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

Question

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.

we need ${\lambda_1,\lambda_2,...,\lambda_n}={\lambda_1^2,\lambda_2^2,...,\lambda_n^2}$, so squaring the eigenvalues must permute them. — The Count, Jul 13 '19 at 01:48

score 18 · Answer 1 · answered Jul 13 '19 at 01:41

18

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

answered Jul 13 '19 at 01:41

carmichael561

54,993

3

Holy crap that's brilliant. – The Count Jul 13 '19 at 01:42
How did you get this answer? will you please elaborate? – Ri-Li Jul 13 '19 at 01:46
11

Sure: if $\lambda,\mu$ are the eigenvalues of $A$ then your conditions force $\lambda^2=\mu$ and $\mu^2=\lambda$, hence $\lambda^4=\lambda$. – carmichael561 Jul 13 '19 at 01:47

score 4 · Accepted Answer · edited Jul 13 '19 at 04:19

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 $$A=\begin{bmatrix}\omega&0\\0&\omega^2\end{bmatrix}$$

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

2 Answers2