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# Understand the problem

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?

##### Source of the problem
TIFR GS 2019, Part B Problem 4
Linear algebra
Hard
##### Suggested Book
Linear Algebra; Hoffman and Kunze

Do you really need a hint? Try it first!

Think in terms of eigenvalues. 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. Can you think from here? What about $a+b$ and $a+1/a$?

$a+b=a^2+b^2\implies a+1/a=a^2+1/a^2\implies a^4-a^3-a+1=0$ $\implies (a-1)(a^3-1)=0\implies a=1$,$\omega$,$\omega^2$

Now, what are the cases?

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 $A=\begin{bmatrix}\omega&0\\0&\omega^2\end{bmatrix}$.   Hence, the statement is false.

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