I cannot find an appropriate counterexample.
Is there a counterexample? Or is $A$ indeed Hermitian?
2026-02-22 19:35:41.1771788941
If a $2 \times 2$ matrix $A$ satisfies $A^2=I$, then is $A$ necessarily Hermitian?
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The matrix $$A = \begin{bmatrix}1&1\\0&-1\end{bmatrix}$$ is not Hermitian, but $$A^2 = \begin{bmatrix}1&1\\0&-1\end{bmatrix} \begin{bmatrix}1&1\\0&-1\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = I.$$