I having been chipping away at understanding contravariant vs covariant vectors, linear functionals, 1-forms, dual space, dual basis, etc towards the goal of finally understanding multilinear algebra and tensors (and ultimately general relativity).

Bowen’s excellent text Introduction to Vectors and Tensors: Linear and Multilinear Algebra just threw a wrench in things and has me confused.

In section 14 he carefully introduces the concept of a “reciprocal basis” which was new to me but initially made sense. Every basis $\{\mathbf e_i\}$ of an $N$-dimensional inner product vector space has a unique reciprocal basis $\{\mathbf e^i\}$ such that $${\mathbf e^i}\cdot{\mathbf e_j}=\delta^i_j$$ He provides a nice graphical example in $\Bbb R^2$. He goes on to describe components under the basis and reciprocal basis as “contravariant” and “covariant” components of the same vector. These notions make sense to me since the fact that the scalar product of matching pairs of basis/reciprocal basis vectors is one assures that that they vary in opposite directions when scaled or rotated.

The confusion arose when in Section 31 he launches into a traditional description of a dual basis of the dual space of linear functionals (1-forms, covectors) on a vector space. Here contravariant and covariant vectors are entirely different objects in different vector spaces and a bilinear operation between elements of the basis and the dual basis (not an inner product defined in either vector space) returns the Kronecker delta.

How should I reconcile these superficially similar descriptions of what seem to be fundamentally different things? Are they linked by an isomorphism between the dual basis in the dual space and the reciprocal basis in original vector space? Does this relate to the canonical isomorphism (independent of basis) between a vector space and the dual of its dual (something I am still trying to get my head around)? Thanks!

(I did read this question and it’s answers but was looking for more depth)

  • $\begingroup$ "Reciprocal basis" vectors are just dual basis vectors relative to the inner product under a slightly more abstract, symmetric definition of duality than the one given in your book. I personally prefer the symmetric definition because it tends to eliminate the sort of confusion seen in your question, but it is decidedly unpopular. $\endgroup$
    – blargoner
    Dec 30, 2022 at 1:37

3 Answers 3


The reciprocal basis and dual basis are essentially the same. Here is why.

To explicate the "abstract, symmetric definition of duality" mentioned in the comments, the idea here is that an inner product space is canonically isomorphic to its dual under the map $v \mapsto \langle v, \_\rangle$. Since we are in $\mathbb R^n$, there is a canonical inner product given by the dot product (in general, this comes from the Rietz Representation Theorem).

Thus, in an inner product space $V$ with inner product $\langle \_, \_ \rangle$ and orthonormal basis $e_i$ (meaning $\langle e_i, e_j \rangle = \delta_i^j$, i.e. take a set of basis vectors, make them orthogonal via Gram-Schmidt, then normalize) we get a dual basis $e^i$ in $V^*$ given by $e^i = \langle e_i, \_ \rangle$. Indeed, $e^i(e_j) = \langle e_i, e_j \rangle = \delta_i^j$ as desired.

Thus the dot product gives us an isomorphism from $\mathbb R^n$ to its dual, meaning $e^i$ should themselves be a basis for $\mathbb R^n$. As such, we can calculate $e^i$ using a change of basis matrix.

While this was a bit abstract, let us now think about how to represent dual vectors a bit more concretely in $\mathbb R^n$. To start, we usually start by writing a vector in a finite dimensional space as a column vector, e.g. in $\mathbb R^3$ we write $e_2 = [0,1,0]^T$. We then write dual vectors as row vectors, since it is forced by compatibility with the identification of linear maps with matrices. (Indeed, this is essentially forced from the fact we want to write $Ax$ for a matrix $A$ and vector $x$ instead of $xA$.) Recall the standard representation of a linear map from $\mathbb R^n$ to $\mathbb R^m$, or more generally from an $n$ dimensional vector space to an $m$ dimensional vector space, is given by an $m \times n$ matrix (whose columns are the images of the basis vectors). Since a dual vector is a linear map from $\mathbb R^n$ to $\mathbb R$, it must be represented by a $1 \times n$ matrix, i.e. a row vector.

