PYTHEGOREAN THEOREM AND DETERMINANTS The Pythagorean theorem is an application of the determinant properties in the plane. With the formalism of complex numbers, these can be stated as follows: det: C x C --> R det(1,i) = 1 (normalization) det(k·a,b)=det(a,k·b)=k·det(a,b) (multiple areas ... and negative ones) det(a,b)=det(a,b+ka))det(a+kb,b) (parallel gliding) Using these properties one brings the unit square to a coordinate rectangle and then (leaving the area unchanged) to a parallelogram. In this applet (move the point c horizontally to the right): h = ax - ay·bx/by so that the product h·by, which is the signed area of the parallelogram defined by a and b, is det(a,b) = ax·by - ay·bx. If b is the orthonormal n of a , i.e. b = n = (-ay , ax), we get : det(a,n) = ax² + ay² As an easy consequence of last equality, the versor vers(a) of a non-zero vector a is the positive multiple of the vector a whose sqare is 1. This means: det( vers(a) , ort( vers(a) ) ) = 1 [ where ort(v) is the orthonormal of v, i.e. ( -vy , vx ) ]. So vers(a) = k·a (with k>0) and det( k·a , ort(k·a) ) = 1, hence det(k·a , k·ort(a)) = 1 and then: k²·det(a , ort(a)) = 1, so that k = 1 / v(ax² + ay²) . This method can easily be estended, in a very direct way, to the n-dimensional real space, through n-dimensional determinants. First of all, some background considerations about determinants det: (Rn)n --> R : The 3) basic properies of determinants: a) det(coordinate_basis) = 1 b) det(... , k·u, ...) = k·det(... , u , ...) c) det(..., u ,..., v ,...) = det(..., u ,..., v + k u ,...) = det(..., u + k v ,..., v ,...) easily yield the following three more properties: a') det( ... , u , ... , u , ...) = 0 b') det( ... , u , ... , v , ...) = - det( ... , v , ... , u , ...) c') det( ... , u + u' , ...) = det( ... , u , ...) + det( ... , u' , ...) These properties univocally define this function det: (Rn)n --> R. Now, consider the following geometrically intuitive hypotheses: i) for any positive integer n, each coordinate vector in Rn made of only one component equal to 1 and all other components equal to 0 is a unitary (i.e of length 1) vector in Rn ii) for any positive integer n, each coordinate vector in Rn along the axis x_i is orthogonal to any of the vectors in the subspace x_i=0 iii) for any positive integer n, a unitary vector in Rn yields a unitary vector of Rn+1 by adding a 0 component in some position as (n+1)th component. By this adding of 0, also orthogonal vectors of Rn yield orthogonal vectors of Rn+1 iv) if u and v are unitary mutually orthogonal vectors, then the vectors h·u+k·v and (-k)·v+h·u (both vectors of the (u,v)-subspace) are mutually orthogonal, and if h·u+k·v then also (-k)·v+h·u is unitary v) if a vector is orthogonal to some vectors, it is also orthogonal to a linear combination of these vectors vi) for any positive integer n, any n-tuple of mutually orthogonal unitary vectors in Rn gives a determinant having absolute value equal to 1. Theorem: If i) ... vi) are true, then, for any positive integer n, every unitary vector u = (u1, ... ,un) in Rn has the two following properties: 1) u can be "completed" with n-1 more unitary mutually orthogonal vectors v1, ... , vn-1 , all orthogonal to the vector u, to build a n-dimensional determinant det( u , v1, ... , vn-1 ) whose absolute value is 1 2) Σi=1,...,n ui ² = 1 Proof: By induction suppose that in Rn every unitary vector u=(u1, ... ,un) has properties 1) and 2). (this is true for n=2, where v = (- u2 , u1) or its opposite vector) Consider a unitary vector a=(a1,...,an+1) in Rn+1 and suppose, that it is not along the fist axis x_1 (i.e. a2,...,an+1 are not all equal to 0; otherwise the vector a would be the first coordinate vector and properties 1) and 2) would trivially follow). So, the "unheaded" vector a* = (a2,...,an+1) is a non-zero vector of Rn. Put r = √ ( Σi=2,...,n+1 ai ² ) , s = a1 and u = a*/r so that u is a unitary vector of Rn and the inductive hypotheses are verified. Let's call v1, ... , vn-1 the n-1 completing unitary vectors of Rn given by property 1) and let's call v1*, ... ,vn-1* the same vectors with a zero prefixed as first coordinate ( so they are all vectors of Rn+1 ). The (n+1)-dimensional vectors (0,a1,...,an) (let's denote it by the same symbol a*), the vector u* = (0,u1, ... ,un) and the completing vectors v1*, ... ,vn-1* are all in the coordinate (n+1)-dimensional subspace x1=0 and the vector a is a linear combination of u* and the (n+1)-dimensional vector w = ( 1 , 0 , ... , 0 ) having all null components but the first one which is equal to 1; actually, it is: a = a1·w + a** = s·w + r·u* , u* and w being unitary mutually orthogonal vectors, so that the vector a' = -r·w + s·u* is orthogonal to the vector a. The function (w1 , ... , wn) --> det(w , w1* , ... , wn*), where each wi* is the vector of Rn+1 obtained from the n-dimensional vector wi by prefixing 0 as a first component, has properties a), b), c) of the n-dimensional determinant, then it coincides with the n-dimensional determinant , so it is: det( w , u* , v1*, ... ,vn-1* ) = det( u , v1, ... ,vn-1 ) = ± 1 = d. Now we consider the vectors d·a' = ± a' and v1*, ... ,vn-1* as the n "completing" mutually orthogonal unitary vectors of the unitary vector a (all of them are orthogonal to the vector a). We also have: det( a , d·a' ,v1*, ... ,vn-1* ) = d·det( s·w+r·u*, -r·w+s·u* , v1*, ... ,vn-1* ) = d·det( s·w , -r·w+s·u* , v1*, ... ,vn-1* ) + d·det(r·u*, -r·w+s·u* , v1*, ... ,vn-1* ) = s²·d·det( w , u*, v1*, ... ,vn-1* ) + r²·d·det( u* , - w , v1*, ... ,vn-1* ) = d²·(s² + r²) = s² + r² . Since |det( a , d·a' ,v1*, ... ,vn-1* )| = 1 , because of (vi), we deduce that s² + r² = |s² + r² | = 1, so that 2) is verified for the vector a. Then det( a , d·a' ,v1*, ... ,vn-1* ) = 1, and so also property 1) is verified. The best way to understand this process is to develop the step from n=2 to n+1=3. The induction could also be started, in a more subtle and elegant way, at n=1 instead of n=2 (for n=1 properties 1) and 2) are trivially satisfied, because the determinant of a number is the number itself, the "unitary vectors" are the numbers 1 and -1, and the "completing" (n-1)-tuple of mutually orthogonal unitary vectors is void).