2

I am currently going through a book, "Tensor Algebra and Tensor Analysis for Engineers" by Itskov, and I am trying to understand what is meant by dual basis for a vector space. The author states that an arbitrary vector in a Euclidean space can be represented as:

$\textbf{x}=x^i\textbf{g}_i=x_i\textbf{g}^i$

In trying to understand this, I went to the Wikipedia page hoping for a concrete example, and it states that a dual basis for the Euclidean space of column vectors is the corresponding basis row vectors. But if the $\textbf{x}$ vector in the above equation is a column vector, and so the $\textbf{g}_i$ are basis column vectors, then how can it possibly be represented by a basis made of row vectors as in $\textbf{g}^i$? No combination of scalars times basis row vectors equals a column vector! If the Wikipedia article just has it wrong, then what is a good example to illustrate what the author saying with the above equation?

Link to Wiki:

https://en.m.wikipedia.org/wiki/Dual_basis#Examples

user3113647
  • 302

1 Answers1

1

The only reason why we represent elements in the dual as row vectors is so the product of elements in the dual with an element in the original space is given by matrix multiplication. If we don't care about this representation, there's no "row vector" and no "column vector." There's simply a vector.

In fact, if you look at the next page, it continues, for fixed $i,$ $$(x^j\mathbf{g}_j) \cdot \mathbf{g}^i = x^j\delta_j^i = x^i = \mathbf{x}\cdot \mathbf{g}^i \\ (x_j\mathbf{g}^j)\cdot \mathbf{g}_i = x_j\delta_i^j = x_i = \mathbf{x} \cdot \mathbf{g}_i$$ so they have the same coordinates. (Note: I've changed the order of the equalities to be more logical and to emphasize what the author is showing)

Brian Moehring
  • 21,088
  • Hmm. Thank you but if that is true then I am really confused. The author states that every basis in a Euclidean space has a unique dual basis. I thought this made sense because, for example, the unique dual of the unit column vector basis could be their transposes, i.e., row vectors. But now it seems that's not the case? If we have [1,0]' and [0,1]', say, then the unique dual is not the transpose of each? Confused. – user3113647 Aug 17 '19 at 00:09
  • @user3113647 Your last question is answered immediately before the equation you were having problems with. Specifically, an orthonormal basis is self-dual, so the dual basis of [1,0], [0,1] is itself... the dual to [1,0] is [1,0], and the dual to [0,1] is [0,1]. – Brian Moehring Aug 17 '19 at 00:21
  • OK. I may have to look in a different book for a concrete example. The row/column vector thing threw me off and now it seems very confusing. Can you recommend a book that explains duals with examples of what they are and, perhaps as important, why we care about them or need to define them? Thanks – user3113647 Aug 17 '19 at 00:32
  • @user3113647 If you're thinking about real space with respect to the standard basis (i.e. so vectors are identified with xyz-coordinates), then the dual basis of some basis $P, Q, R$ can be found element-wise. Specifically, the basis element dual to $P$ would be the unique vector $V$ which is perpendicular to the plane containing $Q,R$ and whose inner product with $P$ equals $1$. Then you do similarly for $Q$ and $R$. Here, note that we never mentioned any vector is a row vector or a column vector, because these concepts don't have any meaning in this context. – Brian Moehring Aug 17 '19 at 00:32
  • So concretely, if V•(QxR) =0 and V •P =1, then V is dual to P when P, Q, R are a basis? Am I on the right track? – user3113647 Aug 17 '19 at 00:40
  • @user3113647 I don't know what $W$ means there. Just using the definition, though, you get that $V$ is in the dual basis of ${P,Q,R}$ and is dual to $P$ means exactly that $$V \cdot Q = V \cdot R = 0 \iff V = t(Q\times R) \ V \cdot P = 1.$$ -- As for your question in the second comment, I haven't studied as an engineer to know why they would care. In infinite dimensional spaces, though, the dual is more commonly called the set of linear functionals, which importantly includes definite integrals, so it gives mathematics a good context for studying integrals, among other things. – Brian Moehring Aug 17 '19 at 00:49
  • OK. I'm not sure what W you mean, but I do appreciate your time with me very much! Can I summarize by saying the dual to a basis for some space is simply a certain other basis that is related to the first basis by such a rule (like P•V =1 above)? Is that accurate? – user3113647 Aug 17 '19 at 01:00