6
$\begingroup$

Is the dual space of a subspace of a vector space simply its orthogonal complement? My professor seems to use the terms interchangeably, but from Wikipedia they seem to be quite distinct concepts (like dual spaces are much more complicated and involve functionals).

$\endgroup$
  • 7
    $\begingroup$ The orthogonal complement is a kind of dual, but should not be confused with the dual. "Dual" is one of those heavily overloaded words in mathematics and only context can make it clear what it means. $\endgroup$ – Qiaochu Yuan Apr 6 '12 at 0:44
  • $\begingroup$ you're probably thinking of something like en.wikipedia.org/wiki/Riesz_representation_theorem $\endgroup$ – yoyo Apr 6 '12 at 0:49
  • $\begingroup$ Suppose $V\subseteq W$ with $W$ an euclidean space of finite dimension. Then we have a map $W\rightarrow V^*$ given by $w\mapsto \langle w,-\rangle$. An element belong to the kernel of this map iff it belongs to $V^\perp$ and so by dimension $V^*$ can be identified with any complement of $V^\perp$ (i.e, a subspace $V'$ of $W$ of dimension $\text{dim}(V)$ such that $V'\cap V^\perp=\{0\}$). In particular we can identified $V^*$ with $V$ itself (in the style of Riesz theorem). $\endgroup$ – yamete kudasai May 10 '18 at 14:57
6
$\begingroup$

If $W \subset V$ is a subspace and $W^\perp$ its orthogonal complement, we have a direct sum decomposition $$V\cong W\oplus W^\perp.$$

This is true for any subspace $W'$ complementary to $W$, by definition. Since they are two pieces that together combine to make the whole, and the orthogonal complement in particular is an intuitive choice of complement, it is in some sense true that $W^\perp$ is "dual" to $W$.

However, there is already a vector space construction called the "dual vector space," denoted $V^\vee$.

This is the vector space of linear maps $V\to k$ (where $k$ is the field $V$ is defined over). The reason this is a vector space is, for any $\alpha,\beta\in k$ and $\mu,\lambda\in V^\perp$, we find $\alpha\mu+\beta\lambda\in V^\perp$ is a linear map also.

It is called the dual because there is a categorically natural isomorphism $V\cong V^{\vee\vee}$ (on the right is the dual of the dual space). The details of category theory might be out of place here, but suffice it to say there is no way to construct an isomorphism $V\cong V^{\vee}$ without a coordinate system for $V$, whereas it is easy to exhibit such an isomorphism for the double dual. For any $v\in V$, we can define the "evaluation map" that sends the linear functional $\lambda:V\to k$ to $\lambda(v)$ (that is, we interchange the role of function and argument: vectors act on functionals rather than vice versa.)

Given a basis $\{e_i\}$ for $V$ (finite-dimensional) however, we may construct an isomorphism $V\cong V^\perp$ using effectively the dot product. We can send a $v\in V$ to the linear functional $w\mapsto v\cdot w$. In the other direction, if $\lambda(w)$ is a linear functional it is determined by the values of $\lambda_i:=\lambda(e_i)$, because by linearity $\lambda\big(\sum_i w_i e_i\big)=\sum_i w_i\lambda(e_i)$, thus writing $\vec{\lambda}=(\lambda_1,\cdots,\lambda_n)$, we find that $\lambda(w)=\vec{\lambda}\cdot w$.

This is really an expansion of Qiaochu's comment, with a quick description of the dual space.

$\endgroup$
  • $\begingroup$ Great answer, thanks :D $\endgroup$ – badatmath Apr 6 '12 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.