# Changing the bilinear form on a Euclidean space with an orthonormal basis.

I'm having trouble getting my head around how Euclidean spaces, bilinear forms and dot product all link in with each other. I am told that on a Euclidean space any bilinear form is denoted by $$\tau(u,v) = u\cdot v$$ and in an orthonormal basis we have $$u \cdot v = \underline{u}^T \underline{v}$$ but what, say, if we have an orthonormal basis on a vector space together with a positive definite symmetric bilinear form (so a Euclidean space) then $$\tau(u,v) = \underline{u}^T \underline{v}$$ but what now if we keep the same vector space, the same orthonormal basis, and the same vectors $u,v$ but we change the positive definite symmetric bilinear form $\tau$ surely the computation $\underline{u}^T \underline{v}$ will be the same but the computation $\tau(u,v)$ will surely change? Can someone please explain this?

Orthonormal is defined with respect to $\tau$. That is, with no positive definite symmetric bilinear form $\tau$ around, the statement "$\{v_1,...,v_n\}$ is a an orthonormal basis" is meaningless.

Once you have such a $\tau$, then you can say $\{v_1,...,v_n\}$ is an orthonormal basis with respect to $\tau$." This means that $\tau(v_i,v_i) = 1$ and $\tau(v_i, v_j) = 0$ when $i\neq j$.

Often, once a $\tau$ has been chosen, one doesn't write "with respect to $\tau$", but technically it should always be there.

So, when you say

...what now if we keep the same vector space, the same orthonormal basis, and the same vectors u,v but we change the positive definite symmetric bilinear form τ surely the computation $\underline{u}^T\underline{v}$ will be the same but the computation $\tau(u,v)$ will surely change

(emphasis mine) you have to be careful because you can't stick with the same orthonormal basis. When you change $\tau$, this changes whether or not your orthnormal basis is still orthonormal. So before computing $\underline{u}^T\underline{v}$, you must first find a new orthonormal basis, then compute $u$ and $v$ in this basis, and then compute $\underline{u}^T\underline{v}$.

• Ah! Now I understand, thank you – user26069 Apr 19 '12 at 17:53