jordan block question

**transgalactic** · October 4th 2011, 07:25 AM

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

**FernandoRevilla** · October 4th 2011, 09:56 AM

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

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

$J$ is the Jordan form of $A$ , $K$ is a diagonal block matrix similar to $A$ but it is not its Jordan form.

**transgalactic** · October 4th 2011, 01:08 PM

because of the 2 ,3 on the smaller diagonal
?

**FernandoRevilla** · October 4th 2011, 10:50 PM

Originally Posted by transgalactic

because of the 2 ,3 on the smaller diagonal?

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

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