Let $A\in$ $M_{2015x2015}(\mathbb{R})$ prove that $A^2-A\neq I$

Hint: Show that $A$ does not have eigenvalues over $\mathbb{R}$ if $A^2-A = I$



closed as off-topic by Najib Idrissi, Watson, JKnecht, John B, gebruiker Apr 17 '16 at 20:35

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  • 1
    $\begingroup$ Could the downvoters provide a short explanation so I could edit properly? $\endgroup$ – Zoltán Apr 17 '16 at 15:51
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    $\begingroup$ I didn't downvote but my guess is that you just stated this as "do this problem for me" without showing any kind of effort whatsoever. I have to say though that 4 downvotes seem like an overraction, especially to a new user $\endgroup$ – Ovi Apr 17 '16 at 16:14
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    $\begingroup$ @Ovi Thank you, will pay more attention next time. $\endgroup$ – Zoltán Apr 17 '16 at 16:16
  • $\begingroup$ @Tamir Why next time? You already indicated that you are aware of the fact that you can edit. So why not edit the question to include your thoughts/efforts? $\endgroup$ – gebruiker Apr 17 '16 at 20:35

A diagonal matrix with $\frac{\sqrt{5}+1}{2}$'s on the diagonal is a counterexample.


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