So, we know that there is no "natural" isomorphism between a finite dimensional vector space $X$ and its dual $X^{\vee}$. However, given a basis for $X$ $(e_1, \dots, e_n)$ we can always construct a (dual) basis $(e^1, \dots, e^n)$ for $X^{\vee}$ that satisfies $e^i(e_j) = \delta^i_j$. It follows that $V \approx V^{\vee}$ simply because they have the same finite dimension $n$. Now, it seems to me that we can construct an isomorphism by assigning $$ e_i \mapsto e^i $$ and extending by linearity. That is, for an arbitrary vector $v \in X$ given by $v = \sum v^ie_i$ we can specify a linear map

$$\phi:V \rightarrow V^{\vee}$$


$$ \phi(\sum v^ie_i) := \sum v^ie^i $$ It is a quick exercise to check that $\phi$ is an isomorphism. Now, although this construction seems correct, there is something that doesn't quite sit right with me about it. For one thing, there is no way to really make sense of the summation convention as the right side of the function will always contain two superscripts thus requiring an explicit summation sign. The other thing that feels rather off is actually taking a lower index and moving it to an upper index, i.e., $e_i \mapsto e^i$ (which, of course, is the reason that the summation convention doesn't work).

So, with this background my questions are

  1. Is there any sense in which the above isomorphism is "favored" or "canonical"?

  2. Is there another way to construct an isomorphism between a vector space and its dual, that would "conserve indexes", for lack of a better way to state it. My guess here is that the answer is no but becomes possible if one assumes $V$ has an inner product.

  • 5
    $\begingroup$ A vector space which has a specified ordered basis is indeed canonically isomorphic to its dual, by the construction you outlined. However, many vector spaces do not come equipped with a specified ordered basis. $\endgroup$ Feb 4, 2012 at 1:01
  • 1
    $\begingroup$ For the summation convention, you could write $\phi = \delta_{ij}e^i \otimes e^j$, so $\phi(v) = \delta_{ij}e^i \otimes e^j (v^k e_k) = v^k \delta_{ij}\delta^j_k e^i = v^k \delta_{ik} e^i = v_i e^i$, the last equality defining $v_i$. $\endgroup$
    – yasmar
    Feb 4, 2012 at 12:28

1 Answer 1


Question 1: The problem is that if you change the basis, then the dual basis also changes, but in a different way. More precisely, suppose we decide to use the basis $f_i = M e_i$ instead of the basis $e_i$. Then the corresponding dual basis $f_i^{\ast}$ (thinking of $e_i^{\ast}$ as a linear functional $V \to k$) is given by $f_i^{\ast} = e_i^{\ast} M^{-1}$. Indeed the defining property of the dual basis, namely that $$e_i^{\ast}(e_j) = \delta_{ij}$$

(where $\delta_{ij} = 1$ if $i = j$ and is $0$ otherwise) is satisfied here, since $$f_i^{\ast}(f_j) = e_i^{\ast} M^{-1} M e_j = e_i^{\ast} e_j = \delta_{ij}.$$

Note that $M^{-1}$ acts on the right instead of on the left, so if one insists on writing linear transformations as matrices acting on the left, then we need in addition to take the transpose.

This is a reflection of the fact that taking dual spaces is a contravariant functor rather than a covariant one.

Question 2: For $V$ finite-dimensional, specifying an isomorphism $V \to V^{\ast}$ is equivalent to specifying a nondegenerate bilinear form $V \times V \to k$. This form need not be symmetric or an inner product in general. (In fact the notion makes sense over an arbitrary field, and an arbitrary field doesn't have a notion of positivity.)

  • $\begingroup$ This made me think that any finite dim vectorspace could be equipped with an inner prodouct and hence become a Hilbert space. Or must one always also impose the norm on the space that this innerprodct indues? We could just use the inner product to characterise the functionals for instance. I cant figure out why this woult work $\endgroup$
    – user123124
    Oct 18, 2015 at 15:54
  • $\begingroup$ @User2313: yes we could, but such an inner product would not be unique; you'd need to specify it as extra data. $\endgroup$ Oct 18, 2015 at 17:08
  • $\begingroup$ Do you mean $ (\mathbb{R}^{n}, \mid \mid * \mid \mid , <*, * > ) $ and stressing that <*,*> has nothing to do with he norm, as "specify as extra data" ? $\endgroup$
    – user123124
    Oct 18, 2015 at 17:18
  • $\begingroup$ @User2313: yes, but I don't see what a norm has to do with this discussion. $\endgroup$ Oct 18, 2015 at 17:20
  • $\begingroup$ Im sorry, because we already have one, "any finite dim. normed vector space". I just want an nice way to charactrise the functionals $\endgroup$
    – user123124
    Oct 18, 2015 at 17:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .