I was trying to disprove the following thoerem:
If $A$ and $B$ are subspaces of $\mathbb{R}^n$ then $(A\cap B)^\bot = A^\bot \cup B^\bot$.
I know that the theorem is not true, but my book gave the following example (other language):
Consider $A=Sp\{(1,0)\}$ and $B=Sp\{(0,1)\}$. We get $A\cap B=\emptyset$ and $(A\cap B)^\bot =\mathbb{R}^2$. We also get $A^\bot = B$ and $B^\bot = A$ so $A^\bot \cup B^\bot = A\cup B \neq \mathbb{R}^2 $ (for example $(1,1)\not \in U\cup W$).
I don't understand why this example works. Isn't $A\cup B=Sp\{(0,1),(1,0)\}=\mathbb{R}^2$? then we get that it does works.
I also read the previous thread on the subject (link).
The point is that the intersection of subspaces is always a subspace, but the (set theoretic) union of subspaces is only very rarely a subspace (e.g. if one subspace contains the other), it rather mostly looks like a cross, just as $A\cup B$ in the given example.
Put that formally, we simply don't have $\rm{span}(A\cup B)=A\cup B$ while we do have $\rm{span}(A\cap B)=A\cap B$ if $A$ and $B$ are subspaces.
However, the formula can be corrected, at least in a finite dimension space, this does hold: $$(A\cap B)^\perp=\rm{span}(A^\perp\cup B^\perp)\,.$$