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).