Dual space/Introduction/Section

Definition

Let ${}K$ be a field and let ${}V$ denote an ${}K$-vector space. Then the space of homomorphisms

${}{V}^{*}:=\operatorname {Hom} _{K}{\left(V,K\right)}\,$

is called the dual space of ${}V$.

Addition and scalar multiplication are defined as in the general case of a space of homomorphisms, thus ${}(f+g)(v):=f(v)+g(v)$ and ${}(sf)(v):=s\cdot f(v)$. For a finite-dimensional ${}V$, we obtain, due to fact, that the dimension of the dual space ${}{V}^{*}$ equals the dimension of ${}V$.

Definition

Let ${}V$ denote a finite-dimensional ${}K$-vector space, endowed with a basis ${}v_{1},\ldots ,v_{n}$. Then the linear forms

${}v_{1}^{*},\ldots ,v_{n}^{*}\in {V}^{*}\,,$

defined by

${}v_{i}^{*}(v_{j})=\delta _{ij}={\begin{cases}1,{\text{ falls }}i=j\,,\\0,{\text{ falls }}i\neq j\,,\end{cases}}\,$

are called the dual basis

of the given basis.

Because of fact, this rule defines indeed a linear form. The linear form ${}v_{i}^{*}$ assigns to an arbitrary vector ${}v\in V$ the ${}i$-th coordinate of ${}v$ with respect to the given basis. Note that for ${}v=\sum _{j=1}^{n}s_{j}v_{j}$, we have

${}v_{i}^{*}(v)=v_{i}^{*}{\left(\sum _{j=1}^{n}s_{j}v_{j}\right)}=\sum _{j=1}^{n}s_{j}v_{i}^{*}(v_{j})=s_{i}\,.$

It is important to stress that ${}v_{i}^{*}$ does not only depend on the vector ${}v_{i}$, but on the basis. There doe not exist something like a "dual vector“ for a vector. This looks different in the situation where an inner product is given on ${}V$.

Example

For the standard basis ${}e_{1},\ldots ,e_{n}$ of ${}K^{n}$, the dual basis consists in the projections onto some component, that is, we have ${}e_{i}^{*}=p_{i}$, where

$p_{i}\colon K^{n}\longrightarrow K,\left(x_{1},\,\ldots ,\,x_{n}\right)\longmapsto x_{i}.$

This basis is called the standard dual basis.

Lemma

Let ${}V$ be a finite-dimensional ${}K$-vector space, endowed with a basis ${}v_{1},\ldots ,v_{n}$. Then the dual basis

${}v_{1}^{*},\ldots ,v_{n}^{*}\in {V}^{*}\,$

is a basis of the

dual space.

Proof

Suppose that

${}\sum _{j=1}^{n}a_{j}v_{j}^{*}=0\,,$

where ${}a_{j}\in K$. If we apply this linear form to ${}v_{i}$, we get directly

${}a_{i}=0\,.$

Therefore, the ${}v_{1}^{*},\ldots ,v_{n}^{*}$ are linearly independent. Due to fact, the dual space has dimension ${}n$, thus we have a basis already.

$\Box$

Lemma

Let ${}V$ be a finite-dimensional ${}K$-vector space, endowed with a basis ${}v_{1},\ldots ,v_{n}$, and the corresponding dual basis

${}v_{1}^{*},\ldots ,v_{n}^{*}\in {V}^{*}\,.$

Then, for every vector

${}v\in V$, the equality

${}v=\sum _{i=1}^{n}v_{i}^{*}(v)v_{i}\,.$

holds. The linear forms ${}v_{i}^{*}$ yield the scalars (coordinates)

of a vector with respect to a basis.

Proof

The vector ${}v$ has a unique representation

${}v=\sum _{j=1}^{n}s_{j}v_{j}\,$

with ${}s_{j}\in K$. The right hand side of the claimed equality is therefore

${}\sum _{i=1}^{n}v_{i}^{*}(v)v_{i}=\sum _{i=1}^{n}v_{i}^{*}{\left(\sum _{j=1}^{n}s_{j}v_{j}\right)}v_{i}=\sum _{i=1}^{n}s_{i}v_{i}=v\,.$

$\Box$

Lemma

Let ${}V$ be a finite-dimensional ${}K$-vector space. Let ${}v_{1},\ldots ,v_{n}$ be a basis of ${}V$ with the dual basis ${}v_{1}^{*},\ldots ,v_{n}^{*}$, and let ${}w_{1},\ldots ,w_{n}$ be another basis with the dual basis ${}w_{1}^{*},\ldots ,w_{n}^{*}$, and with

${}w_{r}=\sum _{k=1}^{n}a_{kr}v_{k}\,.$

Then

${}w_{j}^{*}=\sum _{i=1}^{n}b_{ij}v_{i}^{*}\,,$

where ${}{\left(b_{ij}\right)}_{ij}={{\left(A^{-1}\right)}^{\text{tr}}}$ is the transposed matrix of the inverse matrix of

${}A={\left(a_{kr}\right)}_{kr}$.

Proof

We have

${}{\begin{aligned}{\left(\sum _{i=1}^{n}b_{ij}v_{i}^{*}\right)}(w_{r})&={\left(\sum _{i=1}^{n}b_{ij}v_{i}^{*}\right)}{\left(\sum _{k=1}^{n}a_{kr}v_{k}\right)}\\&=\sum _{1\leq i,k\leq n}b_{ij}a_{kr}v_{i}^{*}(v_{k})\\&=\sum _{i=1}^{n}b_{ij}a_{ir}.\end{aligned}}$

Here, we have the "product“ of the ${}j$-th column of ${}B$ and the ${}r$-th column of ${}A$, which is also the product of the ${}j$-th row of ${}{B^{\text{tr}}}=A^{-1}$ and the ${}r$-te column of ${}A$. For ${}r=j$, this is ${}1$, and for ${}r\neq j$, this is ${}0$. Therefore, the given linear form coincides with ${}w_{j}^{*}$.

$\Box$

With the help of transformation matrices, this can be expressed as

${}M_{\mathfrak {v^{*}}}^{\mathfrak {w^{*}}}={({\left(M_{\mathfrak {v}}^{\mathfrak {w}}\right)}^{-1})^{\text{tr}}}={(M_{\mathfrak {w}}^{\mathfrak {v}})^{\text{tr}}}\,.$