Let $ \ V \ $ be a finite dimensional vector space over the field $ \ \mathbb{C} \ $ and $ \ T: V \to V \ $ be a linear transformation.
Suppose $ \ \lambda \ $ is a Eigenvalue of $ \ T \ $.
Let the characteristics polynomial and minimal polynomial of $ \ T \ $ be respectively given by $ \ \chi(\lambda)=(z-\lambda)^m \ $ and $ \ m(\lambda)=z-\lambda \ $ .
If $ \ \vec v \ $ be the Jordan basis of $ \ T \ $ , then find $ \ dim \ E(\lambda , \ \vec v) \ $ or the dimension of the Eigen-space corresponding to the Eigen value $ \ \lambda \ $ with respect to $ \ \vec v \ $.
Also find the Jordan matrix $ \ \mathcal{M} (T) \ $
Answer:
Since the characteristic polynomial of $ \ T \ $ is $ \ \chi(\lambda)=(z-\lambda)^m \ $ , we have $ \ dim (V)=m \ $ ,
Now the algebraic multiplicity of $ \ \lambda \ $ is $ \ m \ $.
Since the minimal polynomial of $ \ T \ $ is $ \ m(\lambda)=z-\lambda \ $ , the geometric multiplicity of $ \ \lambda \ $ is also $ \ m \ $.
Thus the dimension of the Eigen space $ \ E(\lambda) \ $ of $ \ \lambda \ $ is given by
$ dim \ E(\lambda)=m \ $
Am I right so far?
So there is a jordan block of size $ \ m \ $ as follows:
$$ \mathcal{M}(T)=\begin{pmatrix} \lambda & 0 & 0 & \cdots & 0 \\ 0 & \lambda & 0 & \cdots & 0 \\ \vdots & \vdots & & & \vdots \\ 0 & 0 & & \cdots & \lambda \end{pmatrix} $$
Am I right so far?
Yes, you are right. For justification, see Wikipedia subtopic "Complex matrices" in the "Jordan normal form" article.