# Notes - Linear Algebra MT23, Annihilators > Source: https://ollybritton.com/notes/uni/part-a/mt23/linear-algebra/notes/annihilators/ · Updated: 2023-11-23 · Tags: uni, notes - [Course - Linear Algebra MT23](https://ollybritton.com/notes/uni/part-a/mt23/linear-algebra/) - [Notes - Linear Algebra MT23, Dual spaces](https://ollybritton.com/notes/uni/part-a/mt23/linear-algebra/notes/dual-spaces/) ### Flashcards Let $U \subseteq V$ be a subspace of $V$. Give both the technical definition of an annihilator of $U$ and the intuitive definition.:: - $U^0 = \{f \in V' \mid f(u) = 0 \text{ for all } u \in U\}$ - The set of all linear functionals that vanish on $U$. ^annihilator-of-subspace-definition Suppose $U \subseteq V$ is a subspace of $V$. How does $U^0$ relate to $V'$, and quickly prove that this is the case?:: $$ U^0 \le V' $$ 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 $U^0 \ne \varnothing$. ^annihilator-is-subspace-of-dual Suppose $U \subseteq V$ is a subspace of $V$. How do the dimensions of $U^0, U, V$ relate?:: $$ \dim U^0 = \dim V - \dim U $$ ^annihilator-dimension-formula Suppose $U, W \le V$. If $U \subseteq W$, what can you say about the relationship between $U^0$ and $W^0$?:: $$ U \subseteq W \implies W^0 \subseteq U^0 $$ ^annihilator-reverses-inclusions What does it mean to say "annihilators reverse inclusions"?:: If $U, W \le V$, then $$ U \subseteq W \implies W^0 \subseteq U^0 $$ ^annihilators-reverse-inclusions-meaning Quickly prove that if $U, W$ are subspaces of $V$, then $$ U \subseteq W \implies W^0 \subseteq U^0 $$ :: $$ \begin{aligned} f \in W^0 &\implies \forall w \in W \text{ } f(w) = 0 \\ &\implies \forall u \in U \subseteq W \text{ } f(u) = 0 \\ &\implies f \in U^0 \end{aligned} $$ ^annihilator-reverses-inclusions-proof How can you rewrite $(U + W)^0$?:: $$ U^0 \cap W^0 $$ ^annihilator-of-sum How can you rewrite $U^0 \cap W^0$?:: $$ (U + W)^0 $$ ^annihilator-intersection-as-sum Quickly prove that if $U, W$ are subspaces of $V$, then $$ (U + W)^0 = U^0 \cap W^0 $$ :: $$ \begin{aligned} f \in (U + W)^0 &\iff \forall u \in U,\text{ } \forall w \in W, f(u + w) = 0 \\ &\iff \forall u \in U,\text{ } f(u) = 0 \text{ and } \forall w \in W,\text{ } f(w) = 0 \\ &\iff f \in U^0 \cap W^0 \end{aligned} $$ ^annihilator-of-sum-proof We have the result that $$ (U + W)^0 = U^0 \cap W^0 $$ What is the relationship between $U^0 + W^0$ and $(U \cap W)^0$?:: $$ U^0 + W^0 \subseteq (U \cap W)^0 $$ and equal if $V$ is finite dimensional. ^annihilator-sum-and-intersection What isomorphism $V \to V''$ also maps $U \to U^{00}$?:: $$ v \to E_v $$ ^double-annihilator-via-natural-iso Suppose $U \le V$. What natural isomorphism establishes $$ U^0 \cong (V / U)' $$ ?:: $$ f \to \bar f $$ where $$ \bar f(v + U) := f(v) $$ ^annihilator-iso-quotient-dual Suppose: - $U \le V$ - $\Phi : f \mapsto \bar f$ where $\bar f(v + U) := f(v)$ Quickly prove that this is well defined and we have $$ U^0 \stackrel{\Phi}{\cong} (V/U)' $$ Show that this result does not depend on $V$ being finite dimensional.:: 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 $f|_U = 0$. - 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 $\overline{\hat g} = g$. ^annihilator-iso-quotient-dual-proof What is the canonical projection $\pi$ from $V$ to $V/U$?:: $$ \pi(v) = v + U $$ ^canonical-projection-to-quotient Quickly prove that if $U, W$ are subspaces of $V$, then $$ U^0 + W^0 \subseteq (U \cap W)^0 $$ and are equal if $V$ is finite dimensional. :: $$ \begin{aligned} f \in U^0 + W^0 &\implies \exists g \in U^0 \text{ and } h \in W^0,\text{ } f = g + h \\ &\implies \forall x \in U \cap W, \text{ } f(x) = g(x) + h(x) = 0 \\ &\implies f \in (U \cap W)^0 \end{aligned} $$ Then if $V$ is finite dimensional, we can show they have the same dimension $$ \begin{aligned} \dim(U^0 + W^0) &= \dim(U^0) + \dim(W^0) - \dim(U^0 \cap W^0) \\ &=\dim(U^0) + \dim(W^0) - \dim(U + W)^0 \\ &=(\dim(V) - \dim(U)) + (\dim(V) - \dim(W)) - (\dim(V) - \dim(U + W)) \\ &=\dim(V) - \dim(U) - \dim(W) + \dim(U) + \dim(W) - \dim(U \cap W) \\ &=\dim(V) - \dim(U \cap W) \\ &=\dim((U \cap W)^0) \end{aligned} $$ ^annihilator-sum-and-intersection-proof Suppose $U$ has basis $\{e_1, \cdots, e_k\}$ in an $n$-dimensional vector space. How can you write $U^0$ as the span of some set?:: $$ U^0 = \langle e_{k+1}', \cdots, e_n'\rangle $$ ^annihilator-as-dual-basis-span Suppose: - $V$ is a finite-dimensional vector space - $\Phi : V \to V''$ is the natural map given by $\Phi(v)(f) = f(v)$ Quickly prove that $U$ is mapped isomorphically to $U^{00}$. :: Note that $\Phi(v) \in U^{00}$ since $U^{00} = \{G \in V'' \mid \forall f \in U^{0}, G(f) = 0\}$ (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. ^u-maps-isomorphically-to-double-annihilator-proof Suppose: - $V$ is a vector space (not neccesarily finite dimensional) - $T : V \to V$ What relationship do we have between $\ker T, \mathrm{Im}\, T, \ker T', \mathrm{Im}\,T'$, and which one is difficult to prove in the infinite dimensional case?:: $$ \begin{aligned} \ker(T') &= (\mathrm{Im}(T))^0 \\ \mathrm{Im}(T') &= (\ker(T))^0 \end{aligned} $$ $\mathrm{Im}(T') = (\ker(T))^0$ is difficult to prove in the infinite dimensional case. ^dual-map-kernel-image-relations Suppose: - $V, W$ are vector spaces - $W$ is finite dimensional - $T : V \to W$ Quickly prove that $$ \dim \mathrm{Im}(T) = \dim \mathrm{Im}(T') $$ :: Note that $$ \ker(T') = (\mathrm{Im}(T))^0 $$ A quick proof: $$f \in \ker T' \iff \forall v \in V, f(Tv) = 0 \iff \mathrm{Im}(T) \subseteq \ker f \iff f \in \mathrm{Im}(T)^0$$ Then $$ \begin{aligned} \dim \mathrm{Im}(T) &= \dim W - \dim(\mathrm{Im}(T))^0 \\ &= \dim W' - \dim \ker T' \\ &= \dim \mathrm{Im}(T') \end{aligned} $$ ^dual-map-rank-equality-proof Suppose: - $V$ is a vector space - $T : V \to V$ Quickly prove that: $$ \begin{aligned} \ker(T') &= (\mathrm{Im}(T))^0 \\ \mathrm{Im}(T') &= (\ker(T))^0 \end{aligned} $$ For $\mathrm{Im}(T') = (\ker(T))^0$, give two proofs: one under the assumption that $V$ is finite dimensional, and one where $V$ might be infinite dimensional. In the infinite dimensional case, you may assume that for any subspace $U$ of $V$, we can find $W$ such that $V = U \oplus W$ (This is effectively the infinite-dimensional analogue of extending a basis of $U$ to a basis of $V$). :: **For $\ker(T') = (\mathrm{Im}(T))^0$**: $$ \begin{aligned} f \in \ker T' &\iff \forall v \in V, \text{ } f(Tv) = 0 \\ &\iff \mathrm{Im}(T) \subseteq \ker f \\ &\iff f \in \mathrm{Im}(T)^0 \end{aligned} $$ **For $\mathrm{Im}(T') = (\ker(T))^0$**: This is a bit tricker. For one direction: $$ \begin{aligned} f \in \mathrm{Im}(T') &\implies \exists h \in V' \text{ s.t. } f = h \circ T \\ &\implies \forall v \in \ker T, f(v) = h(Tv) = h(0) = 0 \\ &\implies f\in \ker(T)^0 \end{aligned} $$ In the finite dimensional case, we are done, since we have that $$ \mathrm{Im}(T') \subseteq \ker(T)^0 $$ and then since $\dim \mathrm{Im}(T') = \dim \ker(T)^0$, 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 $$ V = \mathrm{Im}(T) \oplus U $$ Then by defining a function $h : \mathrm{Im}(T) \to \mathbb F$, we can extend the definition to all of $V$ by taking $$ g := h \circ \pi $$ where $\pi : V \to \mathrm{Im}(T)$ is the canonical projection from $V$ to $\mathrm{Im}(T)$. 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 $f \in (\ker T)^0$, $f(r_1 - r_2) = 0$, 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 $v \in V$ $$ \begin{aligned} (T' g)v &= (T'(h \circ \pi))v \\ &= h(\pi(T(v))) \\ &= h(T(v)) &&\text{(since } Tv \in \mathrm{Im}(T)) \\ &= f(v) &&\text{(by the definition of }h) \end{aligned} $$ Thus $f = T'g$, so $f \in \mathrm{Im}(T')$. (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 $\mathrm{Im}(T)$). ^dual-map-kernel-image-relations-proof ### Proofs Suppose that $U \subseteq V$ is a subspace of $V$. Quickly prove that $$ \dim U^0 = \dim V - \dim U $$ :: **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). ^annihilator-dimension-formula-proof --- Olly Britton — https://ollybritton.com. Machine-readable index: https://ollybritton.com/llms.txt