Next, note the way matrix multiplication works means a row vector acting a column vector is basically just a dot product. Indeed, $[a_1, \ldots, a_i][b_1, \ldots, b_i]^T = a_1 b_1 + a_2 b_2 + \cdots + a_i b_i = (a_1, \ldots, a_i)\cdot (b_1, \ldots, b_i)$. We can thus think of vectors as the things on the "right" of a dot product, and dual vectors as the things on the "left". Thinking back to our earlier isomorphism $e^i = \langle e_i, \_ \rangle$, since our inner product is here given by the dot product, the dual vector $e^i$ is simply given by $e^i(v) = e_i \cdot v$

The reason this is called the "reciprocal basis" comes from how we actually calculate the dual basis from the change of basis matrix. Indeed, given some basis $\{b_i\}$ of a vector space $V$ of dimension $n$ we can form the matrix $P = [b_1, \ldots, b_n]$ and the matrix $B = [b_1^*, \ldots, b_n^*]$, where $\{b_i^*\}$ is the dual basis (to $\{b_i\}$) which we seek (the choice of the letter $P$ comes from it being a change of basis matrix, explained below). We now consider the transpose matrix $B^T$ whose rows are the vectors $b_i^*$. Since matrix multiplication takes rows on the left and dot products them with columns on the right and $b_i^* \cdot b_j = \delta^i_j$ per our earlier discussion, we find: $$B^T P = [b_1^*, \ldots, b_n^*]^T [b_1, \ldots, b_n] = \begin{bmatrix} b_1^* \cdot b_1 & b_1^* \cdot b_2 & \cdots & b_1^* \cdot b_n \\ b_2^* \cdot b_1 & b_2^* \cdot b_2 & \cdots & b_2^* \cdot b_n \\ \vdots & \vdots & \ddots & \vdots \\ b_n^* \cdot b_1 & b_n^* \cdot b_2 & \cdots & b_n^* \cdot b_n \end{bmatrix} = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix}$$

i.e. $B^T P = I$ so $B = (P^{-1})^T$. In other words, the dual basis we seek is given by the columns of $(P^{-1})^T$, or equivalently (and more in line with our view of dual vectors as row vectors) is given by the rows of $P^{-1}$. Since taking the inverse of a real number is often called taking its "reciprocal", by analogy these dual vectors are sometimes called the "reciprocal basis" since they come from taking the inverse.

Finally, note this is the same thing we would get from the change of basis matrix idea mentioned earlier, and to get there we need a few facts. First, recall that if $T \colon A \to B$ is a linear map with matrix $M$ then the tranpose of $T$ is the map $T^* \colon B^* \to A^*$ given by $f \mapsto f \circ T$ whose matrix representation in the dual basis is given by $M^T$ (note the transpose matrix "flips" the vector spaces along with dualizing them). Second, recall that if $T$ is invertible then $M^{-1}$ is the matrix associated to $T^{-1} \colon B \to A$ (note the inverse also "flips" the spaces). Finally, between any two bases $B_1$ and $B_2$ on a vector space $V$, there exists a change of basis matrix $P$ whose columns are vectors $B_1$ expressed in terms of $B_2$. Multiplication on the left by $P$ takes a vector expressed in $B_1$ coordinates and returns the same vector expressed in $B_2$ coordinates, so $P$ is referred to as the identity map from $(V, B_1)$ to $(V, B_2)$

Proceeding, let $\scr E$ be the canonical basis $\{e_i\}$ on $V$ with dimension $n$ and $\scr B$ be the given basis $\{b_i\}$ at hand. Since we seek the coordinates for the dual basis $\{b_i^*\}$ in terms of the canonical dual basis $\{e_i^*\}$, what we need to calculate is the identity map from $(V^*,\scr B^*)$ to $(V^*, \scr E^*)$. To this end, note the the identity map from $(V,\scr B)$ to $(V, \scr E)$. is given by $P := [b_1, \ldots, b_n]$, i.e. the columns are the vectors $b_i$. We would like the dualize immediately via the transpose, but this would flip the spaces; to fix this, we flip them an extra time via the inverse. Indeed, $P^{-1}$ is the identity map from $(V,\scr E)$ to $(V, \scr B)$ so $(P^{-1})^T$ is the identity map from $(V^*,\scr B^*)$ to $(V^*, \scr E^*)$. We once again conclude the columns of $(P^{-1})^T$ are the dual basis to $\{b_i\}$, i.e. that the rows of $P^{-1}$ are the dual basis to $\{b_i\}$.

