Confusions with quadratic forms and basis of vector space

Question

Suppose $V$ is a Euclidean finite-dimensional space. Let $Q(x)=\sum \limits_{i,j=1}^na_{ij}x_ix_j$ be a quadratic form on $V$.

We note that that we can write $Q(x)$ in this way $$Q(x)=X^TAX,$$ where $A=(a_{ij})$ is $n\times n$ matrix and $X=(x_1,\dots,x_n)^T$ is the column-vectors of coordinates of $x\in V$.

My question may sounds very stupid but let me ask it because I am very confused with this question:

Here $x_i$ are coordinates of $x\in V$, i.e. $x=\sum \limits_{i=1}^nx_ie_i$. But what is $\{e_1,\dots, e_n\}$ here? I know that this is a basis of $V$ but is it some fixed basis or arbitrary basis or orthonormal basis?

I am agree that my question may sounds quite unclear but I would be happy if anyone can explain to me this from the scratch, please!

EDIT: Here is the source of my question: yesterday I have learnt some theory about that for any symmetric bilinear function $\mathcal{B}:V\times V\to \mathbb{R}$ there is an orthonormal basis such that the matrix of $\mathcal{B}$ is diagonal.

And I want to solve the following problem: Find an orthogonal transformation which reduces the following quadratic function to canonical form and find its canonical form: $x_1^2+x_2^2+x_3^2+4x_1x_2+4x_1x_3+4x_2x_3$.

Actually I know how to solve this problem but I was wondering in what basis this quadratic form is given?

Answer 1

Even if $\{e_1,\dots,e_n\}$ is not an orthonormal basis, the formula $Q(x)$ defines a quadratic form on $V$. That said, when $A$ is described as being "the" matrix associated with a quadratic form, the basis relative to which the quadratic form is defined is assumed to be orthonormal.