# Inner product and the choice of coordinate frames

How do I prove that dot product of two free vectors ( inner product in 2D and 3D vectors) does not depend on the choice of frames in which their coordinates are defined?

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1) What do you mean by a "frame" ? 2) What do you mean by a "dot" product? –  Braindead Feb 3 '11 at 11:26
frame = basis (it's a more physical terminology) and dot product is scalar product. –  Raskolnikov Feb 3 '11 at 11:47
You should state what definition of dot product you are using. –  Eric O. Korman Feb 3 '11 at 12:49
3) What do you mean by a 'free vector'? –  wildildildlife Feb 3 '11 at 14:52
probably linearly independent, and frame probably means orthonormal basis. –  Soarer Feb 3 '11 at 15:05

## 2 Answers

Since I guess it's homework, I'll just give a hint. A change of basis (or 'frame') is a linear mapping, i.e. applying an certain matrix $A$, sending each vector $v \mapsto A.v.$ What properties does $A$ have?

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Yes, indeed it is a HW problem. And A can be a Rotational Matrix –  Ender Feb 4 '11 at 0:58

The dot-product is in the first place a bilinear function of two vector variables with the additional property that ${\bf x}\bullet{\bf x}>0$ for ${\bf x}\ne{\bf 0}$. For any basis $({\bf e}_i)_{1\leq i\leq n}$ of your space the dot-product is encoded in a certain symmetric matrix $G:=(g_{ik})_{1\leq i\leq n,1\leq k\leq n}$. This matrix changes with the basis in a characteristic way. When a dot product is given then there are certain distinguished bases, namely the orthonormal ones. For these the matrix $G$ is just the identity matrix, and in the corresponding coordinates the dot-product computes as $(*)\ {\bf x}\bullet{\bf y} =\sum_{i=1}^n x_i y_i$. In particular, if one starts with the "standard space" ${\mathbb R}^n$ then the standard basis is considered orthonormal, whence one has the standard formula $(*)$ there.

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