Is the following Proof Correct?

Theorem. Given that $U$ and $V$ are finite dimensional vector spaces and $S\in\mathcal{L}(V,W)$ and $T\in\mathcal{L}(U,V)$. Show that $$\dim\operatorname{range}ST\leq\min\{\dim\operatorname{range}S,\dim\operatorname{range}T\}$$

The argument below makes use of the following Lemmma.

Lemma. If $V$ and $W$ are vectorspaces where $V$ is finite-dimensional and $v_1,v_2,...,v_n$ is a linearly independent list in $V$ and $T\in\mathcal{L}(V,W)$ then $U = \dim\operatorname{span}(Tv_1,Tv_2,..,Tv_n)\leq n$.

Proof. Let $u_1,u_2,...,u_{\dim U}$ be the basis for the finite dimensional vector space $U$. Let us now proceed by considering the following cases. $$\dim\operatorname{range}T = \min\{\dim\operatorname{range}S,\dim\operatorname{range}T\}\tag{1}$$ $$\dim\operatorname{range}S = \min\{\dim\operatorname{range}S,\dim\operatorname{range}T\}\tag{2}$$

Let us first examine case $(1)$. We begin by letting $Tu_1,Tu_2,...Tu_{\dim\operatorname{range}T}$ be the basis for $\operatorname{range}T$. It is then not difficult to see that $$\operatorname{range}ST = \operatorname{span}(STu_1,STu_2,...,STu_{\dim U}) = \operatorname{span}(STu_1,STu_2,...STu_{\dim\operatorname{range}T})$$ Furthermore from the lemma above it is evident that $\dim\operatorname{span}(STu_1,STu_2,...STu_{\dim\operatorname{range}T})$ and by extension $\dim\operatorname{span}(STu_1,STu_2,...,STu_{\dim U})$ is at most $\dim\operatorname{range}T$ consequently $\dim\operatorname{range}ST\leq\dim\operatorname{range}T$.

For case $(2)$. The observation that $\operatorname{range}ST$ is a subspace of $\operatorname{range}S$ implies that $\dim\operatorname{range}ST\leq\dim\operatorname{range}S$.



1 Answer 1


Your proof is correct! I think the argument could be simplified by the restriction to subspaces, denoted by $|_{subspace}$.

$$rangeST|_U = rangeS|_{rangeT}\subset rangeS|_V$$

So it follows $$dim(rangeST) \leq dim(rangeS)$$ $$dim(rangeST)=dim(rangeT) - dim(NullS|_{rangeT})\leq dim(rangeT)$$

  • $\begingroup$ @ Thanks for your answer i really needed to get this checked out. $\endgroup$ Commented Mar 26, 2018 at 13:17

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