Compositions of transforms :  and  have the same eigenvalues geometrically If $S$ and $T$ are both linear maps then $ST$ and $TS$ have the same eigenvalues. I understand the proof (which can be found at $ST$ and $TS$ have the same eigenvalues.). But why must this be the case intuitively/geometrically? Is there a geometric picture for this?
The statement says that $ST$ and $TS$ stretch the space by exactly the same amount but only in different directions(so just rotating ST by correct amount will give be TS?) So somehow the magnitude gets preserved but not direction. 
I am thinking if this tells us that the reason matrix multiplication is not communicative is only due to directional changing. 
--Clarification---
Assume T,S are linear operators that map $R^N$ to $R^N$. They are square and invertible.
 A: A partial answer in a special case: If we're talking about square matrices and one of $S$ or $T$ is invertible then $ST$ and $TS$ are similar.
The fact that similar matrices have the same eigenvalues is possibly somewhat "geometric"; the two matrices represent the same transformation in different coordinate systems.
I've been asked to elaborate on "similar matrices represent the same transformation in different coordinate systems". All the assertions below are elementary, and straightforward for many readers; others can find proofs in a book on linear  algebra:
Say $C=(b_1,\dots,b_n)$ is an ordered basis for $V$. If $x\in V$ then $[x]_C$ is the coordinate vector for $x$  wrt $C$: $$[x]_C=(c_1,\dots,c_n)$$where $$x=\sum c_jb_j.$$
Now if $T:V\to V$  is linear then $[T]_C$ is the  matrix representing $T$ wrt $C$, which by definition means $$[T]_C[x]_C=[Tx]_C\quad(x\in V).$$


True Fact:  Suppose $A$ and $B$ are $n\times n$  matrices over the field $F$. Then $A$ is similar to  $B$ if and only iif there exists a basis $C$ for $F^n$  with $A=[B]_C$.


