The Unapologetic Mathematician

Mathematics for the interested outsider

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 dth 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.

February 18, 2009 - Posted by | Algebra, Linear Algebra

2 Comments »

  1. […] the dimension of is , and so by this point the sequence of images of powers of has stabilized! That […]

    Pingback by The Multiplicity of an Eigenvalue « The Unapologetic Mathematician | February 19, 2009 | Reply

  2. […] the last equality holds because the dimension of is , and so the image has stabilized by this point. Thus we can choose so that . And […]

    Pingback by The Multiplicity of an Eigenpair « The Unapologetic Mathematician | April 8, 2009 | Reply


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