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I am reading about the Jordan Normal Form for matrices, and am struggling with how to show that a null space is always a cyclic module - at least in this particular example. By Theorem 9.6 Chapter 5 in Serge Lang's Undergraduate Algebra:

"Let V be a finite dimensional space over the algebraically closed field K, and $V\neq{0}$. Let $A:V\rightarrow{V}$ be an endomorphism. Then V is the direct sum of A-invariant subspaces $V=V_1\oplus{...}\oplus{V_k}$ (define $V_\lambda=Null[(A-\lambda{I})^{r_i}]$ for $\lambda$ eigenvalue) such that each $V_i$ is cyclic, generated over $K[t]$ by an element $v_i\neq{0}$ and the kernel of the map $f(t)\mapsto{f{(A)}v_i}$ is a power $(t-\lambda_i)^{r_i}$ for some positive integer $r_i$ and $\lambda_i\in{K}$."

I understand everything in this theorem statement apart from the fact that each $V_i$ is cyclic. Here, Lang views vector spaces as modules over $K[t]$ (useful as principal ring), defining $f(t)v=f(A)v$, and says a module $V_A$ is cyclic iff there exists an element $v\in{V}$, $v\neq{0}$ such that every element of V is of the form $f(A)v$ for some polynomial $f(t)\in{K[t]}$. How would I show that this is the case for each $V_i$ in the theorem. Cheers.

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You define each $V_i$ to be cyclic. Actually perhaps a better way to phrase this is the following : given an eigenvalue $\lambda$, denote by $V_\lambda$ the associated space (with respect to the characteristic polynomial, not the minimal polynomial).

Then $A-\lambda I$ is nilpotent on $V_\lambda$ and so $V_\lambda$ can be decomposed as $\bigoplus_i V_\lambda^i$ where on each $V_\lambda^i$, $A-\lambda I$ is nilpotent of maximal rank. Then on each of these, $A$ has a matrix representation with $\lambda$'s on the diagonal and $1$'s right behind them. It's easy to see then that on each of these spaces, $A$ is cyclic.

In general, $V_\lambda$ need not be cyclic : take $A=\lambda I$ as a counterexample. So it's important to take a further decomposition.

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