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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?

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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...

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  • $\begingroup$ Thank you for your answer. Yes, I see what you mean. But in this case, what is being represented is a map and not a matrix. So wouldn't anything that is a multiple of the identity still be a part of the identity map? If not, would simply listing, for example, $X$ and $Y$ as the $2$ by $2$ matrices with $2's$ at the top left and bottom right suffice? If not, how exactly would I represent this matrix linear transformation? Thank you! $\endgroup$ – sktsasus Nov 2 '18 at 2:54
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    $\begingroup$ That one would be determined by $T(a)=2a$, and $T(b)=2b$... Or just $T(x)=2x$. $\endgroup$ – Chris Custer Nov 2 '18 at 3:00
  • $\begingroup$ Right. Thank you! $\endgroup$ – sktsasus Nov 2 '18 at 3:06
  • $\begingroup$ Only one matrix needs to be a multiple of the identity; the second can be anything. $\endgroup$ – Chris Custer Nov 2 '18 at 3:08
  • $\begingroup$ Yeah, makes sense. Thank you! $\endgroup$ – sktsasus Nov 2 '18 at 3:09
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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.

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