

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"Then $B=A^{1002}$ is a real 2x2 matrix such that.."-Why that $A^{1002}$ will be well defined? –  user6163 Aug 19 '11 at 10:13
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@Nir: Because the first line says "Suppose...". So we assume that there is such an $A$, and then of course $A^{1002}$ also exists. –  Hans Lundmark Aug 19 '11 at 10:16
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Ok, I got that, Thanks. –  user6163 Aug 19 '11 at 10:36
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Or more compactly: $A^{2004} = A^{1002}A^{1002}$ but negative definite matrices do not admit a square root in $M_2(\mathbb{R})$. Contradiction. –  user13838 Aug 19 '11 at 11:17
    
Dear @Peter: thank you for the answer, when I'll be able to start a bounty and award you, I'll do that. I believe it will be available tomorrow. –  user6163 Aug 19 '11 at 12:47

If $\lambda$ and $\mu$ are the eigenvalues of $A$ then $\lambda^{2004}=-1$ and $\mu^{2004}=-2$. This implies that $\lambda$ and $\mu$ are complex, non-conjugate eigenvalues of $A$. But for any real matrix all non-real eigenvalues come in conjugated pairs.

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"If $\lambda$ and $\mu$ are the eigenvalues of $A$ then $\lambda^{2004}=-1$ and $\mu^{2004}=-2$"- Why this is true? why that A will be diagonal matrix? –  user6163 Aug 19 '11 at 8:52
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@Nir Because if $Av=\lambda v$, then by induction, $A^n v = \lambda ^n v$. This doesn't assume anything about $A$ being diagonal. This argument does tell you that, if your matrix is diagonalizable, then the corresponding eigenvalue multiplicities will be the same. If you're not diagonalizable, something like Jordan normal form will give what is needed. –  Aaron Aug 19 '11 at 9:33
    
As Aaron indicates I am not assuming that $A$ is diagonal. To elaborate somewhat any matrix $A$ can be written as $S^{-1}TS$ where $T$ is a triangular matrix and $S$ is some invertible matrix. (There are many ways of proving this and you may have seen a proof under the name of the Jordan normal form, or the Schur normal form, which are particular ways of writing $A$ in such a way.) The diagonal elements of $T$ are the eigenvalues of $A$. Furthermore $A^{2004}=S^{-1}T^{2004}S$.(Try a smaller power than 2004 to be convinced of this.) Now see what the diagonal elements of $T^{2004}$ are. –  Johan Aug 19 '11 at 11:37

We assume that such a $A$ exists. We have $A^2-\mathrm{Tr}(A)A +(\det A)I_2= 0$ hence $A^{2004}$ can be written as $\alpha I_2+\beta A$ where $\alpha$ and $\beta$ are real, since $\mathrm{Tr}(A)$ and $\det A$ are real (and $\beta\neq 0$ because $A^{2004}\neq \alpha I_2$). Since $\beta A =A^{2004}-\alpha I_2$, $A$ is diagonal with real entries, namely $A= \begin{pmatrix}d_1&0\\0&d_2\end{pmatrix}$ but since $d_1$ is real we can have $d_1^{2004}=-1$. We can take $2p$ where $p\in\mathbb N$ instead of $2004$.

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Nice argument without using eigenvalues. –  user1551 Aug 19 '11 at 9:32
    
@user1551 No direct eigenvalues (though they are sort of there in the description of $A$), but it does require Cayley-Hamilton, which seems slightly more advanced than eigenvalues. –  Aaron Aug 19 '11 at 9:41
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@Aaron: Yes, but the general statement of the C-H theorem is not needed here because the 2-by-2 case can be verified manually. I think this proof has a certain "wow" factor to the novices. Those proofs that make straightforward uses of eigenvalues are just a bit too boring. –  user1551 Aug 19 '11 at 9:57

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