For example, $\left( \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right)$ has eigenvalues $1\pm i$ over $\mathbb{C}$ but no eigenvalue over $\mathbb{R}$. It is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$. But I wonder how to prove the general case. It seems that the argument of algebraic multiplicity equal geometric multiplicity for each eigenvalue does not quite work here…because the matrix has no eigenvalue if considered to be over $\mathbb{R}$.
Prove that a real matrix with a non-real eigenvalue is not diagonalizable over the reals
linear algebra
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Best Answer
Assume $A$ has no eigenvalues. If it were diagonalizable, then $A = PDP^{-1}$ with a diagonal matrix $D$. This implies $AP = PD$. Let $e_1 = (1,0,0,...,0)^T$ and put $v = Pe_1$. Then $$ Av = APe_1 = PDe_1 = P(d_1e_1) = d_1Pe_1 = d_1v $$ with $d_1$ being the first entry of $D$ on the diagonal. Hence, $A$ has the eigenvalue $d_1$, contrary to the assumption.