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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

  • 11
    For some reason, I have the feeling that this is a seven-year old problem...2011-08-19
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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... ;) )2011-08-19
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    I hadn't realized the ambiguity, even though I like these linguistic jokes. Nice catch, J.M.!2011-08-19
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    People here can amaze me, @Georges, does this Cynicism made you feel good? now you're OK? you had your math-snob daily moment?! COOL!2011-08-19
  • 12
    @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?2011-08-19
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    Read carefully and you'll find it.2011-08-19
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    Take care man, you, J.M and your linguistic jokes.2011-08-19
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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]"2011-08-19
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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.2011-08-19
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    @Nir: so now you are threatening us? This is grotesque.2011-08-19
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    Threat? Lol! take care? Is that a threat?2011-08-19
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    @Georges: we've had our laughs, and now we ought to let this go I think. I don't think reasoning would work here in the OP's current state...2011-08-19
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    Your laughs.. this is so sad :-)2011-08-19
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    Dear J.M., I agree with your wise proposition.2011-08-19
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    @Nir: Georges has already explained the completely benign meaning of his original comment: the occurrence of the number 2004 most likely means that this problem was written 7 years ago, because it is common practice for a problem-writer to include the current year in their problems. **Neither Georges or J.M. was insulting you**, so you have become very upset over nothing, and you have incorrectly accused them of "cynicism" and "math snobbery". I recommend that you apologize.2011-08-19
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    @Georges: As to the "threat" aspect: I honestly did not interpret "take care" as a threat. I understand that it could be interpreted as "be careful" (is this what you were thinking?), but I initially read it as "goodbye", as in, Nir was attempting to walk away from what he recognized as an inflamed situation. So unless "take care" is some cultural reference I'm not getting in which it implies a threat, or there is some other good argument for interpreting it that way, I won't treat it as such.2011-08-19
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    @Zev: Thank you for the comment. See, Fisrt I want to say that a lot of people here, Especially users with a lot of point, tend to be very unpleasant sometimes, or impatient, when they think that the problem is not clever enough, and I'm aware of the fact that the problem is not the best one that I can pulled out, but I had couple of doubts about that, so I decided to post that. There were more than four or five cases, and you can see it in my questions, that users acted simply vicious and said really insulting things, I can give you links, but I won't do than now.2011-08-19
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    Regarding this one, I have to admit that first I really didn't completely understand what Gerorges mean, but you have to admit that J.M Did mean in his comment that the question belongs to 7 years ols. he DID understand but decided to comment this way- and this is not nice and kind of insulting. He admits in his comment that he know what George meant and still he emphasizes the fact that he might meant something else- cause in his opinion.. I don't really need to tell you what he meant. So this innocence from both of them is really funny, they could have been say sorry or something.2011-08-19
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    My initial comment was too offensive, I'm aware of that but J.M did mean it,and I should have refereed it to him, and not to George, and I ask for George's forgiveness because of that. Regarding the threat, come on.. As I said many times- you think that my questions are stupid, don't fit your intellectual level?don't even read it, more than that-don't even bother to comment.I'm sick of permanent snobbery from users here.There are such nice people here that always help and explain everyting with a lot patince and there are others,who don't donate nothing but side-annoying-comments I'm done :-)2011-08-19
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    The problem might be for 7 years old kids, but see how many people saw it, commented, took part, voted, and it maybe because of the points, but it made people think, discusses- and this is what the site for, and the OP got this answer. Now I'm really done2011-08-19
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    Dear Zev : yes, I thought that "take care man" was a threatening phrase; even if I'm wrong, "man" is no way to address users. As I wrote to J.M., I don't want to be further involved in this exchange. And, last not least, I'm quite grateful to you, Zev, for taking the time and energy to address this little incident.2011-08-19
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    @Georges: I fear you got caught in the splash of a different altercation. I don't know if you can "see" deleted answers/question. [Here](http://math.stackexchange.com/questions/57435/finding-the-minimal-polynomial-of-a-a-given-a-diagonal-matrix-equivalent-to-x) and [here](http://math.stackexchange.com/questions/57885/v-m-3k-t-v-to-v-so-that-tb-bt-find-ts-minimal-polynomial-eigensp). I also *think* I was being blamed for [this](http://meta.math.stackexchange.com/questions/2799/how-can-i-see-who-voted-down-20-25-questions-of-mine-yesterday).2011-08-19
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    @Arturo Magidin: Dear Arturo, thanks for this information : I had no idea why my little joke about the question having been posed seven years ago was met with such vehemence, even after explicit clarification. The picture is now clearer. Unfortunately I get a "page not found" message when trying to connect to your second link. Also, the OP seeems to have changed his user name recently. To end on a cheerful note, I'm glad to have the opportunity to tell you how much I appreciate your activity here: this is the silver lining in this not so pleasant thread!2011-08-19
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    @Georges: The second link is to a deleted question, and I have no problem accesing it (question 57885). Rather, the "change in name" reflects the fact the OP has "left"/deleted his registered account, if I'm not mistaken.2011-08-19
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    @ Zev Chonoles. Yes, you're right."Take care" cannot be considered threatening : I've just asked on the sister site "English Language and Usage". Sorry for my misconception. http://english.stackexchange.com/questions/38575/is-take-care-always-a-friendly-utterance-or-can-it-sometimes-be-considered-thre2011-08-19
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    *Le* \*sigh\*. Just so we're above the board: I was laughing at the different ways one could interpret Georges's original comment (and certainly not at Georges or the OP). I thought it was patently obvious that I knew Georges meant the sense of "this problem is seven years old". Oh well...2011-08-21

3 Answers 3

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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?2011-08-19
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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.2011-08-19
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    Ok, I got that, Thanks.2011-08-19
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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.2011-08-19
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    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.2011-08-19
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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?2011-08-19
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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.2011-08-19
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    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.2011-08-19
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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.2011-08-19
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    @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.2011-08-19
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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.2011-08-19