$Span(A)\cap Span(B\setminus(A\cap B))=\{\vec 0\}\Longrightarrow Span(A)\cap Span(B)=Span(A\cap B)$?

Question

$A$ and $B$ are two linearly independent sets. $A\cap B = \varnothing$ and $A\nsubseteq B$,$B\nsubseteq A$.

Is the following statement true?: $$Span(A)\cap Span(B\setminus(A\cap B))=\{\vec 0\}\Longrightarrow Span(A)\cap Span(B)=Span(A\cap B)$$

Answer 1

Answer for the original post with $A\cap B=\emptyset$

There is some redundancy.

If $A\cap B=\emptyset$, then $B\setminus(A\cap B)=B$, so you are asking whether $$ \operatorname{Span}(A)\cap\operatorname{Span}(B)=\{0\} \implies \operatorname{Span}(A)\cap\operatorname{Span}(B)=\operatorname{Span}(A\cap B) $$ which is obvious because $$ \operatorname{Span}(\emptyset)=\{0\} $$

Answer for the modified question with $A\cap B\ne\emptyset$

Since the intersection of $\operatorname{Span}(A)$ and $\operatorname{Span}(B\setminus(A\cap B))$ is trivial, we know that the sum of the two spaces has the same dimension as the sum of the dimensions. Let's call, for simplicity,

• $U=\operatorname{Span}(A)$
• $V=\operatorname{Span}(B)$
• $W=\operatorname{Span}(B\setminus(A\cap B))$

Note that $U+V=U+W$.

Then we know that $\dim(U+W)=\dim U+\dim W$. By Grassmann's formula, $$ \dim(U+V)=\dim(U+W)=\dim U+\dim W $$ On the other hand, $$ \dim(U+V)=\dim U+\dim V-\dim(U\cap V) $$ so we have $$ \dim W=\dim V-\dim(U\cap V) $$ so $$ \dim(U\cap V)=\dim V-\dim W $$ Now, $\dim V-\dim W=|A\cap B|$. Since, clearly, $$ \operatorname{Span}(A\cap B)\subseteq U\cap V, $$ equality of dimensions make us conclude.