# Let $A\in$ $M_{2015x2015}$ prove that $A^2-A\neq I$ [closed]

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 $A^2-A = I$

Thanks

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, Watson, JKnecht, John B, gebruiker
If this question can be reworded to fit the rules in the help center, please edit the question.

• Could the downvoters provide a short explanation so I could edit properly? – Zoltán Apr 17 '16 at 15:51
• 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 – Ovi Apr 17 '16 at 16:14
• @Ovi Thank you, will pay more attention next time. – Zoltán Apr 17 '16 at 16:16
• @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? – gebruiker Apr 17 '16 at 20:35

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