inequality of dimensions I got this questions on one of my example sheets in a first year algebra course:
"Let $B:W×V→K$ be a bilinear map, where $V,W$ are vector spaces over a a field $K$. Let $U$ be a subspace of $V$. If we denote by $U^{\perp}= \{w \in W|B(w,u)=0,\forall u∈U \},$ prove that $dim(U)+dim(U^{\perp})≥dim(W)$.
I managed to come up with a proof in the case when $V,W$ are finite dimensional by considering a basis of $V$ which contains a basis of $U$, a basis of $W$ and then considering the coordinate matrix of $B$.
I can't think of an argument which works for the infinite dimensional case. Could somebody suggest me something? Thank you
 A: This is false in the sense of cardinal arithmetic if the spaces are allowed to be infinite dimensional. Take $U=V$ any infinite dimensional space, $W=U^*$, and $B$ the natural pairing $(w,u)\mapsto w(u)$. Then $\dim U^\perp=0$ but $\dim U<\dim U^*=\dim W$.
For a concrete example take $W=K[[X]]$ (uncountable dimensional) and $U=K[X]$ (countable dimensional) with the "standard" pairing between these two $K$ vector spaces.
A: The dimension of $V$ appears to be irrelevant, as we can replace it with $U$. So the infinite-dimensional case is really when $\dim \, (W) = \infty$, and so we need to show that one of the quantities on the left-hand side is infinite. Suppose that $\dim  \, U$ is finite. What can you conclude about $\dim \, U^\perp$?
Hint: Since $\dim \, U$ is finite, we can find a basis $e_1, \cdots, e_n$ and construct linear functionals $L_i : W \to K$ defined by $L_i(w) = B(w,e_i)$. Clearly, $U^\perp = \bigcap_{i=1}^n \ker L_i$. For each $L_i$, what are the possible dimensions of $W / \ker L_i$ -- the subspace of $W$ of all vectors not in the kernel of $L_i$? What can you conclude about the dimension of $W/ \bigcap_{i=1}^n \ker L_i$? When $\dim \, W = \infty$, what does that tell you about the dimension of $\bigcap_{i=1}^n \ker \, L_i$, and hence the dimension of $U^\perp$?
