For my exam Algebra I have to study the proof for the Jordan Normal Form in the module theory. It says:
Let $F$ be an algebraically closed field and $V$ a $F$-vector space (dimension is finite), and let $f \space \epsilon \space End_F(V)$ Look at the submodule $ W \cong F[x] / ((x- \lambda)^{\alpha})) $ Then there is a basis from $W$ such that the matrix for the linear operator $f$ over $W$ is a Jordan Matrix.
In the proof it says:
Let $q = (x- \lambda)^\alpha$ and choose $w = 1 + (q) \space \epsilon \space W$ then the elements $w_1 = w$, $w_2 = (x-\lambda)\cdot w$,..., $w_\alpha = (x-\lambda)^{\alpha-1} \cdot w$ are a basis for $W$ over $F$. We also see that $ x\cdot w_i = \lambda w_i + w_{i+1}$. Then the matrix A for $x$ is in Jordan Normal Form.
This is really hard for me to understand, so I have some questions about this:
- Why are $w_1$ etc elements of the basis, so why is it a basis?
- Why is A in Jordan Normal Form?
Note: Sorry for the long message, just been cracking my head on this and just can't figure it out.