If I have real valued matrix $A$, are these two notions of being Diagonalizable and being Normal equivalent?
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2No. Any normal matrix is diagonalizable. The converse is not true. – Artem Jan 07 '13 at 01:19
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1@Artem Isn't $\begin{pmatrix} 0 & 1 \ -1 & 0 \ \end{pmatrix}$ a Normal matrix that isn't diagonalizable over the reals? – Calvin Lin Jan 07 '13 at 01:27
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@CalvinLin, the usual meaning is "diagonalizable over the complex". – DonAntonio Jan 07 '13 at 02:27
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1@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. – Calvin Lin Jan 07 '13 at 03:36
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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 $$D=\left[\begin{array}{cc} -1 &0\\0 & 2\end{array}\right].$$
user7530
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2Then what's the geometrical meanings of real diagonalizable matrices and real normal matrices? – chaohuang Jan 07 '13 at 04:25