= Solution
\b[False; in fact no matrix with this spectrum is diagonalizable over $\mathbb R$.] If a real invertible $P$ gave $P^{-1}AP=D$ with $D$ a real <diagonal matrix>, all <eigenvalues> of $A$ would be the real diagonal entries of $D$. But two of the required <eigenvalues> have nonzero imaginary parts. This contradiction proves
$$
\boxed{A\text{ is not diagonalizable over }\mathbb R}.
$$
The real block form in the initial example is consistent with this: a two-dimensional rotation block is not a real diagonalization.
Back to article page
