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Based off the theory I can’t see any reason that an example would not exist. Specifically, the fact that A matrix is orthogonal only implies that the possible eigenvalues are $\pm 1$. However, we don’t know anything about the sizes of the eigenspaces.

Nonetheless, it is not hard to show that a 2×2 orthogonal matrix must in fact be diagonalizable. So an example has to at least be 3×3.

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If $A$ is real orthogonal, its eigenvalues must have *moluli* $1$, but they are not necessarily $\pm1$. Also, when you mention diagonalisation, *what* matrices are used to diagonalise $A$? If you mean diagonalisation by a *real* invertible matrix, this is not always possible: $P$ is real implies that $P^{-1}AP$ is real. So, if $A$ has nonreal eigenvalues, its eigenvalue matrix cannot be equal to $P^{-1}AP$ for any real $P$. In fact, every $2\times2$ rotation matrix that is not equal to $\pm I$ is *not* diagonalisable over $\mathbb{R}$.

If you mean diagonalisation by a perhaps complex matrix, then it is known that every real orthogonal matrix is diagonalisable. More generally, every real or complex normal matrix (including real orthogonal matrix) is unitarily diagonalisable over $\mathbb{C}$

**Hint:** It depends on the field you are working over. Look again at the $2\times 2$.

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