prove that if W is a subspace of V, then dim(W)+dim(W^0)=dim(V)

Question

Prove that if W is a subspace of V, then $dim(W)+dim(W^0)=dim(V)$, where $S^0$ is the annihilator defined by $S^0={f \in V^* |f(x)=0 \text{for all} x \in S}$

So for this problem, we are going to construct basis for W and $W^0$ such that $W \bigcup W^0$ generates V right?

@lisyarus Seems to me this is $W^\bot$, i.e. the dual space. — Parcly Taxel, Apr 07 '20 at 04:34
Possible duplicate of Proof: Sum of dimension of orthogonal complement and vector subspace — Parcly Taxel, Apr 07 '20 at 04:39
@ParclyTaxel I'm thinking that $W^0$ is the annihilator of $W$, which is the subspace of $V^$ of functionals $f \in V^$ such that $f(w) = 0$ for all $w \in W$. In a Hilbert space, the Riesz representation theorem implies that this is essentially just $W^\perp$, but $W^0$ makes sense in an arbitrary topological linear space (even the topology is unnecessary if you deal with the algebraic dual). — user764828, Apr 07 '20 at 05:12

score 1 · Accepted Answer · answered Apr 07 '20 at 05:17

A basis for $W$ must be formed by, say, some vectors $w_1,\dots,w_k$. Since this can be extended to a basis for $V$ like that $$w_1,\dots,w_k,v_1,\dots,v_n,$$ consider the dual basis $g_1,\dots,g_k,f_1,\dots,f_n$ of it. Clearly the $f_1,\dots,f_n$ belongs to $W^0$ (why?) and we claim that is in fact a basis for $W^0$.

It is easy to see that $f_1,\dots,f_n$ are linearly independent (why?).
If $h \in W^0$ we want to see that $h$ can be written as a linear combination of $f_1,\dots,f_n$. Well, since $g_1,\dots,g_k,f_1,\dots,f_n$ is a basis for $V^*$ and $h$ is one element in there, by the theorem of your last question we know $$h = h(w_1)g_1 + \cdots + h(w_k)g_k + h(v_1)f_1 + \cdots h(v_n)f_n$$ but, since $h$ is in $W^0$, $h(w_1) = \cdots = h(w_k) = 0$, so $$h = h(v_1)f_1 + \cdots h(v_n)f_n$$ and then $f_1,\dots,f_n$ generates $W^0$ since $h$ was arbitrary.

prove that if W is a subspace of V, then dim(W)+dim(W^0)=dim(V)

1 Answers1