Complex diagonalizable matrices with given eigenvalues can be conveniently parametrized as $A=T^{1} \Lambda T$, where $T$ is any invertible matrix, and $\Lambda=diag(\lambda_1,...,\lambda_N)$ with $\lambda_1,...,\lambda_N$ the desired eigenvalues. The same parametrisation is not available for real diagonalizable matrices: If $\Lambda$ has complex conjugate eigenvalues, $T$ cannot be chosen arbitrarily among invertible matrices (e.g. it has to have at least some complex entries such that $T A T^{1}= \Lambda$ is complex). Hence the questionis there an elegant, full parametrisation of real diagonalizable matrices with given eigenvalues?

2$\begingroup$ Yes, there is, and it was known to Frobenius if my memory is correct. Replace $2 \times 2$ diagonal matrix with eigenvalues $re^{i\theta}$ and $r e^{i \theta}$ by the matrix $\left( \begin{array}{clcr} r\cos \theta & r\sin \theta\\r\sin\theta & r\cos \theta \end{array} \right). $ $\endgroup$– Geoff RobinsonMay 4, 2014 at 23:45

$\begingroup$ Thanks, Geoff! Would you remember a reference for this? That matrix looks like a Givens rotation. Can I use Givens rotations for higher dimensional parametrizations? $\endgroup$– OzzMay 4, 2014 at 23:55

$\begingroup$ Yes, I meant to indicate that whenever you have the complex diagonal form of the matrix, arranged so that complex conjugate eigenvalues succeed each other, you can replace each $2 \times 2$ diagonal subblock corresponding to those eigenvalues with the $2 \times 2$ corresponding scaled rotation matrix. This works for diagonalizable matrices of any size. $\endgroup$– Geoff RobinsonMay 5, 2014 at 0:01

1$\begingroup$ I think it was known to Frobenius in the sense that he knew how to reduce to that form directly with real matrices without using complex matrices. If you use complex matrices, it's elementary linear algebra to see that it suffices to do the $2$dimensional case, which is fairly easy to do directly. $\endgroup$– Geoff RobinsonMay 5, 2014 at 0:16

$\begingroup$ @Geoff: Indeed, up to change of orthonormal basis, it suffices to take $\theta=\pi/2$. $\endgroup$– Nate EldredgeMay 5, 2014 at 0:25
1 Answer
This is elementary, but to take a slightly different line from what I said in the comments, the theory of the rational canonical form allows us to diagonalize a real square matrix with multiplicity free minimum polynomial by conjugating by an invertible real matrix $T.$ The only irreducible monic polynomials in $\mathbb{R}[x]$ are linear, or quadratic with a pair of complex conjugate roots. Each irreducible factor of degree $d$ of the characteristic polynomial of the matrix gives rise to $d \times d$ block. If $d = 1,$ the block is obvious. If $d = 2,$ and the monic factor is $x^{2} 2rx \cos \theta + r^{2}$ for positive real $r$ and $0 < \theta < 2 \pi,$ then the corresponding $2 \times 2$ block is $\left( \begin{array}{clcr} 0 & 1\\ r^{2} & 2r\cos \theta \end{array} \right)$ which is a real matrix with eigenvalues $re^{ i \theta}$ and $re^{i \theta}$ and is the companion matrix to the polynomial $x^{2}  2rx \cos \theta + r^{2}.$