As an example, note the claim made on Wikipedia that

$$b^1 = \left(\frac{b_2 \times b_3}{V}\right)^T,\;\; b^2 = \left(\frac{b_3 \times b_1}{V}\right)^T,\;\; b^3 = \left(\frac{b_1 \times b_2}{V}\right)^T.$$

(where $V$ is the volume of the parallelepiped spanned by the $b_i$) simply follows from the following formula for the inverse of a $3 \times 3$ matrix: $$P^{-1} = [b_1, b_2, b_3]^{-1} = \frac{1}{\det P} [b_2 \times b_3, b_3 \times b_1, b_1 \times b_2]^T$$

(Note $V = \det P$ by the standard geometric interpretation of the determinant.) For a proof of this formula you can use the standard expression for the cross product as a determinant, which breaks into a sum of cofactors. When one writes out the formula for the the cofactor matrix, the columns are immediately seen to be the cross products, and the inverse matrix is classically given by the tranpose of the cofactor matrix scaled by the multiplicative inverse of the determinant. Alternatively, a more geometric proof can be found here.


Q) How should I reconcile these superficially similar descriptions of what seem to be fundamentally different things?
A) They are the same thing, depending on what you mean by "same" (equivalent up to some dualities). It is thus expected they have the same description.

Q) Are they linked by an isomorphism between the dual basis in the dual space and the reciprocal basis in original vector space?
A) Yes.

Q) Does this relate to the canonical isomorphism (independent of basis) between a vector space and the dual of its dual (something I am still trying to get my head around)?
A) Not in any meaningful way.

  • 1
    $\begingroup$ @user175324 while there are many reasons for it, the most basic one is that the dual space of a vector space $V$ is never canonically isomorphic to $V$... It's only in the presence of an inner product this is true. Now, all finite dimensional spaces do admit an inner product (as do some infinite dimensional ones), but on most spaces none of these inner products is natural in any sense, and there are plenty to choose from. Even in $\mathbb R^n$ there isn't a unique inner product, just a canonical one coming from the dot product. Wanting to vary inner products is an important idea in... $\endgroup$ Jan 2 at 7:42
  • 1
    $\begingroup$ ... differential geometry where the choice of an inner product on your surface determines lengths, angles, mass, etc. In fact, on a manifold (the technical name for $n$ dimensional surface), even though the tangent space at each point is a vector space of dimension $n$, the choice of inner product on each space usually differs point to point. As a physical example, in General Relativity we must assign to each point of spacetime a metric (inner product) with signature $(-, +, +, +) $, but if the inner product didn't change between points then gravity couldn't arise from bent spacetime! $\endgroup$ Jan 2 at 7:46
  • 1
    $\begingroup$ (I mention these examples since tensors come up a lot in physics). Other reasons include generalizations of vector spaces having duals which are no longer isomorphic to their original space (e.g. Pontrayagin Duality) and infinite dimensional vector spaces being super important but usually not being isomorphic to their duals. Finally, it's also just important to have a notion of linear functionals on a space, so we need to name the space anyway (linear functionals are so important that the field of Functional Analysis is named after them, and is one of the biggest branches of mathematics) $\endgroup$ Jan 2 at 7:47
  • 1
    $\begingroup$ The very brief answer to why tensors involve dual spaces is that tensors arose from the need to express global transformations that locally transform in a way that is partly contravariant and partly covariant, depending on what is being measured. By fixing a vector space $V$ and considering a tensor as a multilinear map from a product of $V$ and $V^*$ to the base field, we get the covariant parts from the copies of $V$ and the contravariant parts from the copies of the dual space. Also, if $T: V\times\cdots \times V \times V^* \times\cdots \times V^* \to F$ is the tensor, then... $\endgroup$ Jan 2 at 7:59
  • 1
    $\begingroup$ the fact $T$ is multilinear means it is entirely determined by how it acts on the basis, i.e. it's entirely determined by the values $T(e_{i_1},\ldots,e_{i_p},e^{j_1},\ldots,e^{j_q})$. If you think about it, this basically says tensors can be though of as a multidimensional array, for the same reason that linear maps can be thought of as matrices (recall the associated matrix to a linear map has as it's columns the images of the basis under the linear map, so knowing the values of a (multi)linear map on the basis determines the associated array) $\endgroup$ Jan 2 at 8:04

