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I don't quite see the use of the fact that $n$ is odd.

Anyway, I give a counterexample: Take a matrix $A$ where $a_{ii}$ are the non-zero rational numbers, and all other entries are zero. Clearly, $A^2 \neq 2I.$ If it was, $a_{ii}$ will be irrational. Contradiction.

If this enough? Or should I prove the statement. If I should prove, please help me!

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  • $\begingroup$ That's just an example, not a counterexample. $\endgroup$ – Robert Israel Oct 16 '13 at 7:14
  • $\begingroup$ yes, I thought that wouldn't be enough! $\endgroup$ – ugstudent1243 Oct 16 '13 at 7:14
  • $\begingroup$ I'm guessing you may want to look at the determinant of A. $\endgroup$ – Mike Oct 16 '13 at 7:19
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    $\begingroup$ This much should be clear: If you do not understand why it is important for $n$ to be odd, then you have not solved the problem. $\endgroup$ – Slade Oct 16 '13 at 7:26
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If $A^2 = 2I$, then the eigenvalues of $A$ lie in the set $\{\pm \sqrt 2 \}$; since $A$ has rational entries, $\det(A)$ is rational; but $-\det(A)$ is the product of these eigenvalues. An odd number of factors of the form $\pm \sqrt 2$ will always be of the form $\pm 2^k \sqrt 2$ for some non-negative integer $k$, not rational. Thus $A^2 = 2I$ may be ruled out. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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If $A^2=2I$ then $\det(A)^2=2^n$ where $n$ is odd. Note that $\det(A)\in\mathbb{Q}$.

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Hint: What could the characteristic polynomial of $A$ be?

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  • $\begingroup$ please give me more hints! I don't find the connection. $\endgroup$ – ugstudent1243 Oct 16 '13 at 7:20
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You've found an example (not counter-, as it does not show the claim is wrong) in favor of the claim. But this is not enough, you've found one matrix $A \in \def\Mat{{\rm Mat}}\Mat_n(\mathbb Q)$ such that $A^2 \ne 2I$, but the claim is that all matrices $A \in \Mat_n(\def\Q{\mathbb Q}\Q)$ for odd $n$ have $A^2 \ne 2I$. Towards your goal: Look at the characteristic polynomial and the eigenvalues.

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