2
$\begingroup$

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 $\mathbb{R}$ ?

$\endgroup$
  • 2
    $\begingroup$ $K$ must be large enough to contain the eigenvalues of $T$. $\endgroup$ – fredgoodman Apr 14 '18 at 18:06
1
$\begingroup$

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 $\mathbb R$.

$\endgroup$

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.