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Let $U,V,W$ be $F$-vector spaces. Let $T$ be a linear mapping from $U$ to $V$. Let $S$ be a linear mapping from $V$ to $W$ where $\dim(\text{range}(T)) > \dim(\text{null}(S))$. Prove that $ST$ does not map every vector in its domain to the $0$-vector.

I tried doing a proof by contradiction. I showed that if $ST$ is the $0$-map, then $\text{range}(T)$ is a subspace $\text{null}(S)$. How do I use the fact that $\dim(\text{range}(T)) > \dim(\text{null}(S))$ to show that I have a contradiction? Thanks in advance!

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    $\begingroup$ If $U$ is a subspace of $V$, then $\text{dim}(U) \leq \text{dim}(V)$... $\endgroup$
    – DKS
    Commented Dec 3, 2016 at 4:40
  • $\begingroup$ Thank you so much. I solved it. $\endgroup$
    – clockadd
    Commented Dec 3, 2016 at 4:55

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From the comment above by @DKS

If $V_1$ is a subspace of $V_2$, then $\dim(V_1) \leq \dim V_2$. This should help complete the proof.

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