# If $S$ is a linear operator on $V$ such that $S^n=0$ and $S^{n-1}\neq 0$, show $\dim(S^k(V))=n-k$

Let $$V$$ be an $$n$$-dimensional vector space over the field $$F$$. Let $$S$$ be any linear operator on $$V$$ such that $$S^n=0$$ but $$S^{n-1}\neq 0$$. Prove that $$\dim(S^k(V))=n-k$$.

An approach I saw but I have some questions about it:

1. Firstly I want to know that this notation $$\dim(S^k(V))$$ means $$\dim(\text{Im}(S^k))=\text{rank}(S^k)$$? If so, I'm going to continue.
2. $$\text{rank}(S^k)$$ cannot be greater than $$n$$ because $$\text{Im}(S^k)\subseteq V$$ and if $$\dim(\text{Im}(S^k))$$ is greater than $$n$$, then $$V\subseteq \text{Im}(S^k)$$ which is a contradiction. Am I right?
3. Suppose $$\text{rank}(S)=n$$, then $$\text{rank}(S^k)=n$$. My first question is how is this concluded? Can we say, because $$\dim(\text{Im}(S))$$ is $$n$$, and $$\text{Im}(S)\subseteq V$$, so $$\text{Im}(S)=V$$. So image of all $$S^k$$ is the whole $$V$$ and $$\text{rank}(S^k)=n$$ which is a contradiction by $$\text{rank}(S^n)=0$$.
4. So $$\text{rank}(S)\leq n-1$$. Now $$\text{rank}(S^2)$$ cannot be greater than $$\text{rank}(S)$$. Because if not, $$\text{Im}(S^2)$$ will be a subset of $$\text{domain}(S^2)$$. Am I right?
5. Suppose $$\text{rank}(S^2)=n-1$$. So $$n-1=\text{rank}(S^2)\leq \text{rank}(S)\leq n-1$$. So $$\text{rank}(S)=n-1$$. Thus $$S$$ is an isomorphism on $$\text{Im}(S)$$. It follows that $$S^k$$ is an isomorphism for $$k\geq 2$$. Thus $$\text{rank}(S^k)=n-1$$ for all $$k\geq 2$$. Which is another contradiction. So $$\text{rank}(S^2)\leq n-2$$ Is the reasoning correct?
6. Now suppose $$\text{rank}(S^3)=n-2$$. We will reach a contradiction like part 5, so $$\text{rank}(S^3)\leq n-3$$. Continuing this proces, we will reach $$\text{rank}(S^{n-1}) \leq n-1$$. As we assumed $$S^{n-1}\neq 0$$ at first, so $$\text{rank}(S^{n-1})=1$$. Using previous reasonings, $$\text{rank}(S^{k})=n-k$$.

Any help and explanations is immensely appreciated! I also would be thankful if you add another method for the proof if you have in mind.

1. Firstly I want to know that this notation $$\dim(S^k(V))$$ means $$\dim(\text{Im}(S^k))=\text{rank}(S^k)$$? If so, I'm going to continue.

Yes that's correct. For a transformation (or more generally a function) $$T$$, $$T(V)$$ denotes the set $$\{T(v):v \in V\}$$, which is to say the image of $$T$$.

1. $$\text{rank}(S^k)$$ cannot be greater than $$n$$ because $$\text{Im}(S^k)\subseteq V$$ and if $$\dim(\text{Im}(S^k))$$ is greater than $$n$$, then $$V\subseteq \text{Im}(S^k)$$ which is a contradiction. Am I right?

The contradiction that you have in mind, I suspect, is $$V \subsetneq S$$, i.e. $$V\subseteq S$$ and $$V \neq S$$. If that's what you mean, then that's correct.

1. Suppose $$\text{rank}(S)=n$$, then $$\text{rank}(S^k)=n$$. My first question is how is this concluded? Can we say, because $$\dim(\text{Im}(S))$$ is $$n$$, and $$\text{Im}(S)\subseteq V$$, so $$\text{Im}(S)=V$$. So image of all $$S^k$$ is the whole $$V$$ and $$\text{rank}(S^k)=n$$ which is a contradiction by $$\text{rank}(S^n)=0$$.

