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I'm trying to find out a normal, real and $\boldsymbol n\times \boldsymbol n$ ($n\ge3$) matrix $A$ which is diagonalize over $C$ but isn't diagonalize over $R$.

I know that the following matrix (a.k.a the rotating matrix) satisfies the conditions above and isn't diagonalize over $R$:

$$A=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$$

But even within this example in my mind, I can't find such a matrix (satisfies the conditions above and isn't diagonalize over $R$) for $n\ge3$.

I'm almost sure there exists such a matrix. Can please someone give me an hint on how to establish such a matrix?

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Hint: come up with a matrix that acts like $A$ on vectors of the form $(x,y,0)$ and does something different for $(0,0,z)$ (you can have $(0,0,z) \mapsto (0,0,z)$ or $(0,0,0)$ or whatever; you don't need to get too fancy here).

$$A = \begin{pmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix} \textrm{ or } A = \begin{pmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}$$ will work

Basically, you can satisfy all the conditions by constructing an anti-symmetric (implies normal, implies diagonalizable over $\Bbb C$) matrix with at least one pair of complex eigenvalues (complex eigenvalues always come in conjugate pairs if the original entries are real; thus in odd dimensions, you are always going to have at least one real eigenvalue, but this is fine).

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  • $\begingroup$ Thank you for the detailed answer. Why anti-symmetric matrix implies normal? $\endgroup$ – SyndicatorBBB Jan 6 '15 at 18:14
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    $\begingroup$ @SyndicatorBBB If $A=-A^T$, then $AA^T=-AA=A^TA$. $\endgroup$ – Algebraic Pavel Jan 6 '15 at 18:16
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Another way how to construct a real normal matrix, which is not diagonalizable over $\mathbb{R}$, can be done as follows: take a block diagonal matrix $D$ $$ D=\pmatrix{D_1&&\\&\ddots&\\&&D_k}, $$ where the real diagonal blocks have either size $1$ or $2$ with the latter of the form $$\tag{1} D_k=\pmatrix{\alpha_k&\beta_k\\-\beta_k&\alpha_k} $$ with $\beta_k\neq 0$. The spectrum of $D$ consists of the union of the spectra of the blocks $D_k$. The $1\times 1$ blocks give the real eigenvalues and the $2\times 2$ blocks (1) give the complex eigenvalues $\alpha_k\pm i\beta_k$. Now "mix" $D$ with any real orthogonal matrix to get a real normal matrix with complex eigenvalues.

In fact, any normal real matrix, which is not diagonalizable over $\mathbb{R}$, can be constructed in this way. That is, given a real normal matrix $A$, there is a real orthogonal matrix $Q$ and a real block diagonal matrix $D$ with $1\times 1$ and $2\times 2$ blocks of the form (1) such that $A=QDQ^T$ (you might search for the term real Schur form).

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