Jordan normal form theorem - a question about the proof I've found

225 Views Asked by At

I've been reading this proof of Jordan's theorem: http://www.cs.uleth.ca/~holzmann/notes/jordan.pdf

and there are a few questions I hope you could answer for me.

Firstly, why $(A_{\lambda})_{\mu _i} = A _{\lambda + \mu _{i}} =: A _{\lambda_{ \ i}}$

and secondly, how do we know that $\dim (W \cap Ker A_{\lambda}) \ge 1$?

The problem in the first question is that I'm not sure what $(A_{\lambda})_{\mu _i}$ means.

Thank you.

I understand that we pick a vector $v\neq 0$ with an eigenvalue $\lambda$. $A$ is $n \times n$

$A_{\lambda} = A - \lambda I \ \ \ \ \ \ \ \ \ A_{\lambda}v=0,$ so we have $r = \dim \ker A_{\lambda}>0$

$\dim \ range A_{\lambda} = n-r<n $ , $W := range \ A_{\lambda} $

I undestand that because $W := range \ A_{\lambda} $, then $ A_{\lambda} (W) \subset W$, so $A_{\lambda} $ induces $T: W \rightarrow W$ and we can apply inductive hipothesis to its matrix. What I don't understand is why all of a sudden there are $\lambda_i$ over the arrows and it is probably because I don't know what $(A_{\lambda})_{\mu _i}$ means

2

There are 2 best solutions below

3
On BEST ANSWER

For the first, these are just successive $B_x:=B-xI$ operations, so $_{\mu_i}$ applied to $A_\lambda$. That $(A_\lambda)_{\mu_i}$ is $A_{\lambda+\mu_i}$ follows from definition.

For the second, I don't think we need that at all (and the author doesn't seem to claim anything of the sort). In fact, it might not be true, as $A_\lambda$ can very well be a projection onto its range. Consider the matrix $$ \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} $$ with $\lambda=1$, for instance.

0
On

According to that paper

$$A_\lambda:=A-\lambda I\implies\left(A_\lambda\right)_{\mu_i}:=(A-\lambda I)_{\mu_i}=(A-\lambda I)-\mu_iI=A-(\lambda+\mu_i)I=:A_{\lambda+\mu_i}$$