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I am studying the spectral theorem for matrices, and the book says that if a $nxn$ matrix A is real and symmetric then its diagonalizable over $\mathbb{R}$. And that this fact is a corollary of the Spectral Theorem for the complex case of normal matrices.

Although I agree that since $A$ is symmetric then $A$ is normal hence it implies that $A$ is diagonalizable over $\mathbb{C}$, and moreover it is easy to prove that all eigenvectors are real. But how can I see that all eigenvectors are also real?

Thanks in advance.

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  • $\begingroup$ $\lambda$ is an eigenvalue iff $\ker (A-\lambda I)$ is non trivial. If$A-\lambda I$ is real, then the kernel contains a purely real element. $\endgroup$ – copper.hat Aug 9 '18 at 23:00
  • $\begingroup$ "IfA−λI is real, then the kernel contains a purely real element." this is precisely what I want to prove. $\endgroup$ – User43029 Aug 9 '18 at 23:11
  • $\begingroup$ If $B$ is real and $B (u+iv) = 0$ where $u,v$ are real, then $Bu = 0 $ **and** $Bv=0$, so you always have a real eigenvector (assuming that $U+iv =neq 0$, of course). $\endgroup$ – copper.hat Aug 9 '18 at 23:12
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Suppose $v$ is a complex eigenvector of $A$, a real symmetric matrix, with corresponding real eigenvalue $\lambda$. Note that \begin{align} \lambda \overline{v}=\overline{\lambda v} = \overline{Av}= A\overline{v} \end{align} then $\overline{v}$ is also an eigenvector. Hence $v+\overline{v}$ is a real eigenvector. So pick this one.

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  • $\begingroup$ great! Can I also see from your proof that these new real eigenvectors are orthogonal with each other? $\endgroup$ – User43029 Aug 9 '18 at 23:06
  • $\begingroup$ From the complex spectral theorem, you can find $\{v_1, v_2, \ldots, v_n\}$ linearly independent complex eigenvectors. Take the real component of each $v_i$. $\endgroup$ – Jacky Chong Aug 9 '18 at 23:08
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Because if $\lambda$ is an eigenvalue of $A$ and $\lambda\in\mathbb R$, then $A$ has an eigenvector in $\mathbb{R}^n$ whose corresponding eigenvalue is $\lambda$. In fact, asserting that $A$ has a real eigenvector corresponding to $\lambda$ is equivalent to asserting that $\lambda$ is a real root of the characteristic polynomial of $A$.

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  • $\begingroup$ I can't see why your first claim is true here. Can you explain it? $\endgroup$ – User43029 Aug 9 '18 at 23:07
  • $\begingroup$ @Charles THat's what I explain in the second sentence. If $f$ a linear map from $\mathbb{R}^n$ into itself, how can you determine whether it has a real eigenvector? You can do it by proving that the characteristic polynomial $P(x)$ of $f$ has a real root $\lambda$, which will be then an eigenvalue of $f$. Then (and only in that case) $f$ will have a real eigenvector corresponding to $\lambda$. $\endgroup$ – José Carlos Santos Aug 9 '18 at 23:12

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