# dual space

Dual of a vector space^{}; dual bases

Let $V$ be a vector space over a field $k$. The *dual* of $V$,
denoted by ${V}^{\ast}$, is the vector space of linear forms^{} on $V$, i.e.
linear mappings $V\to k$.
The operations^{} in ${V}^{\ast}$ are defined pointwise:

$$(\phi +\psi )(v)=\phi (v)+\psi (v)$$ |

$$(\lambda \phi )(v)=\lambda \phi (v)$$ |

for $\lambda \in K$, $v\in V$ and $\phi ,\psi \in {V}^{\ast}$.

$V$ is isomorphic to ${V}^{\ast}$ if and only if the dimension^{} of
$V$ is finite. If not, then ${V}^{\ast}$ has a larger (infinite^{})
dimension than $V$; in other words, the cardinal of any basis
of ${V}^{\ast}$ is strictly greater than the cardinal of any basis of $V$.

Even when $V$ is finite-dimensional, there is no canonical or natural isomorphism $V\to {V}^{\ast}$. But on the other hand, a basis $\mathcal{B}$ of $V$ does define a basis ${\mathcal{B}}^{\ast}$ of ${V}^{\ast}$, and moreover a bijection $\mathcal{B}\to {\mathcal{B}}^{\ast}$. For suppose $\mathcal{B}=\{{b}_{1},\mathrm{\dots},{b}_{n}\}$. For each $i$ from $1$ to $n$, define a mapping

$${\beta}_{i}:V\to k$$ |

by

$${\beta}_{i}(\sum _{k}{x}_{k}{b}_{k})={x}_{i}.$$ |

It is easy to see that the ${\beta}_{i}$ are nonzero elements of ${V}^{\ast}$
and are independent. Thus $\{{\beta}_{1},\mathrm{\dots},{\beta}_{n}\}$ is a basis of
${V}^{\ast}$, called the dual basis^{} of $\mathcal{B}$.

The dual of ${V}^{\ast}$ is called the *second dual* or *bidual* of $V$.
There *is* a very simple canonical injection $V\to {V}^{\ast \ast}$,
and it is an isomorphism^{} if the dimension of $V$ is finite.
To see it, let $x$ be any element of $V$ and define a mapping ${x}^{\prime}:{V}^{\ast}\to k$
simply by

$${x}^{\prime}(\varphi )=\varphi (x).$$ |

${x}^{\prime}$ is linear by definition, and it is readily verified that the mapping
$x\mapsto {x}^{\prime}$ from $V$ to ${V}^{\ast \ast}$ is linear and injective^{}.

Dual of a topological vector space^{}

If $V$ is a topological vector space, the *continuous dual*
${V}^{\prime}$ of $V$ is the subspace^{} of ${V}^{\ast}$ consisting of
the *continuous ^{}* linear forms.

A *normed* vector space $V$ is said to be *reflexive ^{}* if the natural
embedding $V\to {V}^{\prime \prime}$ is an isomorphism. For example,
any finite dimensional space is reflexive, and any Hilbert space

^{}is reflexive by the Riesz representation theorem.

Remarks

Linear forms are also known as linear functionals^{}.

Another way in which a linear mapping $V\to {V}^{\ast}$ can arise is via
a bilinear form^{}

$$V\times V\to k.$$ |

The notions of duality extend, in part, from vector spaces to modules,
especially free modules^{} over commutative rings. A related notion is
the duality in projective spaces.

Title | dual space^{} |

Canonical name | DualSpace |

Date of creation | 2013-03-22 12:16:52 |

Last modified on | 2013-03-22 12:16:52 |

Owner | Daume (40) |

Last modified by | Daume (40) |

Numerical id | 15 |

Author | Daume (40) |

Entry type | Definition |

Classification | msc 15A99 |

Synonym | algebraic dual |

Synonym | continuous dual |

Synonym | dual basis |

Synonym | reflexive |

Synonym | natural embedding |

Synonym | topological dual |

Related topic | DualHomomorphism |

Related topic | DoubleDualEmbedding |

Related topic | BanachSpace |

Related topic | Unimodular |

Related topic | LinearFunctional |

Related topic | BoundedLinearFunctionalsOnLpmu |