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 $R$:


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?


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).

  • $\begingroup$ Thank you for the detailed answer. Why anti-symmetric matrix implies normal? $\endgroup$ – SyndicatorBBB Jan 6 '15 at 18:14
  • 1
    $\begingroup$ @SyndicatorBBB If $A=-A^T$, then $AA^T=-AA=A^TA$. $\endgroup$ – Algebraic Pavel Jan 6 '15 at 18:16

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.