# Dual space/Introduction/Section
## Definition

Let $K$ be a field and let $V$ denote an $K$-vector space. Then the space of homomorphisms
    ${V}^{*}:=\operatorname {Hom} _{K}{\left(V,K\right)}\,$
$V$
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
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

### 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
    $v_{1}^{*},\ldots ,v_{n}^{*}\in {V}^{*}\,.$
$v\in V$, the equality
    $v=\sum _{i=1}^{n}v_{i}^{*}(v)v_{i}\,.$
holds. The linear forms $v_{i}^{*}$ yield the scalars (coordinates)

### 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
    $w_{r}=\sum _{k=1}^{n}a_{kr}v_{k}\,.$
    $w_{j}^{*}=\sum _{i=1}^{n}b_{ij}v_{i}^{*}\,,$
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
    $M_{\mathfrak {v^{*}}}^{\mathfrak {w^{*}}}={({\left(M_{\mathfrak {v}}^{\mathfrak {w}}\right)}^{-1})^{\text{tr}}}={(M_{\mathfrak {w}}^{\mathfrak {v}})^{\text{tr}}}\,.$