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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.

1 Answers 1

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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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    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. – 2017-01-08
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    You can try Gilbert Strang or Hofman or Halmos. – 2017-01-08
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    Thaks. I shall look into these books. – 2017-01-08
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    You are very welcome – 2017-01-08
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    @user398623 I would also personally recommend you spend an evening or two watching [this short course](http://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab) 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) – 2017-01-08
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    Gonna do it tonight. – 2017-01-08
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    Good point @Arthur – 2017-01-08
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    It doesn't touch equations, I think, but it does linear transformations and matrix multiplication really well. – 2017-01-08