How to prove that a matrix is diagnolizable if it has no real eigenvalues?

Question

Given a matrix $A \in Mat_{2 \times 2}(\mathbb{R})$

suppose $A$ has no real eigenvalues, how can i prove that it's diagnolizable under the complex field?

thank you.

Answer 1

Eigenvalues are roots of the characteristic polynomial, which here is real by assumption. Nonreal roots of real polynomials come in conjugate pairs. So the two eigenvalues are distinct which implies diagonalizability.

Answer 2

A square matrix having distinct eigen values (real or non-real) is always diagonalizable. If roots are repeated it may or may not be diagonalizable.