Let $A$ a $n\times n$ matrix such that $$3A^3 = A^2 + A + I$$ Proof that $$\lim_{n\to\infty}A^n$$ Is a idempotent matrix.

My attempt:

Given a eigenvalue $\lambda$ of $A$, we have that $3\lambda^3=\lambda^2+\lambda+1$ so $\lambda\in\{1, z, \overline z\}$, where $z$ is a complex number such that $|z|<1$. So, the eigenvalues of $A^n$ are inside the set $\{1, z^n, {\overline z}^n\}$. For $n\to\infty$ $z^n\to0$ so the eigenvalues of $A$ gonna be $0$ or $1$, what implies $A$ will be a idempotent matrix.

Is this correct?

Thanks for attention.

  • 1
    $\begingroup$ The limit having $0$ and $1$ as it’s eigenvalues is insufficient; such a matrix will be idempotent if and only if it is diagonalizable. $\endgroup$ Commented Aug 23, 2023 at 12:41
  • 1
    $\begingroup$ Note: "proof" and "prove" are different words. $\endgroup$
    – Shaun
    Commented Aug 23, 2023 at 12:41
  • 3
    $\begingroup$ The challenge seems to be the proof of existence of this limit. If $B:=\lim_{n \to \infty}A^n$ exists, then $B=\lim_{n \to \infty}A^{2n} =\lim_{n \to \infty}(A^{n})^2 = B^2$. $\endgroup$
    – Gerd
    Commented Aug 23, 2023 at 12:59

1 Answer 1


Note that if $\lim A^n=B$ exists, then $B=\lim A^n=\lim (A^{2n})=B^2$ by the uniqueness of the limit matrix, so it suffices to prove that $A^n$ converges to some matrix $B$. To this end, we show $A$ is diagonalizable with eigenvalues $|\lambda|<1$ or $\lambda=1$.

By assumption, the polynomial $3x^3-x^2-x-1$ vanishes when $A$ is plugged, that means $m_A(x)$ divides the polynomial: $$3x^3-x^2-x-1=(x-1)(3x^2+2x+1)=(x-1)(x-\frac{-1+\sqrt{2}i}{3})(x-\frac{-1-\sqrt{2}i}{3})$$

So possible eigenvalues of $A$ are $1$ and $\frac{-1\pm\sqrt{2}i}{3}$ (where the last two have absolute value less than 1). Moreover, the minimal polynomial necessarily splits to a product of different linear factors, hence $A$ is diagonalizable. Write: $$A=PDP^{-1}\Rightarrow A^n=PD^nP^{-1}$$ $D^n$ converges to a diagonal matrix $D'$ with $0$'s and $1$'s on the diagonal (as $1^n\to 1$ and the other two eigenvalues set to the n'th power converge to $0$). So $A^n\to PD'P^{-1}$ and we've already showed the limit must be idempotent.


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