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Suppose $R(U)$ and $N(U)$ (the range and null space of a linear transformation $U$, respectively) are $T$-invariant ($T$ linear) subspaces of some vector space $V$. Does this imply $UT=TU$?

I've proven the converse, that $UT=TU$ implies the $T$-invariance of the range and null space. However, I'm a bit stumped trying to prove or come up with a counterexample for the above.

For instance, I can see that when something is in the null space of $U$ that $UT=TU=0$ trivially, but does this extend to the range as well?

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    $\begingroup$ If $U$ is invertible, then $R(U)$ (the whole vector space) and $N(U)$ (the zero subspace) are invariant subspaces of any $T$. $\endgroup$
    – user1551
    Commented Nov 15, 2018 at 2:14
  • $\begingroup$ Is the converse of this true? That is, does the fact that $R(U)$ and $N(U)$ are $T$-invariant imply that $U$ is invertible? That would then prove that $UT=TU$, correct? $\endgroup$
    – notadoctor
    Commented Nov 15, 2018 at 2:19
  • $\begingroup$ Cool problem. Where did you get this problem? $\endgroup$
    – user614287
    Commented Nov 15, 2018 at 2:23
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    $\begingroup$ @justadampaul No. Consider $U=\pmatrix{1\\ &2\\ &&0},\ T=\pmatrix{0&1\\ 1&0\\ &&1}$. Then $R(U)=\{(\ast,\ast,0)^T\}$ and $N(U)=\{(0,0,\ast)^T\}$ are $T$-invariant, but $U$ is singular and $UT\ne TU$. $\endgroup$
    – user1551
    Commented Nov 15, 2018 at 2:33
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    $\begingroup$ It doesn't have to be non-invertible. user1551's first comment suggests the opposite. Consider two rotations in 3D; both are invertible and have trivial nullspace and range, but rotations are not commutative in general. $\endgroup$
    – mr_e_man
    Commented Nov 15, 2018 at 2:50

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Via user mr_e_man in the comments above: No. Rotations in $R^3$ are not generally commutative, but two rotations $T$ and $U$ have nullspaces and ranges which are equal (and therefore $T$-invariant).

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