Suppose $T : \mathbb R ^ n \to \mathbb R ^ n$ is a linear map which satisfies $T ^ k = O$ for some positive integer $k$. Show that $T ^ n = O$

Question

I'm trying to prove this problem: Suppose $T : \mathbb R ^ n \to \mathbb R ^ n$ is a linear map which satisfies $T ^ k = O$ for some positive integer $k$. Show that $T ^ n = O$ and $O$ is the $0$ map with dimensions $n \times n$.

This is what I have so far:

Let $w + k = n$ and let $w > 0$

Then, $T ^ n = (T ^ k)(T ^ w) = O$

I just can't figure how to prove this problem in the case that $w < 0$.

Thanks!

Answer 1

Let $m$ the smallest number so that $T^m=0$. It's enough to show that $m\le n$. From the definition of $m$, there exists a vector $v$ so that $T^{m-1}v\ne 0$. We will show that the vectors $v$, $Tv$, $\ldots$, $T^{m-1}v$ are linearly independent. Assume the contrary. There exist $a_0$, $a_1$, $a_{m-1}$ not all zero, so that $$a_0 v+ a_1 Tv +\cdots + a_{m-1} T^{m-1} v = 0$$ Let $k$ the smallest index such that $a_k\ne 0$. We have therefore $$a_k T^{k} v = -(a_{k+1}T^{k+1}v + \cdots + a_{m-1}T^{m-1}v)$$ Apply to both sides of the equality above the operator $T^{m-1-k}$. On the left hand side we get $a_k T^{m-1}v \ne 0$, while on the right hand side we get $0$, contradiction. Therefore, the $m$ vectors are linearly independent, and so $m\le n$.

Alternative proof: from the Hamilton-Cayley theorem, we conclude that for some $k \le n$, $T^k$ is a linear combination of higher powers of $T$, and a similar argument as above shows that $T^k = 0$, and so $T^n=0$.

Answer 2

First note that $T$ cannot have a nonzero eigenvalue, else there would be some eigenvector $v$ with $Tv=\lambda v$ and hence $T^kv=\lambda^kv\neq 0$. Thus all eigenvalues of $T$ are zero.

Now we find $T$'s Schur decomposition $T=QUQ^{-1}$, where $U$ is upper triangular with $0$ (the eigenvalues of $T$) on the diagonal. It is possible to do this with $Q,U$ real, since all eigenvalues of $T$ are real. For details of the real Schur decomposition, see here.

Now, $T^n=(QUQ^{-1})^n=QU^nQ^{-1}=Q0Q^{-1}=0$. The key part is that $U^n=0$; each additional power of $U$ has one more superdiagonal all zero.

Answer 3

As $T^k=0$ then the minimal polynomial of $T$ is a factor of $x^k$, so is $x^r$ for some $r$. But the minimal polynomial of a linear map on $\Bbb R^n$ has degree at most $n$, so $r\le n$. Therefore $T^r=0$, and so $T^n=0$.