Notes - Linear Algebra MT23, Annihilators

• $U^0 = \{f \in V’ \mid f(u) = 0 \text{ for all } u \in U\}$
• The set of all linear functionals that vanish on $U$.

\[U^0 \le V'\]

Proof: let $f, g \in U^0$. Then

\[(\lambda f + g)(u) = \lambda f(u) + g(u) = 0 + 0 = 0\]

also $0 \in U^0$ and $U^0 \ne \varnothing$.

\[\dim U^0 = \dim V - \dim U\]

\[U \subseteq W \implies W^0 \subseteq U^0\]

If $U, W \le V$, then

\[U \subseteq W \implies W^0 \subseteq U^0\]

\[\begin{aligned} f \in W^0 &\implies \forall w \in W \text{ } f(w) = 0 \\\\ &\implies \forall u \in U \subseteq W \text{ } f(u) = 0 \\\\ &\implies f \in U^0 \end{aligned}\]

\[U^0 \cap W^0\]

\[(U + W)^0\]

\[\begin{aligned} f \in (U + W)^0 &\iff \forall u \in U,\text{ } \forall w \in W, f(u + w) = 0 \\\\ &\iff \forall u \in U,\text{ } f(u) = 0 \text{ and } \forall w \in W,\text{ } f(w) = 0 \\\\ &\iff f \in U^0 \cap W^0 \end{aligned}\]

\[U^0 + W^0 \subseteq (U \cap W)^0\]

and equal if $V$ is finite dimensional.

\[v \to E_v\]

\[f \to \bar f\]

where

\[\bar f(v + U) := f(v)\]

If we can show that $\Phi$ is:

• well-defined
• linear
• injective

then we are done, since the linearity and injectivity implies $\Phi$ will be surjective and that we can establish the isomorphism in the finite-dimensional case by considering the dimensions of both sides.

• To see well-definedness, note that $f \vert _ U = 0$.
• Linearity follows from the fact $\overline{\lambda f + h} = \lambda \overline f + \overline h$.
• Injectivity: $\overline f = 0 \implies \overline f(v + U) = f(v) = 0$ for all $v \in V$ $\implies$ $f = 0$

Then since

\[\begin{aligned} \dim U^0 &= \dim V - \dim U \\\\ &= \dim(V/U) \\\\ &= \dim((V/U)') \end{aligned}\]

It is an isomorphism. In the infinite-dimensional case, we can also construct an explicit inverse: Let $g \in (V/U)’$. Define $\hat g \in V’$ by $\hat g(v) := g(v + U)$. Then $\hat g$ is linear in $v$. Furthermore, the map $g \mapsto \hat g$ is linear in $g$, with image in $U^0$. Then $\hat{\overline f} = f$ and $\overline{\hat g} = g$.

\[\pi(v) = v + U\]

\[\begin{aligned} f \in U^0 + W^0 &\implies \exists g \in U^0 \text{ and } h \in W^0,\text{ } f = g + h \\\\ &\implies \forall x \in U \cap W, \text{ } f(x) = g(x) + h(x) = 0 \\\\ &\implies f \in (U \cap W)^0 \end{aligned}\]

Then if $V$ is finite dimensional, we can show they have the same dimension

\[\begin{aligned} \dim(U^0 + W^0) &= \dim(U^0) + \dim(W^0) - \dim(U^0 \cap W^0) \\\\ &=\dim(U^0) + \dim(W^0) - \dim(U + W)^0 \\\\ &=(\dim(V) - \dim(U)) + (\dim(V) - \dim(W)) - (\dim(V) - \dim(U + W)) \\\\ &=\dim(V) - \dim(U) - \dim(W) + \dim(U) + \dim(W) - \dim(U \cap W) \\\\ &=\dim(V) - \dim(U \cap W) \\\\ &=\dim((U \cap W)^0) \end{aligned}\]

\[U^0 = \langle e_{k+1}', \cdots, e_n'\rangle\]

Note that $\Phi(v) \in U^{00}$ since $U^{00} = \{G \in U^{00} \mid \forall f \in U^{0}, G(f) = 0\}$ (where $0$ represents the $0$ map) and if $f$ vanishes on $U$, $\Phi(v)(f) = 0$ since we are just evaluating $f$ on elements of $u$. This means that

\[U \cong \Phi(U) \subseteq U^{00}\]

Since $V$ is finite dimensional, we also then have that

\[\begin{aligned} \dim(U^{00}) &= \dim(V') - \dim(U^0) \\\\ &= \dim(V) - (\dim (V) - \dim (U)) \\\\ &= \dim (U) \end{aligned}\]

and hence $\Phi(U)$ and $U^{00}$ must be equal.

\[\begin{aligned} \ker(T') &= (\text{Im}(T))^0 \\\\ \text{Im}(T') &= (\ker(T))^0 \end{aligned}\]

$\text{Im}(T’) = (\ker(T))^0$ is difficult to prove in the infinite dimensional case.

