Why is this a valid definition of the dot product?

Question

$(\vec{u},\vec{v})=u_1v_1+2u_2v_2+3u_3v_3$

I have never seen this definition before. I am used to the dot product looking something like this:

$(\vec{a},\vec{b})=a_1b_1+a_2b_2+a_3b_3$

Where do the coefficients 2 and 3 come from?

Thanks in advance

Answer 1

If $A$ is an $n\times n$ symmetric and positive defined matrix, then you can define a scalar product on $\mathbb{R}^n$ as follows: for $u, v\in\mathbb{R}^n$ $$(u,v) = u^TAv.$$ $1.$ since $A$ is positive defined, then $$(u,u) = u^TAu \ge 0$$ $2.$ from symmetricity $$(u,v) = u^TAv = (u^TAv)^T=v^TA^Tu = v^TAu = (v,u)$$ $3.$ linearity is straightforward: $$(\alpha u+\beta v,w) = \alpha(u,w) + \beta(v,w).$$

In your case $$A= \begin{pmatrix} 1&0&0\\ 0&2&0\\ 0&0&3 \end{pmatrix} $$

Answer 2

It looks like a definition of a dot product in a kernel space: $\langle\vec u, \vec v\rangle_K$ with $K=\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{pmatrix}$.

The matrix $K$ is positive semi-definite, so you can express the dot product like you would normally: $\langle\vec a,\vec b\rangle=a_1b_1+a_2b_2+a_3b_3$ with $\vec a=\vec u K^{1/2}=a_1\sqrt{K_{11}}+a_2\sqrt{K_{22}}+a_3\sqrt{K_{33}}$ and $\vec b=\vec vK^{1/2}$.

Answer 3

Consider the linear transformation given by the matrix $$ T= \begin{pmatrix} 1&0&0\\ 0&\sqrt 2&0\\ 0&0&\sqrt 3 \end{pmatrix} $$ Apply this to both $\vec u$ and $\vec v$ and see what happens to the dot product ...

Answer 4

I would say it is a valid definition of "a" dot product, rather than "the" dot product.

More precisely your formula defines a valid bilinear form, as per this definition: http://en.wikipedia.org/wiki/Bilinear_form.

PS: if you "re-parameterize" your space with $(x,y,z) \mapsto (x,\frac{1}{\sqrt{2}}y,\frac{1}{\sqrt{3}}z)$, you see that both definitions are pretty much equivalent.