$A$ and $A^2$ have same characteristic polynomial

Is it possible to have a non-identity $$2 \times 2$$ diagonalizable, invertible, complex matrix $$A$$ s.t characteristics polynomials of $$A$$ and $$A^2$$ are the same?

I am not getting any hint even how to create one.

I can start with two different eigenvalues but for this, we won't have the same characteristic poly.

I was also trying to play with $$\begin{pmatrix} 0 & i \\ i & 0 \\ \end{pmatrix}$$ Does not help.

• So, a matrix $A$ such that $(A-\lambda I)=(A^2-\lambda I)$? Commented Jul 13, 2019 at 1:38
• Yes, right you are. Commented Jul 13, 2019 at 1:40
• we need $\{\lambda_1,\lambda_2,...,\lambda_n\}=\{\lambda_1^2,\lambda_2^2,...,\lambda_n^2\}$, so squaring the eigenvalues must permute them. Commented Jul 13, 2019 at 1:48

Yes: consider $$A=\begin{bmatrix}\omega&0\\0&\omega^2\end{bmatrix}$$ where $$\omega$$ is a primitive 3rd root of unity.

• Holy crap that's brilliant. Commented Jul 13, 2019 at 1:42
• How did you get this answer? will you please elaborate? Commented Jul 13, 2019 at 1:46
• Sure: if $\lambda,\mu$ are the eigenvalues of $A$ then your conditions force $\lambda^2=\mu$ and $\mu^2=\lambda$, hence $\lambda^4=\lambda$. Commented Jul 13, 2019 at 1:47

Let $$a$$ and $$b$$ are eigenvalues of $$A$$ then eigenvalues of $$A^2$$ are $$a^2$$ and $$b^2$$. You need

$$t^2-(a+b)t+ab=t^2-(a^2+b^2)t+a^2b^2$$

So $$ab=a^2b^2\implies ab=1$$ as $$A^{-1}$$ exists.

And $$a+b=a^2+b^2\implies a+1/a=a^2+1/a^2\implies a^4-a^3-a+1=0$$

$$\implies (a-1)(a^3-1)=0\implies a=1$$,$$\omega$$,$$\omega^2$$

Discard $$a=1$$ as $$A\ne I$$ and diagonalizable. Discard the case of both eigenvalues equal as $$A$$ is not diagonalizable for repeated eigenvalues. The only possibility is $$\{\omega,\omega^2\}$$ as the eigenvalue set for $$A$$. So $$A=\begin{bmatrix}\omega&0\\0&\omega^2\end{bmatrix}$$