3

The following is from the book "Tensor Methods in Statistics", which could be downloaded there

http://www.stat.uchicago.edu/~pmcc/tensorbook/

I have a question regarding section 0.3.1, titled "Duality and dual spaces", there it is said

Let $V$ be a vector space with basis $\{ e_1, \ldots, e_n \}$. In the usual expression for $v \in V$ as a linear combination of the basis vectors, we write $v = v^i e_i$. The notation exhibits a certain formal symmetry between the components $v^i$ and the basis vectors $e_i$, but the interpretation, as usually given, is quite asymmetric: $v^i$ is a 'scalar', whereas $e_i$ is a 'vector' in $V$. However $\{v^i, \ldots, v^n\}$, as linear functionals, form a basis in $V^*$.

Here comes my first question, how could the scalar $v^i$ be regarded as a linear functional, and furthermore, how could $\{ v^i, \ldots, v^n\}$ be interpreted as a basis for $V^*$ when its just a set of scalars?

He continues

Thus, the expression $v^i e_i$ could equally well be interpreted as a point in $V^*$ with components $(e_1, \ldots, e_n)$ relative to the dual basis.

Now he interprets the basis vectors $e_i$ as components in the dual basis, but I thought linear functionals, i.e. element of $V^*$ could be represented equally by $n$-tuples of numbers.

StefanH
  • 17,046
  • 6
  • 41
  • 109
  • Dear Stefan, [This discussion](http://math.stackexchange.com/a/3755/221) is somewhat related. Regards, – Matt E Aug 07 '12 at 11:31

1 Answers1

3

When the author says that $v^i$ can be thought of as a functional, they are not talking about the scalar $v^i$ attached to any particlar vector $v$. Rather, they mean the scalar valued function $v \mapsto v^i$ (sometimes called "projection onto the $i$th component").

This is a scalar-valued function on the set of vectors $\{v \, | \, v = (v^1,\ldots, v^n)\},$ i.e. it is a functional.

Although it is a little confusing to denote this functional by $v^i$, this is similar to the kind of confusion that can arise in calculus, where we might use $x$ to denote both a particular number and a variable, and hopefully won't be hard to get used to.

The resulting basis $v^i$ of $V^*$ is called the dual basis, since the value of $v^i$ on $e_j$ is $\delta_{ij}$. (This is just a fancy way of saying that the $i$th component of $e_j$ is $\delta_{ij}$.)

The expresssion $\sum_i v^i e_i $ then has a symmetry: we can interpret it as originally intended: the $e_i$ are basis vectors for $V$, and the $v^i$ are scalars giving the coordinates of a vector in terms of this basis --- or we can think of the $v^i$ as denoting the dual basis of $V^*$, and the $e_i$ as being coordinates of a vector in $V^*$ in terms of this basis.

How are the $e_i$ coordinates on a vector of $v^*$? Well, we just have to reverse engineer the above process of interpreting the scalar valued function $v^i$ as a vector, namely: given a functional $v^*$, we can evaluate it on the vector $e_i$ to get a scalar, and so we can think of this evaluation process as being a functional on $V^*$. In this way the vector $e_i \in V$ becomes thought of as a functional on $V^*$. (This identification of vectors in $V$ with functionals in $V^*$, via evaluation, is often called double duality.)

It's easy to check that in this way, $e_i$ just becomes the functional on $V^*$ which gives the $i$th coordinate of a functional $v^*$ in terms of the basis $(v^i)$. This explains the author's new interpretation of the expression $\sum_i v^i e_i$.

Matt E
  • 118,032
  • 11
  • 286
  • 447