Complex conjugate of a vector space

In mathematics, one associates to every complex vector space $V\,$ its complex conjugate vector space ${\overline {V}}$, again a complex vector space. One reason for considering the conjugate vector space is that it allows one to think of antilinear maps as linear maps: an antilinear map from $V\,$ to $W\,$ gives rise to a linear map ${\overline {V}}\to W$, and conversely.

The underlying set of vectors and the addition of ${\overline {V}}$ are the same as those of $V\,$, but the scalar multiplication in ${\overline {V}}$ is defined as follows:

to multiply the complex number $\alpha \,$ with the vector $v\,$ in ${\overline {V}}$, take the complex conjugate ${\bar {\alpha }}$ of $\alpha \,$ and multiply it with $v\,$ in the original space $V\,$.

The map ${\overline {(\,\cdot \,)}}:V\to {\overline {V}}$ defined by ${\overline {v}}=v$ for all $v\,$ in $V\,$ is then bijective and antilinear. Furthermore, we have ${\overline {\overline {V}}}=V$ and ${\overline {\overline {v}}}=v$ for all $v\,$ in $V\,$.

An antilinear map $V\to W$, for another vector space $W\,$, is the same thing as a linear map ${\overline {V}}\to W$.

Given a linear map $f:V\to W$, the conjugate linear map ${\overline {f}}:{\overline {V}}\to {\overline {W}}$ is defined by the formula:

${\overline {f}}({\overline {v}})={\overline {f(v)}}$.

The conjugate linear map ${\overline {f}}$ is linear. Moreover, the rules $V\mapsto {\overline {V}}$ and $f\mapsto {\overline {f}}$ define a functor from the category of C-vector spaces to itself.

If $V\,$ and $W\,$ are finite-dimensional and the map $f\,$ is described by the matrix $A\,$ with respect to the bases ${\mathcal {B}}$ of $V\,$ and ${\mathcal {C}}$ of $W\,$, then the map ${\overline {f}}$ is described by the complex conjugate of $A\,$ with respect to the bases ${\overline {\mathcal {B}}}$ of ${\overline {V}}$ and ${\overline {\mathcal {C}}}$ of ${\overline {W}}$.

The vector spaces $V\,$ and ${\overline {V}}$ have the same dimension over C and are therefore isomorphic as C-vector spaces. However, there is no natural isomorphism from $V\,$ to ${\overline {V}}$.