(the numerical analysis tag is temporary, just for revision of [[dual vector space|unisolvency]]) ---- NB. This definition carries over to the discussion and verbiage of dual $R$-[[module|modules]], $R$ a [[commutative ring]], mostly cleanly. > [!definition] Definition. ([[dual vector space]]) > Let $V$ be [[vector space]] over [[field|field]] $\mathbb{F}$. We call [[vector space of linear maps between two vector spaces|the vector sapce]] $\hom(V, \ff)$ the **dual space** of $V$, and denote it $V^*$ or $V^{\vee}$. Its elements are variously called **linear functionals**, **linear forms**, **covectors**, **1-forms**, etc. > > Suppose further that $V$ is finite-dimensional. Upon fixing a [[basis]] $(\boldsymbol e_{1},\dots, \boldsymbol e_{n})$ of $V$, an element $\varphi \in V^{*}$ corresponds to a $1 \times n$ [[matrix]] (a **row vector**) written wrt the [[dual basis]] $(\varphi_{1},\dots, \varphi_{n})$. Under the [[isomorphism]] [[linear maps and basis of domain|determined by]] $\boldsymbol e_{i} \mapsto \varphi_{i}$, we have $\boldsymbol v = \sum_{i} v_{i} \boldsymbol e_{i} \mapsto \sum_{i} v_{i} \varphi_{i}$or as matrices ('in coordinates'): $\begin{bmatrix} v_{1} \\ \vdots \\ v_{n} \end{bmatrix} \mapsto \begin{bmatrix} v_{1} & \cdots & v_{n} \end{bmatrix}.$ Thus, $V \cong V^{*}$ by 'tipping vectors over'. *However*, this identification is not [[natural transformation|natural]]: it depended upon a choice of [[basis]]. > > What *is* [[natural transformation|natural]] is the [[double dual of a finite-dimensional vector space is naturally isomorphic to that space|identification]] $V \cong V^{**}$. The number $\varphi(v) \in \mathbb{F}$ obtained by applying a covector $\varphi$ to a vector $v$ is sometimes denoted by either of the more symmetric-looking notations $\langle \varphi, v \rangle$ or $\langle v, \varphi \rangle$. These expressions can be thought of as the action of $\varphi \in V^{*}$ on $v \in V$. Since $v$ is naturally identified with a [[linear functional]] $\xi(v):V^{*} \to \mathbb{F}$, $v \simeq \xi(v)$, they can also be thought of as the action of $\xi(v) \in V^{**}$ on $\varphi \in V^{*}$. Typically we omit mention of $\xi$ and think of $v$ either as a vector or as a [[linear functional]] $V^{*} \to \mathbb{F}$ depending on the context. > When $V$ is an [[dimension|infinite-dimensional]] [[normed vector space|normed vector space]], commonly one constrains $V^{*}$ to consist of *[[operator norm|bounded]]* linear functionals. (Some call this the **topological dual** of $V$.) - [ ] show the correspondence between bases and dual-dual bases (cf. LeeSM 275) > [!note] Remark. > Note that $\dim V = \dim V^*$ if $V$ is finite-dimensional. Hence $V \cong V^{*}$ per [[the rank theorem for free modules|the dimension theorem]]. ^note > [!justification] > From [[vector space of linear maps between two vector spaces#properties]] we have $\dim V^* = \dim V \cdot 1$, justifying the claim regarding [[dimension]]. ^justification > [!basicproperties] > ^properties For $V$ $n$-dimensional and $\varphi^{1},\dots,\varphi^{n} \in V^{*}$, checking that the $(\varphi^{i})$ form a basis for $V^{*}$ is equivalent to checking the **unisolvency condition** $\varphi^{i}(v)=0 \text{ for }i=1,\dots,n \implies v=0 .$ In other words, $(\varphi^{i})$ forms a basis for $V^{*}$ iff $\overbrace{ \bigcap_{i=1}^{n}\operatorname{ker } \varphi^{i} }^{ = (\varphi^{i})_{\circ} }=\{ 0 \}$. Where $(\varphi^{i})_{\circ}$ denotes the [[orthogonal complement|preannihilator]] of the set $\{ \varphi^{i} \}_{i=1}^{n}$. Define $W^{*}=\span \{ \varphi^{i} \}$, the dual space to some $W \leq V$ by [[double dual of a finite-dimensional vector space is naturally isomorphic to that space|double dual]] considerations. Using $W^{*}=V^{*} / W^{\circ}$, have $\text{dim }W^{*}=\text{dim }V^{*}-\text{dim }W^{\circ}$ [[dual of quotient is annihilator]] $\to.$ Suppose $(\varphi^{i})$ is a [[basis]] for $V^{*}$. Suppose $v \in V$ is such that $\varphi^{i}(v)=0$ for $i=1,\dots,n$. Get $e_{i}$ a basis of $V$ whose dual basis is $\varphi^{i}$. Write $v=v^{j}e_{j}$. Then $\varphi^{i}(v)=\varphi^{i}(v^{j}e_j)=v^{i}=0$ for $i=1,\dots,n$. Hence $v=0$. $\leftarrow$. Suppose we are given elements $(\varphi^{i})$ of $V^{*}$ for which it holds that $\varphi^{i}(v)=0 \text{ for }i=1,\dots,n \implies v=0 .$ By [[linearly independent list of the right length is a basis]], just have to check that $(\varphi^{i})$ is linearly independent. Suppose $a_{i} \in \mathbb{F}$ are such that $\sum_ia_{i}\varphi^{i}=0$, meaning $\sum_{i}a_{i}\varphi^{i}(v)=0$ for all $v \in V$. $\to$. Suppose $(\varphi^{i})_{\circ}=\{ 0 \}$. In finite dimensions [[orthogonal complement|annihilator]] and [[orthogonal complement|preannihilator]] are [[natural transformation|naturally isomorphic]], so $(\varphi^{i})^{\circ}=\{ 0 \}$. Now use [[dual of quotient is annihilator]] to get $\text{dim }\span(\varphi^{i})=\text{dim }V^{*}-\underbrace{ \text{dim }\span(\varphi^{i})^{\circ} }_{ =0 }$, from which the direction follows. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```