# Concerning the dimensions of Compositions of Linear Maps

Is the Following Proof Correct?

Theorem. Given that $V$ and $W$ are finite dimensional vector spaces such that $T\in\mathcal{L}(V,W)$ and $S\in\mathcal{L}(W,U)$ where $\mathcal{L}(A,B)$ denotes the set of all Linear Transformations for the vectors spaces $A$ to $B$, then $$\dim\operatorname{range}ST\leq\dim\operatorname{range}T$$

Proof. Assume for the purpose of contradiction that $\dim\operatorname{range}ST>\dim\operatorname{range}T$. We may then invoke the existence of vectors $\alpha_1,\alpha_2,...,\alpha_m\in\operatorname{range}ST$ and $\beta_1,\beta_2,...,\beta_n\in\operatorname{range}T$ such that the at as basis for $\operatorname{range}ST$ and $\operatorname{range}T$ repectively and where $$\dim\operatorname{range}ST = m>n = \dim\operatorname{range}T\tag{1}$$

furthermore it is evident that $$\forall j\in I = \{1,2,...,m\}\exists\eta_j\in\operatorname{range} T(S\eta_j = \alpha_j)\tag{2}$$ Let us instantiate these vectors $\eta_1,\eta_2,...,\eta_m\in\operatorname{range} T$. We now prove that this list of vectors is linearly independent.

Assume that for some $x_1,x_2,...,x_m\in\mathbf{F}$ it is the case that $$0 = \sum_{j=1}^{m}x_j\eta_j\tag{3}$$

Applying $S$ to both sides of $(3)$ we have $$S(0) = S\left(\sum_{j=1}^{m}x_j\eta_j\right) = \sum_{j=1}^{m}x_jS\eta_j=\sum_{j=1}^{m}x_j\alpha_j = 0\tag{4}$$ but the list $\alpha_1,\alpha_2,...,\alpha_m$ is linearly independent and thus $\forall j \in I(x_j = 0)$, we may therefore conclude that the list $\eta_1,\eta_2,...,\eta_m$ is linearly independent in $\operatorname{range} T$, but $\beta_1,\beta_2,...,\beta_n$ is a basis and by consequence a spanning list for $\operatorname{range} T$ which implies that $m\leq n$ contradicting $(1)$.

$\blacksquare$

The proof is correct. But you don't need to set up the contradiction, because you showed directly $m \leq n$.
By the way, there is a simpler proof if you consider $\text{range} (ST)=\text{range} (S|_{\text{range} T})$, where $S|_{\text{range} T}$ is the restriction of $S$ to the subspace $\text{range} T$. This fact can be shown by considering set inclusions. Then the result follows directly from dimension theorem that $\dim \text{range} (S|_{\text{range} T})\leq \dim\text{range}(T)$.