# How to choose an inner product with respect to a basis in such a way that this basis is orthogonal

Is there a general process for choosing an inner product with respect to a basis (finite or infinite) for a vector space in such a way that this basis is orthogonal?

Explanations. In most linear algebra textbooks, given a non-orthogonal basis for a vector space over a field provided with an inner product, it is usual to turn this non-orthogonal basis into an orthogonal one using the Gram-Schmidt process.

I am not intereseted in changing the basis (therefore Gram-Schmit is not what I am looking for), but rather kind of the opposite. Since orthogonality depends on the inner product, what I am looking for is a process to choose a different inner product, in such a way that the basis is orthogonal.

For example, $B = \{x, \cos{2\pi x} \}$ is a basis for $W$, a subspace of real continuous functions space. We can define the inner product $<f(x),g(x)> = \int_{-1}^1 f(x)g(x)dx$ which has the expected inner product properties and $B$ is orthogonal with respect to this inner product.

The motivation is to study general functions approximations by projection on continuous functions spaces. I am particularly curious about cases other than projection on trigonometric functions series (Fourier series) or polynomials (Taylor series) which I am already familiar with.

• Why not just define the inner product by saying that the basis vectors are orthonormal? Apr 13, 2015 at 21:05

If your space is finite dimensional, then it is basically $\Bbb R^n$ (or $\Bbb C^n$, if it's over the complex numbers).
Suppose that $(v_1,\dots,v_n) \subset \Bbb R^n$ is your "orthonormal" basis of choice. We can define a linear transformation $M$ by $$M:v_i \mapsto e_i$$ where $\{e_i\}$ is the standard basis. We can then define the inner product by $$\langle x, y\rangle_M = \langle Mx, My\rangle = y^T(M^TM)x$$ We have to be a little bit more careful with infinite inner product spaces.
In particular, suppose that our space is a Hilbert space with a countable Schauder basis (so, $V = \ell^2$ up to isomorphism).
Suppose that we have a bounded sequence $(v_i) \subset V$ such that $\inf_i\|v_i\| > 0$. Then we can define a bounded linear map $$M:v_i \mapsto e_i$$ So, we can define the inner product $$\langle x, y\rangle_M = \langle Mx, My\rangle$$ and, because $M$ is an isomorphism of vector spaces, the metric induced by this $M$ inner product is equivalent to the usual metric on $V$.