It's not clear how you made the step from $$\text{Im}(S)=V$$ to "image of all $$S^k$$ is the whole $$V$$". That step seems to be equivalent to the question that you started with. Instead, I would say that if $$\text{rank}(S)=n$$ (equivalently if $$\text{Im}(S)=V$$), then $$S$$ is invertible (i.e. $$S$$ is an isomorphism). Because a composition of invertible operators is invertible, we can conclude that $$S^k = S \circ \cdots \circ S$$ is also invertible and hence has rank $$n$$.

1. So $$\text{rank}(S)\leq n-1$$. Now $$\text{rank}(S^2)$$ cannot be greater than $$\text{rank}(S)$$. Because if not, $$\text{Im}(S^2)$$ will be a subset of $$\text{domain}(S^2)$$. Am I right?

I cannot understand what you have in mind with the sentence "Because if not, $$\text{Im}(S^2)$$ will be a subset of $$\text{domain}(S^2)$$." I suspect that whatever you're trying to say is incorrect, but I don't know what you're trying to say so I can't be sure.

I would say that the general fact being used here is that $$\operatorname{rank}(AB) \leq \text{rank}(A)$$. You could say that this is a consequence of the fact that $$\text{Im}(AB) \subseteq \text{Im}(A)$$.

1. Suppose $$\text{rank}(S^2)=n-1$$. So $$n-1=\text{rank}(S^2)\leq \text{rank}(S)\leq n-1$$. So $$\text{rank}(S)=n-1$$. Thus $$S$$ is an isomorphism on $$\text{Im}(S)$$. It follows that $$S^k$$ is an isomorphism for $$k\geq 2$$. Thus $$\text{rank}(S^k)=n-1$$ for all $$k\geq 2$$. Which is another contradiction. So $$\text{rank}(S^2)\leq n-2$$ Is the reasoning correct?

Yes, that's completely correct.

1. Now suppose $$\text{rank}(S^3)=n-2$$. We will reach a contradiction like part 5, so $$\text{rank}(S^3)\leq n-3$$. Continuing this process, we will reach $$\text{rank}(S^{n-1}) \leq n-1$$. As we assumed $$S^{n-1}\neq 0$$ at first, so $$\text{rank}(S^{n-1})=1$$. Using previous reasonings, $$\text{rank}(S^{k})=n-k$$.

I think that you have the correct idea, but I'm not sure we're on the same page so I'll rephrase what I think you mean. Your logic from earlier steps can be extended to show that $$\text{rank}(S^k) \leq n - k$$ (using the fact that $$\text{rank}(S^k) < \text{rank}(S^{k-1})$$. In order to show that $$\text{rank}(S^k) \geq n - k$$, you show that having $$\text{rank}(S^k) < n - k$$ for some $$k$$ leads to the conclusion that $$S^{n-1} = 0$$, contradicting the given information. Thus, we conclude that $$\text{rank}(S^k) = n - k$$ for all $$k = 1,2,\dots,n$$.

The approach that you've outlined (assuming that I have interpreted it correctly) is the nicest and fastest approach to this problem using only the basics of linear algebra.

It is worth noting that a much faster (though perhaps less insightful) approach exists using Jordan canonical form.

The fact that $$S^n = 0$$ tells you that $$S$$ has $$0$$ as its only eigenvalue. The fact that $$S^{n-1}\neq 0$$ tells you that the Jordan form of $$S$$ has a block of size at least $$n$$. Because this Jordan form is an $$n \times n$$ matrix, we are forced to conclude that the Jordan form of $$S$$ is simply a Jordan block of size $$n$$ associated with eigenvalue $$0$$, from which it follows that $$\operatorname{rank}(S^k) = n-k$$.

The corresponding approach using the rational canonical form (AKA Frobenius normal form) would be essentially the same: to begin, the fact that $$S^{n-1} \neq 0$$ but $$S^n = 0$$ tells us that the minimal and characteristic polynomials of $$S$$ are equal to $$p(x) = x^n$$; from there, the solution would similarly follow.