I saw in the book "Linear algebra done right" (by S. Axler) that all complex operator has a Jordan form. The proof is based on the fact that all complex operator is upper triangular (i.e. there is a basis s.t. the matrix is upper triangular). My questions are the following :

  • 1) Could you give me an example of operator that is not upper triangular (I guess such an operator has no eigenvalue... may be an operator with $x^2+1$ as polynomial characteristic ?)

  • 2) If an operator has at least one eigenvalue, does it has a Jordan form ? (in the proof of Axler it's indeed based on the fact that it has an eigenvalue, but also on the fact that the characteristic polynomial is of the form $(x-\lambda _1)^{m_1}...(x-\lambda _n)^{m_n}$ (btw how do you call such a form ? in french we say "scindé" but I didn't find an english equivalent on wikipedia). So I guess that a characteristic polynomial of the form $x^2+x+1$ will not have a Jordan form even not a upper triangular form... but these are just supposition, and if anyone can confirm or not, I would be very happe :)

  • $\begingroup$ I think that "polynôme scindé" in french means "polynomial splits into linear factors" in englich. $\endgroup$ – user296113 Aug 8 '18 at 20:47
  • $\begingroup$ I guess I would emphasize that a real matrix with complex eigenvalues does possess a Jordan form if we allow complex numbers. This would cover your $x^2 + x + 1$ example, trace $-1$ and determinant $1$ will do it, so $$ \left( \begin{array}{cc} 0 & 1 \\ -1 & -1 \end{array} \right) $$ $\endgroup$ – Will Jagy Aug 8 '18 at 21:10

Consider the map $T\colon\mathbb{R}^2\longrightarrow\mathbb{R}^2$ defined by $T(x,y)=(-y,x)$. There is no basis of $\mathbb{R}^2$ such that the matrix of $T$ with respect to that basis is upper triangular. Note that, as you suspected, $T$ has no (real) eigenvalues.

And the map $U\colon\mathbb{R}^3\longrightarrow\mathbb{R}^3$ defined by $U(x,y,z)=(-y,x,0)$ has one eigenvalue ($0$), but no Jordan form.

  • $\begingroup$ So Will Jaggy was not correct ? (because he look to say the opposite, i.e. that all endomorphism has a jordan form) $\endgroup$ – user352653 Aug 8 '18 at 22:16
  • $\begingroup$ @user352653 It has no Jordan form over the reals. Every complex endomorphism of a finite-dimensional complex vector space has a Jordan form over the complex field. $\endgroup$ – José Carlos Santos Aug 8 '18 at 22:19
  • $\begingroup$ So do you agree with the fact that a matrix with polynomial characteristic $x^2+x+1$ has no jordan form, right ? $\endgroup$ – user352653 Aug 9 '18 at 9:39
  • $\begingroup$ @user352653 It has no Jordan form with real entries. $\endgroup$ – José Carlos Santos Aug 9 '18 at 10:33

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