MA20216: Algebra 2A

5.2 Solution sets and annihilators

Here is one way to think about $V^{*}$: consider the equation

\begin{equation} \label {eq:25} \alpha (v)=0, \end{equation}

for some $\alpha \in V^{*}$ and $v\in V$. If we choose dual bases $\lst {v}1n$ and $\dlst {v}1n$ of $V$ and $V^{*}$, (5.1) reads

\begin{equation*} \lc \alpha {x}1n=0 \end{equation*}

where we have written $\alpha =\lc \alpha {v^{*}}1n$ and $v=\lc {x}v1n$. This is a single homogeneous linear equation.

This gives us the idea of viewing $V^{*}$ as the set of linear equations on $V$. From this point of view, a subspace $E\leq V^{*}$ should be viewed as a system of linear equations and so we should be interested in the solutions of that system:

Definition. Let $E\leq V^{*}$. The solution set of $E$ is
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \sol E:=\set {v\in V\st \text {$\alpha (v)=0$, for all $\alpha \in E$}}=\bigcap _{\alpha \in E}\ker \alpha \leq V. \end{equation*}

Exercise.³ If $\lst \alpha 1k$ span $E$ then
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \sol E=\bigcap _{i=1}^k\ker \alpha _i. \end{equation*}

³ Question 1 on sheet 8.

For finite-dimensional $V$, one might expect each equation in a linear system to reduce the dimension of the solution set by one and this is exactly what happens:

Proposition 5.6. If $V$ is finite-dimensional and $E\leq V^{*}$ then
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \dim \sol E=\dim V-\dim E. \end{equation*}

We say that $E$ and $\sol E$ have complementary dimension.

Proof. Let $\dlst {v}1k$ be a basis of $E$ and extend to a basis $\dlst {v}1n$ of $V^{*}$. Let $\lst {v}1n$ be the dual basis of $V$ provided by Proposition 5.4.

Now $E=\Span {\dlst {v}1k}$ so that $\sol E=\bigcap _{i=1}^k\ker v^{*}_i$. Thus $v=\sum _{j=1}^n\lambda _jv_j$ lies in $\sol E$ if and only if $\lambda _i=v^{*}_i(v)=0$, for $\bw 1ik$. Otherwise said,
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \sol E=\Span {\lst {v}{k+1}n} \end{equation*}

so that
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \dim \sol E=n-k=\dim V-\dim E. \end{equation*}

□

Remark. Here is a slicker argument. Let $\ev ^E:V\to E^{*}$ be the linear map given by
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \ev ^E(v)(\alpha )=\alpha (v). \end{equation*}
- (1) $\im \ev ^E=E^{*}$: for this, you use Theorem 5.5 along with the fact that restriction to $E$ is a surjection from $V^{**}$ to $E^{*}$ thanks to Proposition 2.11.
- (2) $\ker \ev ^E=\set {v\in V\st \text {$\alpha (v)=0$, for all $\alpha \in E $}}=\sol E$.
So rank-nullity tells us that
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \dim \sol E+\dim E^{*}=\dim V \end{equation*}

and, since $\dim E=\dim E^{*}$, we are done.

Corollary 5.7. Let $V$ have dimension $n$ and suppose that $\lst {\alpha }1n\in V^{*}$ are such that
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \bigcap _{i=1}^n\ker \alpha _i=\set 0. \end{equation*}

Then $\lst \alpha 1n$ is a basis of $V^{*}$.

Proof. Let $E:=\Span {\lst \alpha 1n}$. The hypothesis says that $\sol E=\set 0$ so, by Proposition 5.6, $\dim E=n$ whence $E=V^{*}$. Thus $\lst \alpha 1n$ span $V^{*}$ and so are a basis. □

Here is an application:

Example. Let $P_2$ be the vector space of polynomials of degree at most $2$. Thus $\dim P_2=3$.

Define $\alpha _i:P_2\to \R $, $i=1,2,3$, by
$\seteqnumber{0}{5.}{1}$
\begin{align*} \alpha _1(p)&=p(1)\\ \alpha _2(p)&=p(\sqrt {2})\\ \alpha _3(p)&=p(\pi ), \end{align*} for all $p\in P_2$. These are all linear maps so that $\alpha _1,\alpha _2,\alpha _3\in P_2^{*}$. We apply Corollary 5.7 so see that $\alpha _1,\alpha _2,\alpha _3$ are a basis of $P_2^{*}$. Indeed, if $p\in \bigcap _{i=1}^3\ker \alpha _i$ then $p(1)=p(\sqrt {2})=p(\pi )=0$ so that $p$ has three distinct roots and so must vanish since it has degree no more than $2$.

