Why do bases of infinite dimensional spaces need to be orthonormal?

Question

I asked this question following a discussion in my Mathematical Methods course and didn't get a satisfactory answer.

If we have an infinite dimensional Hilbert space, why do we need an orthonormal basis? For finite dimensional spaces, it's fine to just have a linearly independent set that spans the space. Why the extra conditions when we move up to infinite dimensions?

Answer 1

When we're working with Hilbert spaces, we have the extra structure given by the norm: continuous linear maps are a very important subclass of such. And on a Hilbert space our notion of isomorphism is a linear surjection $T: H_1 \to H_2$ that preserves the inner product.

The main value of a basis for a vector space is that we can determine the entire linear map from where the basis maps alone. If we restricting to continuous linear maps, a Schauder basis is all we need to determine the entire linear map; because if $v = \sum_{n=0}^\infty a_n e_n$ (where $e_n$ is a Schauder basis), then - by continuity and linearity - $T(v) = \sum_{n=0}^\infty a_n T(e_n)$.

So a Schauder basis is enough for any Banach space (since the entire above discussion works there, too!). Unfortunately, Banach spaces don't necessarily have a Schauder basis - we need the Hilbert space structure to prove they exist.

We've also got even more structure, and it would be nice if our basis took that into account. So suppose we want to check if a map $T: H_1 \to H_2$ is orthogonal; that is, if $\langle v_1, v_2\rangle = \langle T(v_1), T(v_2)\rangle$. Checking this is easier again if we have an orthonormal basis: you only need to check if $\langle T(e_i), T(e_j)\rangle = 0$ for all $i \neq j$, and $\langle T(e_i), T(e_i)\rangle = 1$ for all $i$. So - since we're often interested in orthogonal operators, say - it's really nice to have an orthonormal basis, since it reduces our workload.

Here's a nice corollary of the existence of orthonormal bases of a Hilbert space.

Any two Hilbert spaces with orthonormal bases of the same cardinality are isomorphic.

Proof: I'll assume the bases are countable for clarity; the same argument works with any cardinality basis. Let $H_1$ have orthonormal basis $e_1, \dots$ and let $H_2$ have orthonormal basis $e'_1, \dots$. Define a map $H_1 \to H_2$ by $$T\left(\sum_{n=0}^\infty a_n e_n\right) = \sum_{n=0}^\infty a_n e'_n.$$ Because these are orthonormal bases, this is a continuous invertible linear map with continuous inverse; and because they're, again, orthonormal bases, $\langle v, w\rangle = \langle T(v), T(w)\rangle$. So this is an isomorphism of Hilbert spaces, as desired.

Orthonormal bases have other nice properties (some of which are mentioned in the comments), so we often like to pick one when we have Hilbert spaces in the first place. They make things easier to work with. But, of course, they're not the only Schauder bases. But if we're picking a basis - we might as well pick a nice one.

Answer 2

I'm not very known in this area, so I hesitate to add an answer, but I will anyway. As I understand it, you can have bases in Hilbert spaces that aren't orthonormal (Schauder Basis). However, when we're dealing with an orthonormal basis a lot of nice results follow.

Let $\mathcal{H}$ be a Hilbert space and $(\mathbf{e}_k)_{k=1}^\infty$ an orthonormal basis. Then

$1$) $\mathbf{v} = \sum\limits_{k=1}^\infty \langle \mathbf{v},\mathbf{e}_k\rangle\mathbf{e}_k\quad\forall\mathbf{v}\in\mathcal{H}$

$2$) $\langle \mathbf{v},\mathbf{w}\rangle = \sum\limits_{k=1}^\infty\langle\mathbf{v},\mathbf{e}_k\rangle \langle\mathbf{e}_k,\mathbf{w}\rangle\quad \forall\mathbf{v},\mathbf{w}\in\mathcal{H}$

$3$) $\sum\limits_{k=1}^\infty|\langle\mathbf{v},\mathbf{e}_k\rangle|^2 = \|v\|^2\quad\forall \mathbf{v}\in\mathcal{H}$

$4$) $\overline{\text{span}}[(\mathbf{e}_k)_{k=1}^\infty] = \mathcal{H}$

$5$) If $\mathbf{v}\in\mathcal{H}$ and $\langle\mathbf{v},\mathbf{e}_k\rangle = 0$ for all $k\in\mathbb{N}$, then $\mathbf{v} = \mathbf{0}$.

I'm sure there are more, but these were the nice results I knew. By the why, these hold only if $(\mathbf{e}_k)_{k=1}^\infty$ is an orthonormal basis.

Answer 3

You have to realize that an orthonormal basis $\mathcal B=(e_i)_i$ in the framework of an infinite-dimensional Hilbert space $H$ is never a basis in the sense of linear algebra because the vectors $e_i$ do not generate $H$ .
More precisely a vector $x\in H$ can be written $x=\sum \lambda_i e_i$ but there are in general an infinite number of non-zero scalars $\lambda_i$ and, more importantly, the symbol $\sum$ does not denote an algebraic sum but is a shorthand for a limiting process involving the topology of $H$ deduced from the inner product provided in the structure of a Hilbert space.
Note carefully that in a vector space without supplementary structure only finite sums make sense.

Summing up, the difficulties stem from the extremely unfortunate terminological clash of two deeply different notions sharing the same denomination "basis".

Answer 4

Let us first consider finite dimensional spaces. In fact let's just consider ordinary vectors in $\mathbb{R}^n$ with the dot product as inner product.

Take any basis $v_1,v_2,...,v_n$ of $\mathbb{R}^n$. Then any vector $v$ can be written as:

$v = \lambda_1 v_1 + \lambda_2 v_2 + ... +\lambda_n v_n$

for some unique $\lambda_1,\lambda_2,...,\lambda_n\in\mathbb{R}$.

Question: How do we find these numbers?

Ok so you can imagine this being a case of solving a set of linear equations. But do we REALLY have to do this? I mean if we use the standard basis (i.e. usual coordinate axes) then it is simple to read off the numbers, they are just the components of the vector.

In some sense the reason why this basis is easier to work with is the fact that it is an orthonormal basis with respect to the dot product.

You see in general if you have an orthogonal basis in an inner product space then you have the formula:

$\lambda_m = \frac{\langle v,v_m\rangle}{||v_m||^2}$

for each $m$. This isn't straight forward for an arbitrary basis!

Of course if we demand further that the basis is orthonormal, i.e. $||v_m|| = 1$ for each $m$ then the formula becomes even simpler, $\lambda_m = \langle v,v_m\rangle$.

So to cut a long story short orthogonal/orthonormal bases are much easier to compute with since you can compute the coefficients in the basis expansion easily.

Nothing changes with infinite dimensional spaces (well except having to be careful with infinite sums and convergence etc). For example the reason the formulae for Fourier series are what they are is exactly an application of the above formulae (since $\{1\}\cup\{\sin{mx},\cos{mx}\,|\,m\in\mathbb{N}\}$ are an orthonormal basis for a certain infinite dimensional space of functions with a certain inner product given by integrals.

Answer 5

We don't need an orthonormal basis but we do know one exists. And if found/derived, it has useful properties. (Such as a simple formula for coefficients.)