Commuting Linear Transformations Let $W$ be a $2$-dimensional vector space and {$a$, $b$} be the basis of $W$. What are two linear transformations, $X$ and $Y$, from $W$ into $W$ such that $X ◦ Y$ is equal to $Y ◦ X$? Neither $X$ or $Y$ can be the identity map on $W$.
From what I have seen, if two linear transformations commute, they must share at least one common eigenvector. But in this case, I am not entirely sure how to algebraically find two such transformations. Obviously if they were in matrix form if would be easier but in this case I am not sure. I would assume if the two transformations are identical then the conditions would be satisfied. But I don't think that is what is supposed to be done here.
So I just want to know if two matrices with the same eigenvectors will satisfy this or is some other type of linear transformations required? If so, what would that be?
Any help?
 A: It isn't true that if two linear transformations commute, they have a common eigenvector. For example, rotations in $\Bbb R^2$ commute but they have no eigenvectors. In addition, it doesn't work in reverse. To see this, let $T$ be some rotation by $\frac\pi 2$ in $\Bbb R^2$ and $S$ a reflection across $y$ (also in $\Bbb R^2$). Now, these don't commute, but they also don't share any eigenvectors. However, now let $X:\Bbb R^3\to\Bbb R^3$ where $(x,y,z)\mapsto (T(x,y),z)$ and $Y:\Bbb R^3\to \Bbb R^3$ where $(x,y,z)\mapsto S(x,y),z)$. In other words, $X$ leaves the third coordinate alone and does $T$ to $x$ and $y$ while $Y$ also leaves $z$ alone but does $S$ to $x$ and $y$ instead. Now, it's clear that $(0,0,1)$ is an eigenvector of both $X$ and $Y$. However, $X$ and $Y$ don't commute as $T$ and $S$ don't commute. In the end, it might be better to consider other relationships between $X$ and $Y$ that could imply commutativity.
A: Hint:  How about considering a linear transformation which has a diagonal matrix with the same entries on the diagonal,  rel $\{a,b\}$ (rel any basis actually)?  That is, a multiple of the identity.   It should commute with anything...