Thus any $\alpha \in P_2^{*}$ is a linear combination of the $\alpha _i$. For example, define $\alpha $ by
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \alpha (p)=\int _0^1p. \end{equation*}

Then there are $\lambda _1,\lambda _2,\lambda _3\in \R $ such that $\alpha =\lambda _1\alpha _1+\lambda _2\alpha _2+\lambda _3\alpha _3$. Otherwise said, we have found clever $\lambda _i$ such that, for all $p\in P_2$,
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \int _0^1p=\lambda _1p(1)+\lambda _2p(\sqrt {2})+\lambda _3p(\pi ). \end{equation*}

Solution sets behave somewhat like orthogonal complements (except that $E$ and $\sol E$ live in entirely different vector spaces):

Proposition 5.8. Let $E,F\leq V^{*}$. Then
- (1) If $E\leq F$ then $\sol F\leq \sol E$.
- (2) $\sol $ swaps sums and intersections:
  $\seteqnumber{0}{5.}{1}$
  \begin{align*} \sol (E+F)&=(\sol E)\cap (\sol F)\\ (\sol E)+(\sol F)&\leq \sol (E\cap F) \end{align*} with equality if $V$ is finite-dimensional.

Proof.
- (1) Let $E\leq F$ and $v\in \sol F$. Then $\alpha (v)=0$, for all $\alpha \in F$ and so, in particular, for all $\alpha \in E$. Thus $v\in \sol E$.
- (2) This is an exercise⁴.
□

⁴ Question 3(a) on sheet 9.

Still thinking of $V^{*}$ as the linear equations on $V$, we can turn things around and ask which equations the elements of a subspace $U\leq V$ satisfy:

Definition. Let $U\leq V$. The annihilator of $U$, denoted $\ann U$ or $U^{\circ }$, is given by:
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \ann U:=\set {\alpha \in V^{*}\st \alpha _{|U}=0}=\set {\alpha \in V^{*}\st \text {$\alpha (u)=0$, for all $u\in U$ }}. \end{equation*}

Exercise.⁵ Show that $\ann U\leq V^{*}$.

⁵ Question 1 on sheet 9.

Annihilators have very similar properties to solution sets. They also have complementary dimension:

Proposition 5.9. Let $V$ be finite-dimensional and $U\leq V$. Then
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \dim \ann U=\dim V-\dim U. \end{equation*}

Proof. This is an exercise⁶ in imitating the proof of Proposition 5.6: start with a basis $\lst {v}1k$ of $U$, extend to a basis $\lst {v}1n$ of $V$ and see that $\ann U=\Span {\dlst {v}{k+1}n}$. Can you find a slick argument? □

⁶ Question 2 on sheet 9.

Again annihilators swap the order of inclusions and sums with intersections:

Proposition 5.10. Let $U,W\leq V$. Then
- (1) If $U\leq W$ then $\ann W\leq \ann U$.
- (2)
  $\seteqnumber{0}{5.}{1}$
  \begin{align*} \ann (U+W)&=(\ann U)\cap (\ann W)\\ (\ann U)+(\ann W)&\leq \ann (U\cap W) \end{align*} with equality if $V$ is finite-dimensional.

Proof. This is an exercise⁷. □

⁷ Question 3(b) on sheet 9.

What is the relation between annihilators and solution sets?

Lemma 5.11. Let $U\leq V$ and $E\leq V^{*}$ then $U\leq \sol E$ if and only if $E\leq \ann U$.

Proof. Both inclusions amount to saying $\alpha (u)=0$, for all $u\in U$ and $\alpha \in E$. □

With this in hand, we have:

Theorem 5.12. Let $U\leq V$ and $E\leq V^{*}$. Then
$\seteqnumber{0}{5.}{1}$
\begin{align*} U&\leq \sol (\ann U)\\ E&\leq \ann (\sol E), \end{align*} with equality if $V$ is finite-dimensional.

Proof. Clearly $\ann U\leq \ann U$ so putting $E=\ann U$ in Lemma 5.11 gives
$\seteqnumber{0}{5.}{1}$
\begin{equation*} U\leq \sol (\ann U). \end{equation*}

Similarly, $\sol E\leq \sol E$ so Lemma 5.11 gives
$\seteqnumber{0}{5.}{1}$
\begin{equation*} E\leq \ann (\sol E). \end{equation*}

If $V$ is finite-dimensional,
$\seteqnumber{0}{5.}{1}$
\begin{equation*} \dim \sol (\ann U)=\dim V-\dim \ann U=\dim U \end{equation*}

so that $U=\sol (\ann U)$. Similarly, $E=\ann (\sol E)$. □

Remark. We can view $\ann $ and $\sol $ as maps:
$\seteqnumber{0}{5.}{1}$
\begin{align*} \ann :\set {\text {subspaces of $V$}}&\to \set {\text {subspaces of $V^{*}$}}\\ \sol :\set {\text {subspaces of $V^{*}$}}&\to \set {\text {subspaces of $V$}}. \end{align*} When $V$ is finite-dimensional, Theorem 5.12 is telling us that these maps are mutually inverse bijections. This has a beautiful application to geometry that you can see in MA30231.