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

Question

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.

Answer 1

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 $T$.

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?

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.

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 $\text{rank}(S^n)=0$.

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 $n$.

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?

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 $\text{Im}(AB) \subseteq \text{Im}(A)$.

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?

Yes, that's completely 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 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 $k = 1,2,\dots,n$.

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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.