Example of an inner product in a real vector space, where the standard basis is not orthogonal?

Question

I'm currently trying to figure out a function for a real coordinate space, which satisfies all the rules for an inner product, but with which the standard basis is not orthogonal anymore. Does such a function even exist? Or does someone here know an example of such a function?

Answer 1

Any bilinear operation $\langle x, y \rangle := x^\top A y$ with $A$ being an SPD (symmetric and positive definite) matrix will define an inner product.

The "standard" inner product uses $A=I$ (the identity), but if you change $A$ to be non-diagonal, this will give you an inner product where the canonical basis is not orthogonal any more.

In particular, if you want that every basis vector is non-orthogonal to every other basis vector ($\langle\mathbf{e}_i, \mathbf{e}_j\rangle \neq 0$ for all $i\neq j$, where $\mathbf{e}_i$ is the $i$-th canonical basis vector), you would require that every off-diagonal entry of $A$ is non-zero. However, this is not very restrictive, and there is much freedom in choosing an SPD-matrix $A$ with those properties.

Answer 2

You can actually construct an inner product for any basis that makes that basis orthogonal. Basically, let $V$ be some finite dimensional real vector space, and let $\{b_1,\ldots,b_n\}$ be a basis. Then you can define the inner product to be $(v_1,v_2) := \sum_{i = 1}^{n}x_iy_i$, where $v_1 = x_1b_1 + x_2b_2 + \ldots + x_nb_n$ and $v_2 = y_1b_1 + y_2b_2 + \ldots + y_nb_n$.

So, to answer your question: just do this construction for some basis that isn't orthogonal with the standard inner product, and you are done.

Answer 3

Let $A$ be an invertible matrix. Then the function $f(v,w)=\langle Av,Aw\rangle$. Now if $A$ sends the standard basis vectors to vectors that are not orthogonal, we will obtain the desired result. For instance, say in $\mathbb{R}^2$ we pick the matrix $\begin{pmatrix}1&1\\0&1\end{pmatrix}$. Then $f(e_1,e_2)=1$.