Suppose that $f: V \to V$ is a $k$-linear transformation such that $f^m = 0$ for some integer $m.$ Prove that $f^n = 0.$

Question

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!

Answer 1

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 $f^n = f^{n-k}f^k=0$.

Answer 2

Just for fun; here is an elementary proof not using any minimal polynomial/characteristic polynomial stuff. We may assume $m>n$. Suppose for contradiction that $f^n(V) \neq 0$. Consider the following sequence of subspaces of $V$: $$f^n(V) \subseteq f^{n-1}(V) \subseteq .. \subseteq f(V) \subseteq V.$$ This sequence consists of $n+1$ subspaces, and the dimension of $f^n(V)$ is at least $1$. So there must be $i \in \{0,.., n-1\}$ such that $f^i(V) = f^{i+1}(V)$ since else the dimensions of the above chain would give $n+1$ distinct numbers in $\{1,..,n \}$. But then clearly $f^i = f^{i+1}$, so by induction $f^j = f^{j+1}$ for every $j \geq i$. In particular, $f^n = f^m = 0$, contradiction.

Answer 3

Here is an alternative way.

Denoting "$\rm null$ space of " by $\rm N$,

$\rm T^m=0\implies N(T^m)=V $.

Using the following lemma, it follows that the given statement is true for $\rm m\lt n$ and trivially true for $\rm m=n$.

Lemma $\rm 1$: $\rm N(T^i)$ grows as $\rm i\in \mathbb N$ increases, i.e,
$\rm N(T)\subset N(T^2)\subset \cdots$

Proof: For any $\rm i<j$, $\rm \forall x\in N(T^i), T^i(x)=0\implies T^{j-i}(T^i(x))=0\implies T^j(x)=0$.

$\rm \implies x\in N(T^j)\implies N(T^i)\subset N(T^j). \quad\quad\square$

Lemma $\rm 2$: If there exists an $\rm i\in \mathbb N$ such that $\rm N(T^i)=N(T^{i+1})$ then, $\rm N(T^i)=N(T^j) \quad\forall j\ge i.$

Proof: Suppose that for some $\rm i\in \mathbb N$, $\rm N(T^i)=N(T^{i+1})$. By lemma $\rm 1$, it suffices to show that $\rm N(T^{j})\subset N(T^i)\quad \forall j\ge i.$

Since lemma $\rm 2$ is true for $\rm j=i,i+1$, suppose that $\rm j>i+1.$

$\rm \forall x\in N(T^{i+2}), 0=T^{i+2}(x)=T(T^{i+1}(x))=T(T^i(x))=T^{i+1}(x)\implies x\in N(T^{i+1})$.

$\rm \implies N(T^{i+2})\subset N(T^{i+1})\implies N(T^{i+2})=N(T^{i+1})$. It follows by induction that $\rm N(T^{i+1+r})=N(T^{i+1})=N(T^i)\quad \forall r\in \mathbb N. \quad \square$

Suppose that $m>n$. Suppose on the contrary that $\rm T^n\not\equiv 0$. It follows that $\rm N(T^n)\ne V$. By lemma $\rm 2, N(T)\subsetneq N(T^2)\subsetneq N(T^2)\subsetneq \cdots \subsetneq N(T^n)$. Choose any $\rm v_1\in N(T)$ and for every $\rm i\gt 1$, choose $\rm v_i\in N(T^i)$ such that $\rm v_i\notin N(T^{i-1}).$

Note that the set of $\rm v_i$'s is a basis of $\rm V$. For any $\rm y\in V$ there exist $\rm k_i$'s in $\rm K$ such that $\rm y=k_1v_1+\cdots +k_nv_n\implies T^n(y)=0\implies T^n\equiv 0$, which contradicts the assumption.

So $\rm T^n\equiv 0$.

Answer 4

Here's another proof without using the minimal polynomial, but only using properties of the characteristic polynomial. First assume $k$ is algebraically closed. Then all eigenvectors of $f$ have eigenvalue $0$, simply because $fv = \lambda v$ implies $f^m v = \lambda^m v$. So the characteristic polynomial of $f$ is just $X^n$. By Cayley-Hamilton the result follows. Now if $k$ is not algebraically closed, we can see $f$ as a map from a vector space over an algebraic closure $K$ of $k$, either by extension of scalars or by choosing a basis and replacing $k^n$ by $K^n$. Under this procedure the minimal polynomial obviously stays the same. Then by the result above we get $f^n = 0$ also.