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If I have real valued matrix $A$, are these two notions of being Diagonalizable and being Normal equivalent?

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    $\begingroup$ No. Any normal matrix is diagonalizable. The converse is not true. $\endgroup$ – Artem Jan 7 '13 at 1:19
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    $\begingroup$ @Artem Isn't $\begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}$ a Normal matrix that isn't diagonalizable over the reals? $\endgroup$ – Calvin Lin Jan 7 '13 at 1:27
  • $\begingroup$ @CalvinLin, the usual meaning is "diagonalizable over the complex". $\endgroup$ – DonAntonio Jan 7 '13 at 2:27
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    $\begingroup$ @DonAntonio I don't necessarily agree that is true. Furthermore, there have been people who argue that since OP specified real-valued, he is working with the reals as the base field. In any case, the real version is still slightly interesting to understand. $\endgroup$ – Calvin Lin Jan 7 '13 at 3:36
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A real matrix is diagonalizable (over $\mathbb{C}$) by a unitary matrix if and only if it is normal.

But there are non-normal matrices that can be diagonalized by non-unitary matrices. For instance

$$\left[\begin{array}{cc}0 & 1\\2 & 1\end{array}\right] = PDP^{-1}$$ where $$P = \left[\begin{array}{cc} -1 & 1\\ 1& 2\end{array}\right]$$ and $$D=\left[\begin{array}{cc} -1 &0\\0 & 2\end{array}\right].$$

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    $\begingroup$ Then what's the geometrical meanings of real diagonalizable matrices and real normal matrices? $\endgroup$ – chaohuang Jan 7 '13 at 4:25

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