# Matrix in infinity is idempotent

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.

• 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. Commented Aug 23, 2023 at 12:41
• Note: "proof" and "prove" are different words. Commented Aug 23, 2023 at 12:41
• The challenge seems to be the proof of existence of this limit. If $B:=\lim_{n \to \infty}A^n$ exists, then $B=\lim_{n \to \infty}A^{2n} =\lim_{n \to \infty}(A^{n})^2 = B^2$.
– Gerd
Commented Aug 23, 2023 at 12:59

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.