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 $\mathbb{R}$ ?

  • 2
    $\begingroup$ $K$ must be large enough to contain the eigenvalues of $T$. $\endgroup$ – 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 $\mathbb R$.


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