Let $A \in \mathbb{R}^{n \times n}$, where $n \geq 2$, be a matrix for which $$A^3+A^2+A=0$$ Prove that rank of $A$ is an odd number.

What I've done so far: $A^3+A^2+A=0 \Rightarrow A(A^2+A+I_n)=0$ , then $A$ is the solution of $x(x^2+x+1)=0$ that means that $x_1=0 \;,\;x_2=-\frac{1}{\sqrt{2}}+\frac{i\sqrt{3}}{2}\;,\;x_3=-\frac{1}{\sqrt{2}}-\frac{i\sqrt{3}}{2}$ are eigenvalues of $A$ and so $A$ is not invertible , also $A\neq 0$ and lastly $A$ must be diagonalizable since it has 3 different eigenvalues.

Any help on how to continue ?
Thank you in advance !

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    $\begingroup$ $$A=\begin{bmatrix}-\frac1{2}&-\frac{\sqrt3}2&0\\ \frac{\sqrt3}2&-\frac12&0\\ 0&0&0 \end{bmatrix}$$ satisfies $A^3+A^2+A=0$ and its rank is $2$. $\endgroup$ Jun 4, 2022 at 17:42
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    $\begingroup$ It seems the matrix is intended to be size $n$ by $n.$ In other places you assume $n=3$ $\endgroup$
    – Will Jagy
    Jun 4, 2022 at 17:43
  • $\begingroup$ This question is ill-posed. As demonstrated above, the statement is not true, nor is it any obvious way of modifying it to be true. Using diagonal matrices, whose entries are the three roots the OP has identified, one may construct matrices of all possible ranks satisfying the given equation. $\endgroup$ Jun 4, 2022 at 17:55
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    $\begingroup$ @Möb On the contrary, I think the real matrix $A$ always has an even order. $\endgroup$
    – Zerox
    Jun 4, 2022 at 18:00
  • $\begingroup$ Yes, @Zerox, my bad. My comment is of course only true when allowing complex matrices. I didn't notice at first that the question was restricted to real matrices. Sorry for the confusion. $\endgroup$ Jun 4, 2022 at 18:04

2 Answers 2


Consider p(t)=$t^3+t^2+t$. Since p(t) annihilates A, the minimal polynomial of A say m(t) is a factor of $t^3 +t^2 +t$.Hence we get three possible cases. Case 1: m(t)=t. Hence A=0 matrix. Zero is even. Case 2: m(t)=$t^2+t+I$ and so characteristic polynomial is $(t^2+t+I)^m$ for some positive integer m and thus rank is even in this case. Case3: m(t)=$t^3+t^2+t$ In this case apart from the zero eigenvalues the non zero eigenvalues occur in cojugate pairs and hence the rank is even in this case also. Thus in all three cases rank of A is even.


Consider the characteristic polynomial $\chi_A$ of A. Since A is real, $\chi_A$ is real too. We can consider A as a complex matrix. As you proved, A is diagonalisable, and its eigenvaleurs are included in the set $\{0,j,\bar{j}\}$, where $j=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$. So : $$\chi_A=X^p(X-j)^q(X-\bar{j})^s$$ Rememeber that $\chi_A$ is real. This is only possible when $q=s$. Let's evaluate degrees of both sides, we find $n=p+2q$ so $n-p=2q$. $p$ is the dimension of the kernel of A, so $n-p$ is the rank of A.


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