For an n x n matrix T, prove that if $[T]^k=[0]_n$, then $[T]^n=[O]_n$

Question

Let O be the zero map $O: \mathbb{R}^n \to \mathbb{R}^n$, and let T be a linear map $T: \mathbb{R}^n \to \mathbb{R}^n$. Show that if $T^k=O$ for some k, then $T^n=O$ $\;$(where the exponentiation means composition $T \circ T \circ ... \circ T$ k times).

I'm trying to show that for some $i$, \begin{align*} &\{\vec{0}\} \subset \text{range}(T^i) \subset \text{range}(T^{i-1}) \subset \dots \subset \text{range}(T^2) \subset \text{range}(T) \subset \mathbb{R}^n\\ & \text{so } i \leq n, \; T^i \circ T^{n-i}=T^n=O\\ \end{align*}

I think this is on the right track... but I'm kinda of lost. I can't assume [T] is nilpotent. And I'm trying to do this without using the characteristic polynomial of the [T]. I tried using this as guidance How to show that the nth power of a $nxn$ nilpotent matrix equals to zero $A^n=0$ but I'm still confused on how to go about this.

Answer 1

We see that $\text{range}(T)$ is a proper subspace of $\Bbb R^n$ or else $T$ is not nilpotent.

Note that for each positive integer $r$, $T(\text{range}(T^r))\subseteq \text{range}(T^r)$ (can you prove this?). We also have $\text{range}(T^{r+1})=T(\text{range}(T^r))$. Hence similarly, for each positive integer $r$, if $\text{range}(T^r)$ is not zero subspace, then $T(\text{range}(T^r))$ must be a proper subspace of $\text{range}(T^r)$ or else it contradicts with the assumption that $T$ is nilpotent.

The descending chain $\Bbb R^n\supsetneq\text{range}(T)\supsetneq\text{range}(T^2)...$ of subspaces has strictly decreasing dimensions until the dimension reaches $0$. The dimension start from $n$ and starts decreasing by at least $1$ each step, and must reach $0$ for $\text{range}(T^n)$. If the dimension reaches $0$ earlier for some $r\lt n$, i.e. $\text{range}(T^r)$ is already dimension $0$, then it is more trivial.

Answer 2

Since $T^k=0$, the mimimal polynomial for $T$ in $\mathbb{R}[x]$ must divide $x^k$, hence must be of the form $P(x) = x^j$, for some $j \le k$.

But the degree of $P$ is at most $n$, hence $$P(T) = 0 \implies T^j = 0 \implies T^n=0$$ Alternatively, you can argue as follows . . .

Suppose $T^k=0$, for some positive integer $k$.

Then $T$ is singular, so $\text{im}(T)$ has dimension at most $n-1$.

Consider the descending chain $$\text{im}(T) \supseteq \text{im}(T^2) \supseteq \text{im}(T^3) \supseteq \cdots$$ of subspaces of $\mathbb{R}^n$.

If at any stage, two consecutive subspaces are equal, the chain becomes constant from that point on, hence must be the constant $0$ (since $T^k = 0$).

But if $T^n$ is nonzero, the chain $$\text{im}(T) \supseteq \cdots \supseteq \text{im}(T^n)$$ must be strictly descending, with all dimensions greater than $0$, which is impossible, since there are $n$ terms and $\text{im}(T)$ has dimension at most $n-1$.

It follows that $T^n=0$.

Answer 3

If $k = n$, the size of $T$, we are done.

If $1 \le k < n$, we have

$T^n = T^{n - k} T^k = T^{n - k}(0) = 0, \tag{0}$

and we are done.

Now if $k > n$, we examine the eigenvalues $\lambda$ of $T$, allowing if necessary $\lambda \in \Bbb C$, the complex number field. If

$T\vec v = \lambda \vec v, \tag{1}$

for some non-zero $\vec v$, where in the case $\lambda \in \Bbb C$ we allow $\vec v \in \Bbb C^n$, then we must have

$\lambda = 0, \tag{2}$

since (1) implies

$T^k \vec v = \lambda^k \vec v \tag{3}$

for any positive integer $k$; note:

$T^2 \vec v = T(T\vec v) = T(\lambda \vec v) = \lambda T \vec v = \lambda^2 \vec v, \tag{4}$

and if

$T^m \vec v = \lambda^m \vec v \tag{5}$

for some positive integer $m$, then

$T^{m + 1} \vec v = T(T^m \vec v) = T(\lambda^m \vec v) = \lambda^m T \vec v = \lambda^m \lambda \vec v = \lambda^{m + 1} \vec v; \tag{6}$

this simple induction establishes (3), and now if

$T^k = 0, \tag{7}$

we have

$\lambda^k \vec v = T^k \vec v = 0, \tag{8}$

forcing

$\lambda^k = 0, \tag{9}$

and in turn

$\lambda = 0; \tag{10}$

so the eigenvalues of $T$ must all be zero.

We now exploit the Schur decomposition of $T$, which asserts that $T$ may be written in the form

$T = U^\dagger R U \tag{11}$

with $U$ unitary and $R$ upper triangular, with the eigenvalues of $T$ along its diagonal. But since the eigenvalues of $T$ are all zero, $R$ is in fact strictly upper triangular, and it is well known, and in fact quite easy to see, that a strictly upper triangular matrix of size $n$, such as $R$, satisfies

$R^n = 0; \tag{12}$

thus, by virtue of the fact that

$U^\dagger U = UU^\dagger = I, \tag{13}$

that is,

$U^\dagger = U^{-1}, \tag{14}$

we conclude that

$T^n = U^\dagger R^n U = 0, \tag{15}$

as was to be proved. Finally, we are done.

Observe we have avoided the use of the minimal polynomial of $T$ or its equivalent.

Note Bene: The Schur decomposition is in fact very easy to establish, as may be seen by visiting the linked citing. The proof is simple and concise, much easier than say the existence of Jordan normal form or the Hamilton-Cayley Theorem.