Infinite Dimensional Vector Space linear operator For a finite dimensional vector space $V$ and a linear operator $T:V → V$ I know that,
$rank(T ◦ T ) ≤ rank (T)$. This is true for $n$ compositions of $T$; so if $r_n$ is the rank of$ n = T ◦ T ◦ · · · ◦ T,  r_{n+1} ≤ r_n$ for all $n$ and the sequence $r_1, r_2, r_3, . . .$eventually becomes constant.
How does this change if $V$ is infinite dimensional? The sequence doesn't necessarily eventually become constant, but what else changes?
 A: You can define $\mathrm{rank}(T)$ to be the dimension of of the image, $T(V)$, which is a cardinal number that may be infinite.  
In the finite dimensional case, because the sequence of ranks $(\mathrm{rank}(T^n))$ eventually becomes constant, the sequence of images $T^n(V)$ is also eventually constant, because if $W_1\subseteq W_2$ are subspaces of the same finite dimension, then $W_1=W_2$.  
One thing that doesn't change in the infinite dimensional case is that the ranks still become eventually constant.  Sets of cardinal numbers are well ordered*, so the descending sequence $(\mathrm{rank}(T^n))$ has a smallest element $\mathrm{rank}(T^N)$ for some $N$, and then $\mathrm{rank}(T^m)=\mathrm{rank}(T^N)$ for all $m\geq N$. 
So what does change?  In the infinite dimensional case, the images can be properly nested forever.  For example, on the space of polynomials $V=k[x]$, $Tp(x) = xp(x)$ has the property that $\mathrm{rank}(T^n)=\aleph_0$, the cardinality of countable infinity, for all $n$, although $T^{n+1}V$ is properly contained in $T^nV$ for all $n$. Also, in the infinite dimensional case, the ranks can be infinite forever, while the intersection of all of the images can be $\{0\}$ (same example) or another finite dimensional subspace, or another space of dimension that is infinite but of smaller cardinality than any of the images.  

*This is according to credible seeming sources I have seen and that are easily searchable, but I am not knowledgeable in the area and do not have proof.  I hope I'm corrected if wrong.
