# Trace $A$ is an integer.

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.

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?

• 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. Commented Jan 8, 2017 at 2:51
• You can try Gilbert Strang or Hofman or Halmos. Commented Jan 8, 2017 at 2:56
• Thaks. I shall look into these books. Commented Jan 8, 2017 at 2:57
• You are very welcome Commented Jan 8, 2017 at 2:57
• @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) Commented Jan 8, 2017 at 3:16