If $A \cap B = \{0\}$ then $A^\perp + B^\perp = V$?

Question

Is this statement true: If $A \cap B = \{0\}$ then $A^\perp + B^\perp = V$.

I am trying to prove a bigger question, and my post with the full problem statement is here. But my proof for that other question comes down to the above statement. If the above is true, then I am done. If the above is false, then back to the drawing board. The above statement seems true, but I'm having trouble proving it. Thanks for any help in advance!

Answer 1

As we know $(X+Y)^\perp\leq X^\perp\cap Y^\perp$. Therefore $$ (A^\perp+B^\perp)^\perp\leq A^{\perp\perp}\cap B^{\perp\perp}=A\cap B=0\ \Rightarrow\ (A^\perp+B^\perp)^\perp=0. $$ It follows that $V=A^\perp+B^\perp$.

Answer 2

No, Let $A$ be the nonegative y-axis and $B$ the nonpositive y-axis in $\bf{R}^2$. Then the complement is the x-axis for both and therefore the sum of their complements is the x-axis, not $\bf{R}^2$. If you assume $A$ and $B$ to be subspaces then yes, as a commenter has pointed out.