I am trying to solve the following exercise.
Let $V$ be a finite-dimensional vector space and $T: V \to V$ a linear transformation. For any $m$, $$ \dim \text{ker}(T^{m+1}) - \dim \text{ker}(T^m) \leq \dim \text{ker}(T^m) - \dim \text{ker}(T^{m-1}). $$
I think I can prove this with a general $m$ instead of using induction (or strong induction). The first thing I thought of was that I need to somehow use the rank nullity theorem. Applying this to the maps $T^m, T^{m+1}, T^{m-1}$: \begin{align*} \dim V &= \dim \text{ker}(T^m) + \dim \text{Img}(T^m) \\ \dim V &= \dim \text{ker}(T^{m+1}) + \dim \text{Img}(T^{m+1}) \\ \dim V &= \dim \text{ker}(T^{m-1}) + \dim \text{Img}(T^{m-1}) \end{align*} So I can solve for the dimension of the kernel in each case: \begin{align*} \dim \text{ker}(T^m) &= \dim V - \dim \text{Img}(T^m) \\ \dim \text{ker}(T^{m+1}) &= \dim V - \dim \text{Img}(T^{m+1}) \\ \dim \text{ker}(T^{m-1}) &= \dim V - \dim \text{Img}(T^{m-1}). \end{align*} The left-hand side of the inequality is: \begin{align*} \dim \text{ker}(T^{m+1}) - \dim \text{ker}(T^{m}) &= \dim V - \dim \text{Img}(T^{m+1}) - \dim V + \dim \text{Img}(T^m) \\ &= \dim \text{Img}(T^m) - \dim \text{Img}(T^{m+1}) \end{align*} This doesn't seem to help. I know that $\text{Im}(T^{m+1}) \subset \text{Im}(T^m)$, so $\dim \text{Im}(T^{m+1}) \leq \dim \text{Im}(T^m)$. Another thing I tried to doing was starting with the inequality and driving something true, which will prove the inequality if all of the steps I made were reversible.