# Does transformation invariance of the range and null space imply commutativity?

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?

• If $U$ is invertible, then $R(U)$ (the whole vector space) and $N(U)$ (the zero subspace) are invariant subspaces of any $T$. Commented Nov 15, 2018 at 2:14
• 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? Commented Nov 15, 2018 at 2:19
• Cool problem. Where did you get this problem? Commented Nov 15, 2018 at 2:23
• @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 $UT\ne TU$. Commented Nov 15, 2018 at 2:33
• 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. Commented Nov 15, 2018 at 2:50

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).