If $A$ is a square matrix of order $n$ with real entries and $A^3 + I =0$. Then $\operatorname{trace}(A)$ is an integer. My attempt: Here $|A| = -1$. And $\operatorname{trace}(A^3) = -n$. Then I tried to draw a contradiction assuming that $\operatorname{trace}(A)$ isn't an integer (using definition of determinant). But nowhere near the solution.
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Hint
From the equation we conclude that the eigenvalues ($k$) will fit the equation:
$$k^3=-1$$
So, $k=-1$ or $k=\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$ are the candidates for eigenvalues.
1) We know that the trace is the sum of the eigenvalues.
2) We also know that once we have a real matrix if we have a complex number as an eigenvalue then his conjugate will also be an eigenvalue.
Can you finish?
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$\begingroup$ Yes. Thank you. Can you suggest me a book where the connection between linear algebra and matrix and equations is clearly described? I don't like the book by Micheal Artin that much. $\endgroup$ Commented Jan 8, 2017 at 2:51
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1$\begingroup$ You can try Gilbert Strang or Hofman or Halmos. $\endgroup$– ArnaldoCommented Jan 8, 2017 at 2:56
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$\begingroup$ Thaks. I shall look into these books. $\endgroup$ Commented Jan 8, 2017 at 2:57
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1$\begingroup$ @user398623 I would also personally recommend you spend an evening or two watching this short course on youtube. It shows you a side of introductory linear algebra rarely touched by books (much because books are printed on paper, which makes it difficult, but also because many books don't even try) $\endgroup$– ArthurCommented Jan 8, 2017 at 3:16