Proving $\dim A<\dim B$ and $\dim C=\dim B$ iff $A\subset B$.

Question

Let A, B be spans: $$ A=\operatorname{span}(u_1,...,u_n)$$ $$ B=\operatorname{span}(v_1,...,v_m)$$ $$ C=\operatorname{span}(v_1,...,v_m,u_1,...,u_n)$$ I have an exercise in which I need to check for a lot of different sets like these if $A\subset B$, $B\subset A$ or $A=B$. I thought about a statement which intuitively makes sense and will help me solve this exercise: $$(\dim A<\dim B)\land (\dim C=\dim B) \iff A \subset B $$ I am unable to prove this statement, can someone help?

Answer 1

Hint:

Consider the short exact sequence \begin{alignat}{3} 0\longrightarrow A\cap B&\longrightarrow &A\times B&\longrightarrow{}& C\longrightarrow 0 \\ x&\longmapsto &(x,x) &\\ &&(u,v)&\longmapsto &\hskip-3.5em u-v \end{alignat} which implies $\;\dim A+\dim B=\dim(A\times B)=\dim(A\cap B)+\dim C$. What can you deduce from the hypothesis on dimensions?

Answer 2

It is clear from the definition that $A\subseteq C$, $B\subseteq C$ and $A+B=C$.

By Grassmann's formula, $\dim C=\dim(A+B)=\dim A+\dim B-\dim(A\cap B)$.

($\Rightarrow$) The condition that $\dim C=\dim B$ forces $B=C$ and $\dim A=\dim(A\cap B)$. Since $A\cap B\subseteq A$, this forces $A\cap B=A$, that is, $A\subseteq B$. Together with $\dim A<\dim B$, we obtain $A\subsetneq B$.

($\Leftarrow$) Conversely, assume $A\subsetneq B$. Then $\dim A<\dim B$; moreover $$ \dim C=\dim(A+B)=\dim A+\dim B-\dim(A\cap B)=\dim A+\dim B-\dim A=\dim B $$