Hoffman and Kunze Section 6.3 Exercise 5 (Assumption of Algebraic Closure?)

Question

I'm having trouble with one of the exercises in Hoffman and Kunze's Linear Algebra (Section 6.4, Exercise 5):

Let $V$ be an $n$-dimensional vector space and let $T$ be a linear operator on $V$. Suppose that there exists some positive integer $k$ so that $T^k=0$. Prove that $T^n=0$.

My initial thought was that the minimal polynomial of $T$ must divide $x^k$, hence must be of the form $x^m$ for some $m\leq k$. Now, since the minimal polynomial and the characteristic polynomial of $T$ share the same roots, the only scalar roots of the characteristic polynomial must be $0$. It follows that the characteristic polynomial of $T$ must be $x^n$. Hence by Cayley-Hamilton, we must have $T^n=0$.

However, the above argument assumes that $V$ is defined over an algebraically closed field. Indeed, the characteristic polynomial of $T$ might be of the form $x^m p(x)$ for some irreducible polynomial $p$. Although one could fix this argument by considering the algebraic closure of the field which $V$ is defined over, I don't think that this is the intended solution, since algebraic closure is mentioned nowhere in the book.

Is there any solution which gets around using the fact that every field has an algebraic closure? If so, I would appreciate a hint nudging me in the right direction. Or perhaps is the question implicitly assuming that $V$ is defined over $\mathbb{R}$? (Hoffman and Kunze mention that $\mathbb{C}$ is algebraically closed.)

Thanks for the help!

Answer 1

Neither the algebraic closure nor the characteristic polynomial are needed here. It suffices to consider the sequence of subspaces $V_i=\ker(T^i)$ for $i=0,1,2,\ldots$. One has ($V_0=V$ and) $V_i\supseteq V_{i+1}$ for every $i$, and it is given that set sequence attains the zero-dimensional subspace at some index $k$; the task is to show that already $\dim V_n=0$. The key point is that $V_i= V_{i+1}$ implies $V_{i+1}= V_{i+2}$ (since $v\in V_{i+1}\setminus V_{i+2}$ would imply $T(v)\in V_i\setminus V_{i+1}$) and therefore that the sequence stabilises from that point on, which would contradict the hypothesis $\dim V_k=0$ unless already $\dim V_i=0$. Then simple consideration of the dimension shows that dimension$~0$ can be obtained in at most $n=\dim V$ steps.

A slightly more refined argument shows that $T$ induces an injective linear map of quotient spaces $ V_{i+1}/V_{i+2}\to V_i/V_{i+1}$, so that the sequence of dimensions is convex (the amount of decrease weakly decreases, whence no term is greater than the average of its neighbours).