Help prove this? I can prove for n=2, but I'm stuck on proving it for general n. Thanks!
My proof for n=2
Forward direction: Consider A and B and A $\cup$ B is a subspace of V. We prove by contradiction by assuming $\exists$ x $\in$ A s.t x $\notin$ B and $\exists$ y $\notin$ s.t y $\in$ B. Consider x+y. Since x $\in$ A and y $\in$ B, if A$\cup$B is a subspace, we expect x+y $\in$ A$\cup$B. However, since y$\notin$A, then x+y $\notin$A because of closure under addition and multiplication of scalar. If, -x+x+y$\in$A then y$\in$A, but that is a contradiction. Thus x+y $\notin$ A and by similar logic, same for B. Since x+y $\notin$ A and x+y $\notin$ B, x+y $\notin$ A $\cup$ B which is a contradiction. Thus one of the subspaces must contain the other
Reverse direction: WLOG, A $\subset$ B so A $\cup$ B is equivalent to B. Since A and B are subspaces, they both have the 0 vector so A$\cup$B also does. For a1, a2 $\in$ A$\cup$B, a1, a2 $\in$ B, so a1+a2 $\in$A$\cup$B-->closed under addition. Consider b$\in$ A$\cup$B and scalar $\lambda$. b$\in$ B, and $\lambda$b $\in$ B so $\lambda$b $\in$ A$\cup$B.--> so closed under scalar multiplication. Thus A$\cup$B is a subspace.
Edit: Does it work that since I proved it for n=2, we can use induction to just look at V$\cup$W for dim(W)=1 as one of the comments pointed out? Doesn't that just reduce it to the 2 subspace case again? And since we know that the union is a subspace, is that too easy?