Here is the question I want to tackle:
Let $k$ be a field and let $V$ be an $n$-dimensional vector space over $k.$ Suppose that $f: V \to V$ is a $k$-linear transformation such that $f^m = 0$ for some integer $m.$ Prove that $f^n = 0.$
Here is a solution I found here:
The minimal polynomial divides $t^m$, hence has the form $t^k$ for some $k$. But the characteristic polynomial has degree $n$ and divides a power of the minimal polynomial, so it has to be $t^n$. By Cayley-Hamilton, $T^n =0$.
But I do not understand it. I know that the minimal polynomial $\mu_{A}(x)$ is the smallest degree monic polyniomial such that if we have $p(x) \in \mathbb C[x]$ such that $P(A) = 0$ where A is the matrix corresponding to the linear transformation then $\mu_A(x) | p(x)$ and I know that the degree f the characteristic polynomial $\chi_{A}(x)$ is $n$ and I know that $\mu_A(x) | \chi_{A}(x)$ and I know that by Cayley Hamilton Theorem $\chi_A(A) =0$ but still I am unable to organize these information in a logical way to come up with the solution. Could some one help me in organizing this ideas to come up with the proof?
Thanks!
Best Answer
The proof you cite looks somewhat awkward to me. Instead I would say (assuming you want to use polynomials): The minimal polynomial divides $t^m$ so is given by $t^k$ for some $k\leq m$. The minimal polynomial also divides the characteristic polynomial which is of degree $n$. Whence $k\leq n$ and $f^n = f^{n-k}f^k=0$.