Suppose $V$ and $W$ are subspaces of a finite-dimensional vector space $U$.

(a) Show that $W \subset V$ if and only if $V^0 \subset W^0$.

(b) Show that $(V \cap W)^0 = V^0 + W^0$. (Hint: Show $\supset$ holds, then use lots of fomulas for dimension.)

I am trying to show this question. My understanding of annihilators is that for a vector space $V$ over $K$, with $S$ being a subset, the annihilator of $S$ is the subspace $S^0$ of linear functions $f$ in $V^*$ so that $f(s)=0$ for every $s$ in $S$. However, I don't know how to use this definition to prove (a) and (b). The hint for (b) suggests I should use formulas such as

$$\dim(V + W) = \dim V + \dim W − \dim(V \cap W)$$

but I don't know how to reach this stage.


For (b) first note that $f_{V^0}+f_{W^0}\in V^0+W^0$ implies $$ (f_{V^0}+f_{W^0})(s)=f_{V^0}(s)+f_{W^0}(s)=0+0=0 $$ whenever $s\in W\cap V$ since $W\cap V$ is a subspace of both $V$ and $W$. This proves that $V^0+W^0\subset(V\cap W)^0$

Now, note that \begin{align*} \dim\bigl((V\cap W)^0\bigr) &= \dim(U)-\dim(V\cap W) \\ &= \dim(U)-\bigl\{\dim(V)+\dim(W)-\dim(V+W)\bigr\} \\ &= \dim(U)+\bigl\{\dim(U)-\dim(U)\bigr\}-\bigl\{\dim(V)+\dim(W)-\dim(V+W)\bigr\} \\ &= \bigl\{\dim(U)-\dim(V)\bigr\}+\bigl\{\dim(U)-\dim(W)\bigr\}-\bigl\{\dim(U)-\dim(V+W)\bigr\} \\ &= \dim(V^0)+\dim(W^0)-\dim(V^0\cap W^0) \\ &= \dim(V^0+W^0) \end{align*} where we have used the identities \begin{align*} \dim(S^0)&=\dim(U)-\dim(S)&(S\cup T)^0=S^0\cap T^0 \end{align*} for any two subspaces $S$ and $T$ of $U$. It follows that $V^0+W^0$ is a subspace of $(V\cap W)^0$ with $$ \dim(V^0+W^0)=\dim\bigl(V\cap W)^0\bigr) $$ Hence $V^0+W^0=(V\cap W)^0$ as advertised.


For a)

Proving $W\subset V =>V^0 \subset W^0 $


so $W\subset V$

Let $\phi \in V^0 $ then, $ \phi(v)=0, \forall v \in V$

But $W \subset V $, so $\phi$ annihilates every element of $W$, meaning $\phi \in W^0$ .

So $V^0 \subset W^0$

  • $\begingroup$ What about the other direction in a), that is if the annihilators are nested, then subspaces must be ? That seems tricky, and the more challenging part. $\endgroup$
    – me10240
    Feb 4 '16 at 20:02
  • 2
    $\begingroup$ For the benefit of other readers: math.stackexchange.com/questions/1641006/… $\endgroup$
    – me10240
    Feb 4 '16 at 23:46

Here is an alternative way of proving (b).

As proved in Brian's answer, $V^0+W^0\subseteq(V\cap W)^0$.

For the other direction, take $f\in(V\cap W)^0$. Let $\left\{\alpha_1,\cdots,\alpha_n\right\}$ be a basis for $V\cap W$, $\left\{\alpha_1,\cdots,\alpha_n,\beta_1,\cdots,\beta_m\right\}$ a basis for $V$, and $\left\{\alpha_1,\cdots,\alpha_n,\gamma_1,\cdots,\gamma_\ell\right\}$ a basis for $W$, where $n=\dim(V\cap W),\,n+m=\dim V,$ and $n+\ell=\dim W$.

The the assumption means $f(\alpha_i)=0,\,\forall i$.

Define $g\in V^0$ by $g(\alpha_i)=g(\beta_j)=0,\,\forall i,j$, and $g(\gamma_k)=f(\gamma_k),\,\forall k=1,\cdots,\ell$. And define $h\in W^0$ by $h(\alpha_i)=h(\gamma_k)=0,\,\forall i,k$, and $h(\beta_j)=f(\beta_j),j=1,\cdots,m$.

Then clearly $f=g+h\in V^0+W^0$.

Hope this helps.

  • $\begingroup$ HOW $g(\gamma_k)=f(\gamma_k),\,\forall k=1,\cdots,\ell$? $\endgroup$ Jul 1 '20 at 6:29
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    $\begingroup$ We define $g$ to have the same value as $f$ on $\gamma_k,\,k=1,\cdots,\ell$. Are you asking why is this possible because perhaps $\gamma_k=\beta_j$ for some $j$? If that is the case, then notice that every $\gamma_k$ is linearly independent with the set $\left\{\alpha_1,\cdots,\alpha_n,\beta_1,\cdots,\beta_m\right\}$, otherwise $\gamma_k\in V\cap W$, contradicting our choice. $\endgroup$
    – awllower
    Jul 1 '20 at 9:13

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