# Change basis to get Jordan canonical form

• Dec 9th 2010, 07:21 AM
davesface
Change basis to get Jordan canonical form
In some basis $\displaystyle e_1,e_2,...,e_n$ the matrix of an operator A is an n x n lower triangular matrix with diagonal entries $\displaystyle \lambda$, subdiagonal entries 1, and 0 for all other entries. In what basis does it have Jordan canonical form?

It seems like you could just reverse the order of the basis (ie, $\displaystyle e_n,e_{n-1},...,e_1$). Is it enough to use the fact that $\displaystyle A^T$ is in the required form to prove that this is the correct basis?
• Dec 9th 2010, 08:40 AM
FernandoRevilla
Quote:

Originally Posted by davesface
It seems like you could just reverse the order of the basis (ie, $\displaystyle e_n,e_{n-1},...,e_1$). Is it enough to use the fact that $\displaystyle A^T$ is in the required form to prove that this is the correct basis?

For example,

$\displaystyle B=(e_1,e_2,e_3),\quad B^*=(e_3,e_2,e_1)$

then,

$\displaystyle \begin{Bmatrix}Ae_1=\lambda e_1+e_2\\Ae_2=\lambda e_2+e_3\\Ae_3=\lambda e_3\end{matrix}\Rightarrow [A]_B=\begin{bmatrix}{\lambda}&{0}&{0}\\{1}&{\lambda} &{0}\\{0}&{1}&{\lambda}\end{bmatrix}$

$\displaystyle \begin{Bmatrix}Ae_3=\lambda e_3\\Ae_2=e_3+\lambda e_2\\Ae_1=e_2+\lambda e_1\end{matrix}\Rightarrow [A]_{B^*}=\begin{bmatrix}{\lambda}&{1}&{0}\\{0}&{\lam bda}&{1}\\{0}&{0}&{\lambda}\end{bmatrix}$

It is easy to generalize.

Regards.

Fernando Revilla