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Let $A$ be an $n \times n$ real matrix with $A^3 + A = 0.$ Can we say that $\text {tr}\ (A) = 0\ $?

I think it's true but can't prove it. Any help will be highly appreciated.

Thanks in advance.

Eric Wofsey
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math maniac.
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3 Answers3

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The minimal polynomial of $A$ must divide $x^3+x$. Then the real Jordan form of $A$ can have, consequently, two kinds of blocks:

  • 0 blocks of dimension 1
  • 2-dimensional blocks associated to rotations of $\pi/2$

In both cases the trace is equal to 0.

Recalling that the trace of $A$ is invariant for conjugation you have done.

Observe that this is not true over the complex numbers: $A=i I$ satisfies $A^3+A=0$ but $trA \neq 0$.

Sabino Di Trani
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  • It is possible to find an analogue of classical complex Jordan form over reals, allowing as diagonal blocks some rotations 2 -dimensional blocks associated to the complex eighenvalues. – Sabino Di Trani Jul 16 '20 at 04:58
  • It is described, for example, here https://en.m.wikipedia.org/wiki/Jordan_normal_form in the section about real matrices. – Sabino Di Trani Jul 16 '20 at 05:00
  • How can $2$-dimensional blocks associated to rotations of $\frac {\pi} {2}$ be a Jordan block? All such blocks are of the form $\begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix}.$ This is not looking like a Jordan block as the Jordan blocks are upper triangular matrices with super-diagonal entries all equal to $1.$ – math maniac. Jul 17 '20 at 06:35
  • The matrix of rotation through an angle $\frac {\pi} {2}$ is clearly not upper triangular. – math maniac. Jul 17 '20 at 06:41
  • I think the Jordan canonical form of a matrix can only make sense if all it's eigenvalues are lying inside the ground field from which the entries of the matrix are taken. – math maniac. Jul 17 '20 at 06:43
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    Please, read the Wikipedia article: it is well explained. – Sabino Di Trani Jul 17 '20 at 15:37
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    Thanks. I have accepted your answer after reading this article in wikipedia from the link provided by you in one of your comments above. – math maniac. Jul 17 '20 at 19:19
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The most straightforward way to approach the proof is to use the minimal polynomial of $A$. If $K$ is the minimal polynomial of $A$, then any other polynomial $Q$ with $Q(A)=0$ is a multiple of $K$. Hence, the eigenvalues of $A$ are either $0$, $i$ or $-i$. Moreover, as $A$ is a real matrix, the sum of its eigenvalues must be a real number. In order to clarify this, consider a characteristic equation as below:

$$ a_1 x^n+a_2x^{n-1}+ ...+a_{n+1}=0$$ Then, the sum of the eigenvalues is $\frac{-a_2}{a_1}$.

To conclude the given fact, it's enough to observe that the only real number made by adding $0,-i,i$ together is $0$. So, we are done since $tr(A)$ is the sum of its eigenvalues.

PS: the link below may be useful.

https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)

Reza Rajaei
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    Trace of a matrix is sum of its eigenvalues not sum of its distinct eigenvalues. So considering $T(A^3)=a_1^3+a_2^3$, is not right, I think. –  Jul 16 '20 at 06:20
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the trace of the matrix is the sum of its eigenvalues, Here $i,-i,0$ are the eigenvalues of the given equation $A^3+A=0$ and the sum of its eigenvalues is zero, So the trace of the matrix is $0$.

A matrix with real entries often has non-real eigenvalues.

Anonymous
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  • we take only real entries... – Anonymous Jul 16 '20 at 05:01
  • @JCAA thank you for correcting , now I edited my answer – Anonymous Jul 16 '20 at 05:13
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    You should mention it clearly as to why $i$ and $-i$ are appearing as eigenvalues of $A$ over $\Bbb C$ with same multiplicities. Your argument is not yet complete and precise in my view @Venkat. You should argue as "Either $A = 0$ or $i$ and $-i$ should appear as roots of the characteristic polynomial of $A$ with same multiplicities (this time with proper justification) for otherwise the characteristic polynomial of $A$ will not lie in $\Bbb R[X].$" – math maniac. Jul 16 '20 at 05:25