Let $V$ be an $n$-dimensional. If $T: V\rightarrow V$ is diagonalizable with $n$ distinct eigenvalues, then $V$ is cyclic as a $K[x]$ module. Problem: Let $V$ be an $n$-dimensional vector space over a field $K$. Suppose that $T:V \rightarrow V$ is diagonalizable with $n$ distinct eigenvalues. Then I wish to show that $V$ is cyclic as a $K[x]$ module.
I know that by Cayley-Hamilton that $p(char(x))(v)=0$ for all $v \in V$ and that for any $v_i$ that $Ann(v_i) \subset Ann(V)$ but I am not sure how to get the desired conclusion. Hints appreciated.
Edit: In particular it seems that I would have to show that $Ann(v_i)$ could not have all have degree less than $n$  as well.
 A: Let $v_i$ be an eigenvector for $\lambda_i$ then $\sum v_i$ will generate $V$ as a $K[x]$ module. This is because the action of $x^k$ sends this vector to 
$$\sum \lambda^k_iv_i$$ and these vectors are linearly independent (for $0\leq k\leq n-1$) from the Vandermond matrix, here we use the fact that the eigenvalues are distinct.
A: One way to prove this is by using the invariant factor decomposition. The minimal polynomial is the largest invariant factor, while the characteristic polynomial is the product of the invariant factors; since both polynomials are equal here (both are $(X-\lambda_1)\ldots(X-\lambda_n)$) there is just one invariant factor and the module is cyclic.

If you want to avoid this somewhat heavy theorem, another way is to classify the submodules of this $K[X]$ module (in other words the $T$-invariant subspaces). The minimal polynomial of a submodule must be a divisor of the global minimal polynomial $(X-\lambda_1)\ldots(X-\lambda_n)$, from which it follows that any submodule is the sum of a certain subset of the eigenspaces (they are precisely classified by the $2^n$ monic divisors of $(X-\lambda_1)\ldots(X-\lambda_n)$). Then any vector with a nonzero projection on each of the eigenspaces is outside any proper submodule; it follows that it is a cyclic vector.