$ \newcommand\span{\mathop{\mathrm{span}}} $You defined reciprocal basis using the dot product; such a dot product is extra structure on top of a vector space. Without it, you cannot define a reciprocal basis, whereas dual bases are always defined in any vector space $V$.

More generally, rather than a "dot product", we can consider a bilinear form $B : V\times V \to K$ where $K$ is the field of scalars. Convince yourself that there is a 1-to-1 correspondence between bilinear forms and linear transformations $\xi : V \to V^*$; call such a transformation a correlation.

We are usually interested in symmetric bilinear forms, where then in finite dimensional spaces the induced correlation is an isomorphism if and only if the form is non-degenerate, i.e. $B(v,w) = 0$ for all $v$ implies $w$ is $0$. If $\mathscr B = \{e_i\}_{i=1^n} \subseteq V$ is a basis with dual $\{e^i\}_{i=1}^n \subseteq V^*$, you will find that $\{\xi^{-1}(e^i)\}_{i=1}^n$ is exactly the reciprocal of $\mathscr B$ with respect to $B$.

This has a geometric interpretation. Just as a vector $v \in V$ represents a line $\span\{v\}$, a covector in $\alpha \in V^*$ represents a hyperplane $\ker(\alpha)$, i.e. a subspace of dimension $n-1$. Now choose some $e_i$; the span of all the other basis vectors is a hyperplane, and that hyperplane is none other than $\ker(e^i)$! The condition $e^i(e_i) = 1$ then gives us a normalization so that $e^i$ is fully defined. Now, a symmetric $B$ gives us a notion of orthogonality, so we take the line orthogonal to the hyperplane $e^i$ represents, and this line is exactly the span of the reciprocal of $e_i$.

To summarize: a dual basis is the basis of hyperplanes you can form from the vector basis of lines. The reciprocal basis is then the basis of normal vectors of those hyperplanes.

  • $\begingroup$ Thanks. Now understand that the reciprocal basis arises from the fact that the inner product is a simple dot product instead of a more general form. I saw the hyperplane concept on Wikipedia and elsewhere but still trying to get my head around it. $\endgroup$
    – user175324
    Jan 1 at 19:56

You must grasp that a covector in a vector space $V$ is a linear functional $$V\to\mathbb R,$$ and the set of all of them is a vector space symbolically denoted by $V^*=\{V\to\mathbb R: \mbox{which are linear}\}$

And in the case $V=\mathbb R^n$ the reciprocal basis $\mathbf e^i$ allows to map $$v\mapsto\mathbf e^i\cdot v.$$

Call these maps $\varepsilon^i$. They comply \begin{eqnarray*} \varepsilon^i(v)&=&\varepsilon^i(v^s\mathbf e_s),\\ &=&v^s\mathbf e^i\cdot \mathbf e_s,\\ &=&v^s\delta^i_s,\\ &=&v^i. \end{eqnarray*}

These are used to give a basis to the dual $V^*$:

If $f\in V^*$ and by getting $f_k=f(\mathbf e_k)$ then one can see $$f=f_s\varepsilon^s$$ satisfy $f(v)=f_sv^s$. Here you can see how any $f$ is a linear combination of the $\varepsilon^i$ for which is easy to prove that they are linearly independent.

Since both $V$ and $V^*$ have the same dimension they will be isomorphic.

  • $\begingroup$ had you ever seen that $$v^s\mathbf e_s=v^1\mathbf e_1+\cdots+v^n\mathbf e_n$$? This is the Einstein's sum convention in action $\endgroup$
    – janmarqz
    Dec 30, 2022 at 1:22

You must log in to answer this question.

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