# Jordan canonical form of matrix over arbitrary field

Let V be a finite dimensional vector space over a field $\mathbb{K}$, and $T \in \text{End}(V)$. For an ordered basis $\mathcal{B}$, does the matrix $[T]_\mathcal{B}$ always have a Jordan form, or must $\mathbb{K}$ required to be algebraically closed?

If not, for $\mathbb{K}=\mathbb{R}$ and $V=\mathbb{R}^2$, what would be the Jordan form of a rotation matrix

$R = \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\\ \end{pmatrix}$

as it has no eigenvalues in $\mathbb{R}$ ?

• $K$ must be large enough to contain the eigenvalues of $T$. – fredgoodman Apr 14 '18 at 18:06

It is easy to check that a Jordan form has its eigenvalues on its diagonal. And that the Jordan form of an $n\times n$ matrix $A$ has the same eigenvalues as $A$. In conclusion, to have a Jordan form, you need all the eigenvalues to exist in field. More explicitly, you need to be able to factor the characteristic polynomial of $A$ as $p(t)=(t-\lambda_1)\cdots(t-\lambda_n)$.
As you mention, the matrix $$\begin{bmatrix}0&1\\-1&0\end{bmatrix}$$ has no Jordan form over $\mathbb R$.