Note that

\[\ker(T') = (\text{Im}(T))^0\]

A quick proof:

\[f \in \ker T' \iff \forall v \in V, f(Tv) = 0 \iff \text{Im}(T) \subseteq \ker f \iff f \in \text{Im}(T)^0\]

Then

\[\begin{aligned} \dim \text{Im}(T) &= \dim W - \dim(\text{Im}(T))^0 \\\\ &= \dim W' - \dim \ker T' \\\\ &= \dim \text{Im}(T') \end{aligned}\]

For $\ker(T’) = (\text{Im}(T))^0$:

\[\begin{aligned} f \in \ker T' &\iff \forall v \in V, \text{ } f(Tv) = 0 \\\\ &\iff \text{Im}(T) \subseteq \ker f \\\\ &\iff f \in \text{Im}(T)^0 \end{aligned}\]

For $\text{Im}(T’) = (\ker(T))^0$: This is a bit tricker. For one direction:

\[\begin{aligned} f \in \text{Im}(T') &\implies \exists h \in V' \text{ s.t. } f = h \circ T \\\\ &\implies \forall v \in \ker T, f(v) = h(Tv) = h(0) = 0 \\\\ &\implies f\in \ker(T)^0 \end{aligned}\]

In the finite dimensional case, we are done, since we have that

\[\text{Im}(T') \subseteq \ker(T)^0\]

and then since $\dim \text{Im}(T’) = \dim \ker(T)^0$, these must actually be equal.

Now assume that $V$ is infinite dimensional. For the other direction, suppose $f \in (\ker T)^0$. We want to find some $g \in V’$ such that $f = T’g$.

Ideally we would be able to define

\[g(v) := f(T^{-1}v)\]

so that

\[f(v) = f(T^{-1}T v) = g(Tv)\]

But there are two issues with this: $v$ is not necessarily in the image of $T$, so there might not be any inverse, and if there is an inverse, it won’t necessarily be unique.

We can address the issue of $v$ not being in the image of $T$ by considering that there exists a subspace $U$ of $V$ such that

\[V = \text{Im}(T) \oplus U\]

Then by defining a function $h : \text{Im}(T) \to \mathbb F$, we can extend the definition to all of $V$ by taking

\[g := h \circ \pi\]

where $\pi : V \to \text{Im}(T)$ is the canonical projection from $V$ to $\text{Im}(T)$.

To address the issue that the inverse is not necessarily unique, we can actually show that we can take any inverse $r \in T^{-1}v$ (i.e. any $r$ such that $v = Tr$) and it will give the same answer.

Suppose $r _ 1$ and $r _ 2$ are two inverses, so that $v = T(r _ 1) = T(r _ 2)$. Then $r _ 1 - r _ 2 \in \ker T$, and so crucially by the assumption that $f \in \ker T$, $f(r _ 1 - r _ 2) = 0$, hence $f(r _ 1) = f(r _ 2)$.

(Also, $h$ is linear since $f$ and $T$ are linear and it is well-defined).

Now for arbitrary $v \in V$

\[\begin{aligned} (T' g)v &= (T'(h \circ \pi))v \\\\ &= h(\pi(T(v))) \\\\ &= h(T(v)) &&\text{(since } Tv \in \text{Im}(T)) \\\\ &= f(v) &&\text{(by the definition of }h) \end{aligned}\]

Thus $f = T’g$, so $f \in \text{Im}(T’)$.

(note that this proof also gives an explicit inverse in the finite dimensional case, since the projection $\pi$ could be viewed as the matrix that $0$s out the entries of $v$ in the subspace of $V$ that is not spanned by $\text{Im}(T)$).

Proof sketch: Consider a basis of $U$, extend to a basis of $V$. It turns out that the duals of the new basis vectors in the extension form a basis for $U^0$ (this can be shown by double inclusion).