0

Suppose you have a vector space $V$ (without an inner product) with a basis $B = \{ \vec{v}_i \}$. Does there always exist a choice of inner product $I$ with which you can endow $V$ that will render the basis orthonormal, i.e. $I(\vec{v}_i, \vec{v}_j) = \delta_{ij}$ for all $\vec{v}_i, \vec{v}_j \in B$?

I think that this is the answer, but please let me know if I'm getting anything wrong:

  1. If $V$ is finite-dimensional, then the answer is yes. Here's an explicit construction of the inner product $I$. For any two vectors $\vec{a}, \vec{b} \in V$, decompose them into linear combinations of basis vectors $\vec{a} = \sum_{i=1}^n \alpha_i \vec{v}_i$ and $\vec{b} = \sum_{i=1}^n \beta_i \vec{v}_i$. This decomposition is unique, even if there isn't an inner product. (But unlike in the case of an orthonormal basis for an inner product space, each coefficient $\alpha_i$ won't just depend on the vectors $\vec{a}$ and $\vec{v}_i$, but on $\vec{a}$ and the entire basis $B$). Then define $$I(\vec{a}, \vec{b}) := \sum_{i=1}^n \alpha_i^* \beta_i.$$ I haven't worked it out, but I think that it's fairly straightforward to show that this form satisfies all the requirements for an inner product, and clearly $B$ is orthonormal with respect to it. (Is there a quicker construction for $I$?)

  2. If $V$ is infinite-dimensional, then the answer is no. There are several different notions of "basis" that we need to distinguish, primarily a Hamel basis and an orthonormal basis for a Hilbert space. Without any preexisting inner product structure, there's no meaningful sense of an orthonormal basis, or even a "pre-orthonormal basis" that just needs a simple additional inner product. So we must be talking about a Hamel basis. In this case, (I believe that) we can't always construct an inner product $I$ w.r.t. which the Hamel basis $B$ is orthonormal, because the series defined above doesn't necessarily converge, and no other construction works.

Does this sound right?

tparker
  • 6,756
  • 4
    Even if the space is infinite-dimensional, part of the definition of basis is that each element can be written as a linear combination of a finite set of basis elements. Then you can just define the inner product of two basis elements to be zero if the elements are different, one if they are the same, and extend by linearity to an inner product on the space. – Gerry Myerson Jul 06 '21 at 12:39
  • @GerryMyerson So you're saying that my formula above still works for the infinite-dimensional case, as long as the sum is interpreted as not necessarily being countably indexed, but simply representing a sum over all the basis vectors? Because there will always be only a finite number of nonzero terms in the sum, which will be less than or equal to the max of the (finite) numbers of nonzero terms in the decompositions of $\vec{a}$ and $\vec{b}$ as linear combinations of basis vectors? – tparker Jul 06 '21 at 13:11
  • I'm not sure what sum you have in mind when you write, " a finite number of nonzero terms in the sum," but yes I think the formula works. – Gerry Myerson Jul 06 '21 at 13:17
  • 1
    See also https://math.stackexchange.com/questions/976740/does-every-vector-space-have-an-inner-product or https://math.stackexchange.com/questions/247425/is-there-a-vector-space-that-cannot-be-an-inner-product-space – Gerry Myerson Jul 06 '21 at 13:21
  • @GerryMyerson The sum that I have in mind is a sum over all of the basis vectors in the Hamel basis (with the understanding that the coefficient of any basis vector that doesn't appear in a given linear combination equals 0). Right? – tparker Jul 06 '21 at 13:47
  • For a Hamel basis your sum is always finite because both $\ \vec{a}\ $ and $\ \vec{b}\ $ can be expressed (uniquely) as a finite linear combination of elements of the basis. – lonza leggiera Jul 06 '21 at 15:10
  • "sum over all of the basis vectors in the Hamel basis...." Sum of what over all those basis vectors? – Gerry Myerson Jul 06 '21 at 23:36
  • @GerryMyerson Well, to be completely explicit: let the scalar $f(\vec{a}, B, \vec{v} \in B)$ be defined to be the coefficient of the basis vector $\vec{v} \in B$ when the vector $\vec{a}$ is (uniquely) expressed as a linear combination of basis vectors in the $B$ basis. Then let $\alpha_i$ be shorthand for $f(\vec{a}, B, \vec{v}_i)$ and $\beta_i$ be shorthand for $f(\vec{b}, B, \vec{v}_i)$ (with the basis $B$ implicit). Then we have $$ I(\vec{a}, \vec{b}) = \sum_i \alpha_i^* \beta_i = \sum_i f(\vec{a}, B, \vec{v}_i)^* f(\vec{b}, B, \vec{v}_i),$$ where the sum on $i$ is understood to not ... – tparker Jul 07 '21 at 03:00
  • necessarily imply that $B$ is countably indexed. Right? – tparker Jul 07 '21 at 03:02
  • The way I'd say it is, if ${v_i}_{i\in I}$ is a basis, and $a=\sum_ia_iv_i$ and $b=\sum_ib_iv_i$ where all but finitely many of the $a_i$ and $b_i$ are zero, then $\sum_ia_ib_i$ defines an inner product on the space. – Gerry Myerson Jul 07 '21 at 04:34

1 Answers1

1

Based on Gerry Myerson's comments to my question, I now believe that the answer is actually "yes" in both the finite- and infinite-dimensional cases. The construction that I described for the finite-dimensional case actually carries over for the infinite-dimensional case as well. (Strictly speaking, the "$\sum_i$" now becomes a sum over the not-necessarily-countable set of Hamel basis vectors, but since only a finite number of summands are nonzero, there aren't any convergence issues.)

Gerry Myerson
  • 185,413
tparker
  • 6,756