1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This statement is not true in real. $\endgroup$ – Daniel Li Sep 8 '19 at 0:24
  • $\begingroup$ look the proof in the accepted answer of the statement it does not use complex numbers. $\endgroup$ – Tsemo Aristide Sep 8 '19 at 0:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.