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?
Inner product and the choice of coordinate frames
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0Because the dot product is defined in a coordinate-independent fashion: $\mathbf x \cdot \mathbf y = |\mathbf x| |\mathbf y|\cos\left(\angle\mathbf x\mathbf y\right)$. – 2011-02-03
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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0Yes, indeed it is a HW problem. And A can be a Rotational Matrix – 2011-02-04
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.