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I am reading Hoffman and Kunze linear algebra book. In chapter $8$ there is a theorem that states as " Let $V$ be a finite-dimensional complex inner product space and $T$ a normal operator on $V.$ Then $V$ has an orthonormal basis consisting of characteristic vectors for $T".$ Which also says that every normal matrix is unitarily diagonalizable. The proof is totally depends on the fact that matrix or operator over complex field have characteristic vectors. Now my question is that, if $A$ is real normal matrix with real eigen values then, is it true that $A$ is diagonalizable over $\mathbb{R}?$ Please suggest me. Thanks in advance.

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  • $\begingroup$ this matrix does not have real eigen values... $\endgroup$ – neelkanth Sep 14 '15 at 2:35
  • $\begingroup$ my question is real normal matrix with real eigen values $\endgroup$ – neelkanth Sep 14 '15 at 2:36
  • $\begingroup$ oh, i over saw that. sorry. $\endgroup$ – user251257 Sep 14 '15 at 2:36
  • $\begingroup$ The proof outlined in this answer should work for you in any field. $\endgroup$ – Chappers Sep 14 '15 at 2:40
  • $\begingroup$ its mean it will be diagonalizable over R?...am i ok? $\endgroup$ – neelkanth Sep 14 '15 at 2:42
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If $A$ is normal, it is diagonalizable over $\mathbb C$. Hence its minimal polynomial $m(x)$ is a product of distinct linear factors over $\mathbb C$. Yet by assumption, all eigenvalues of $A$ are real. Therefore $m(x)$ is also a product of distinct linear factors over $\mathbb R$. Consequently $A$ is diagonalisable over $\mathbb R$.

(In fact, if $A$ is a normal matrix with a real spectrum, it must be real orthogonally diagonalisable and hence real symmetric, but that's another story.)

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  • $\begingroup$ Nice answer...thanks a lot... $\endgroup$ – neelkanth Sep 14 '15 at 2:47

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