Proof equivalence of algebraic and geometric definition of dot product

1. Proof equivalence of algebraic and geometric definition of dot product

This is a quick proof on the equivalent definitions for the dot product. Just for me to test on how integrate org roam with latex tikz and sage plugins, I generate this HTML page with the Emacs function org-html-export-as-html(with minor edits),

1.1. Observation

Algebraic and geometric definitions of dot product are equivalent this is: x\cdot y = (x_1,x_2) \cdot (y_1,y_2) = x_1y_1 + x_2y_2 = \Vert x\Vert \Vert y\Vert \cos{\theta} where \theta is the angle between x and y

1.1.1. Proof

We will proove it for the 2 dimention case, it can be easily extended to n dimention by embedding the 2 vectors in a plane… Lets consider x and y as in the definition, notice that the dot product is invariant over chage of orthonormal bases, this because given a 2×2 matrix M we have that Mx \cdot My = (m_{11}x_{1}+m_{12}x_{2}, m_{21}x_{1} + m_{22}x_{2}) \cdot (m_{11}y_{1}+m_{12}y_{2}, m_{21}y_{1}+m_{22}y_{2}) = m_{11}^{2} x_{1} y_{1} + m_{21}^{2} x_{1} y_{1} + m_{11} m_{12} x_{2} y_{1} + m_{21} m_{22} x_{2} y_{1} + m_{11} m_{12} x_{1} y_{2} + m_{21} m_{22} x_{1} y_{2} + m_{12}^{2} x_{2} y_{2} + m_{22}^{2} x_{2} y_{2} Notice that it can be refactored as x_1y_1(m_{11}^2 + m_{21}^{2}) + (x_1y_2 + x_2y_{1})(m_{11}m_{12}+m_{21}m_{22}) + x_2y_2(m_{22}^2 + m_{12}^{2}) and as M is an orthogonal matrix then m_{11}m_{12} + m_{21}m_{22} = 0 as that matrix has as columns orthogonal vectors and this expression is its dot product also notice the m_{11}^2 + m_{21}^2 = \Vert m_{1}\Vert ^{2} and m_{22}^2 + m_{12}^2 = \Vert m_{2}\Vert^{2} where [m_{1} | m_{2}] := m this is are the orthnormal vectors of the base, and as they are unitary then both previous expresions are equal to one so finally Mx \cdot My = x_1y_1 + x_2y_2 = x \cdot y

So to proof x_1y_1 + x_2y_2 = \Vert x\Vert \Vert y\Vert \cos{\theta} we will change the base of x and y from the standard base i, j to a new base u := \hat x and v := \hat x ^{\perp} which are the unitary vector in the x direction and the perpendicular vector (so that we have a right coordinate) to the unitary x vector respectively

Following the diagram of the two bases described in the las parragraph 

\begin{tikzpicture}
  \draw[<->] (-1,0)--(1,0) node[right]{$i$};
  \draw[<->] (0,-1)--(0,1) node[above]{$j$};
  \draw[<->,red] (45:-1)--(45:1) node[above]{$u := \hat x$};
  \draw[<->,red] (135:-1)--(135:1) node[above]{$v := \hat x ^\perp$};
\end{tikzpicture}

To simplifly calculaitions notice that v = (-x_{2}, x_{1}) that way we can express the matrix of base change from (u,v) to (i,y) is the matrix m in then code below. Also recall that \left| x \right| \left| y \right| \cos \theta can be seen geometrically as \left| x \right| \text{Proj}_{x}y this is the norm of x times the projection of y over x Now if we change to the base (u,v) conveniently the projection of y over x will be just be the component in the \hat x direction (first coordinate). 

var('x1 x2 y1 y2');
x = vector([x1,x2]);
y = vector([y1,y2]);
xhat = x/norm(x)
m = matrix([ [xhat[0],-xhat[1]],[xhat[1],xhat[0]] ])
print(f'm = ', end='');
mylatex(factor(m))
print(f'X := m^-1*x = ', end="");
X = (m.inverse()*x).apply_map(factor)
mylatex( X )
print(f'Y := m^-1*y = ', end="");
Y = (m.inverse()*y).apply_map(factor)
mylatex( Y )

m = \left(\begin{array}{rr} \frac{x_{1}}{\sqrt{x_{1}^{2} + x_{2}^{2}}} & -\frac{x_{2}}{\sqrt{x_{1}^{2} + x_{2}^{2}}} \\ \frac{x_{2}}{\sqrt{x_{1}^{2} + x_{2}^{2}}} & \frac{x_{1}}{\sqrt{x_{1}^{2} + x_{2}^{2}}} \end{array}\right)
X := m-1*x = \left(\sqrt{x_{1}^{2} + x_{2}^{2}},\,0\right)
Y := m-1*y = \left(\frac{x_{1} y_{1} + x_{2} y_{2}}{\sqrt{x_{1}^{2} + x_{2}^{2}}},\,-\frac{x_{2} y_{1} - x_{1} y_{2}}{\sqrt{x_{1}^{2} + x_{2}^{2}}}\right)

And according to our last discussion then \text{Proj}_{X}Y = \frac{x_{1} y_{1} + x_{2} y_{2}}{\sqrt{x_{1}^{2} + x_{2}^{2}}} and finally we have that x \cdot y = X \cdot Y = \left| X \right| \text{Proj}_XY = \sqrt{x_1^2 + x_2^2}\ \frac{x_{1} y_{1} + x_{2} y_{2}}{\sqrt{x_{1}^{2} + x_{2}^{2}}} = x_{1} y_{1} + x_{2} y_{2} As stated

Author: Diego Lopez Rodriguez

Created: 2024-10-11 Fri 16:46