# jordan block question

• Oct 4th 2011, 07:25 AM
transgalactic
jordan block question
what is the difference between the jordan form of matrices A
and diagonal block matrices similar to A
?
• Oct 4th 2011, 09:56 AM
FernandoRevilla
Re: jordan block question
Quote:

Originally Posted by transgalactic
what is the difference between the jordan form of matrices A and diagonal block matrices similar to A?

Difference, in what terms? For example

$\displaystyle A\sim J=\begin{bmatrix}{\lambda}&{1}&{0}\\{0}&{\lambda}& {1}\\{0}&{0}&{\lambda}\end{bmatrix}\Rightarrow A\sim K=\begin{bmatrix}{\lambda}&{2}&{0}\\{0}&{\lambda}& {3}\\{0}&{0}&{\lambda}\end{bmatrix}$

$\displaystyle J$ is the Jordan form of $\displaystyle A$, $\displaystyle K$ is a diagonal block matrix similar to $\displaystyle A$ but it is not its Jordan form.
• Oct 4th 2011, 01:08 PM
transgalactic
Re: jordan block question
because of the 2 ,3 on the smaller diagonal
?
• Oct 4th 2011, 10:50 PM
FernandoRevilla
Re: jordan block question
Quote:

Originally Posted by transgalactic
because of the 2 ,3 on the smaller diagonal?

It is irrelevant, only an example. The important thing is that $\displaystyle \dim \ker (K-\lambda I)=1$ so, the Jordan form of $\displaystyle K$ is $\displaystyle J$ .