Suppose we are given a real matrix $A$. We know that we may regard it as a complex matrix (with real entries). Every complex matrix is similar to a Jordan form where all eigenvalues are placed on diagonal. Suppose the Jordan form $\Lambda$ of $A$ is real and diagonal. This implies $A$ is diagonalizable on $\mathbb{C}$: $PAP^{-1}=\Lambda$. The problem I 'm having is that $P$ may be complex so it does not directly imply $A$ is similar to diagonal $\Lambda$ even though $\Lambda$ is real--when we are now working with real instead of just complex field. If all diagonal entries on $\Lambda$ are distinct reals, this would imply diagonalizable over the real trivially. But what if there are eigenvalues with larger than 1 multiplicity?
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Similarity of matrices never changes if you extend the field. See https://math.stackexchange.com/questions/57242/similar-matrices-and-field-extensions – Eric Wofsey Sep 08 '19 at 00:41
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If $A$ is diagonalizable over the complex numbers and its eigenvalues are real, its minimal polynomial is $(X-c_1)...(X-c_p)$ where $c_1,...,c_p$ are the eigenvalues. Apply
Prove that T is diagonalizable if and only if the minimal polynomial of T has no repeated roots.
Tsemo Aristide
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look the proof in the accepted answer of the statement it does not use complex numbers. – Tsemo Aristide Sep 08 '19 at 00:27