Does every finite dimensional real vector space have a canonical basis?

Question

I was trying to prove the following theorem:

Theorem: Let $\phi$ be symmetric bilinear form in $V$, where $V$ a real inner product vector space with dimension $n$. Then there exists a orthonormal basis $B$ of $V$, such that the matrix associated to $\phi$ in that basis is diagonal.

This is the way I "proved" it:

Let $C$ be the canonical basis of $V$, $C'$ the canonical basis of $\mathbb{R}^n$, $A=M_C(\phi)$ the matrix associated to $\phi$ in that basis and $T$ an operator in $\mathbb{R}^n$defined as $T(v)=Av$. We have that $[T]_{C'}=A$ is a symmetric real matrix, so $T$ is selfadjoint and hence there exists some orthonormal basis $B$ of $\mathbb{R}^n$ for which the matrix associated to $T$ is a diagonal matrix $D$. So $$D=Q^{-1}AQ,$$ where $Q$ is the change of basis matrix from $B$ to $C'$. Since $C'$ is the canonical basis, the columns of $Q$ are just the vectors of $B$, so $Q$ is an orthogonal matrix and hence $D$ is not just similar but also congruent to $A$.

At this point I would like to conclude "since $A$ is congruent to $D$ then $D$ is the matrix associated to $\phi$ in an orthogonal basis". The problem is that $B$ is not a basis of $V$ but a basis of $\mathbb{R}^n$, so how can I define an orthonormal basis of $V$ in terms of $B$? I think I'm looking for some kind of isomorphism between $V$ and $\mathbb{R}^n$ preserving orthonormality (an isometry?), is there a sandard one? Also does it make sense to talk about a canonical basis in an arbitrary real inner product space with finite dimension? I'm quite confussed, maybe my "proof" doesn't even make sense.

Answer 1

Does every finite dimensional real vector space have a canonical basis?

For any reasonable definition of "canonical" (and certainly for the usual definition from category theory), "(emphatically) no". There's not even a way to pick a single non-zero element in a way that's "equivariant with respect to vector space isomorphisms".

Instead, you'll need to start with existence of some basis of your finite-dimensional real inner-product space and use this basis to construct an orthonormal basis, such as by applying the Gram-Schmidt algorithm.

Answer 2

I wouldn't call the standard basis of $\mathbb{R}^n$ a canonical basis - I call it standard basis. For me, a canonical object is something which is uniquely determined by a property. I can't see that property for the standardbasis of $\mathbb{R}^n$. Now let's get to an arbitrary finite dimensional (dimension $n$) vector space $V$. As you said correctly, there is an isomorphism to $\mathbb{R}^n$. Now the problem is: This isomorphism is not uniquely determined. In fact, every automorphism of $\mathbb{R}^n$ induces another isomorphism between $V$ and $\mathbb{R}^n$.