Vtyjkyuk

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 :)

Consider the map $TcolonmathbbR^2longrightarrowmathbbR^2$ defined by $T(x,y)=(-y,x)$. There is no basis of $mathbbR^2$ 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 $UcolonmathbbR^3longrightarrowmathbbR^3$ defined by $U(x,y,z)=(-y,x,0)$ has one eigenvalue ($0$), but no Jordan form.