Question: My Linear Algebra book says the following:

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> Let $V$ be a vector space and $T$ be a transformation, which commutes with another transformation $U$. Then the kernel and range of $U$ are invariant under $T$. 
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I can understand why the kernel must be invariant. But I don't understand why the range must follow this. Any help in understanding this would be great!

Answer: 
Suppose that $U(T(v))=T(U(v))$ for all $v\in V$. Now suppose that $w=U(v)$ for some $v$. Then $T(w)=T(U(v))=U(T(v))$. Thus $T(w)$ is also in the range of $U$.
Let $y$ be in the range of $U$. Then there is $x \in V$ such that $y = Ux$. Note that $$Ty = TUx = UTx \in \operatorname{range}(U).$$ So $T(\operatorname{range}(U)) \subseteq \operatorname{range}(U)$.