# The Unapologetic Mathematician

## Images of Powers of Transformations

For some technical points, it’s going to be useful to have a sort of dual to the increasing chain of subspaces we found yesterday. Instead of kernels, we’ll deal with images.

Specifically, if $w\in\mathrm{Im}(T^{k+1})$ then I say $w\in\mathrm{Im}(T^k)$. Indeed, the first statement asserts that there is some $v$ so that $w=T^{k+1}(v)$. But then $w=T^k(T(v))$, and so it’s the image of $T(v)$ under $T^k$ as well. So we have a decreasing sequence

$\displaystyle V=\mathrm{Im}(T^0)\supseteq\mathrm{Im}(T^1)\supseteq\mathrm{Im}(T^2)\supseteq...$

Just like last time, these stabilize by the time we get to the $d$th power, where $d=\dim(V)$. Instead of repeating everything, let’s just use the rank-nullity theorem, which says for each power $n$ that $d=\dim\left(\mathrm{Ker}(T^n)\right)+\dim\left(\mathrm{Im}(T^n)\right)$. Now if $m>d=\dim(V)$ then we calculate

\displaystyle\begin{aligned}\dim\left(\mathrm{Im}(T^m)\right)&=d-\dim\left(\mathrm{Ker}(T^m)\right)\\&=d-\dim\left(\mathrm{Ker}(T^d)\right)\\&=\dim\left(\mathrm{Im}(T^d)\right)\end{aligned}

where in the second line we used the stability of the sequence of kernels from yesterday. This tells us that $\mathrm{Im}(T^m)=\mathrm{Im}(T^d)$ for all these higher powers of $T$.