The determinant is the essential algebraic tool which provides a way to give a unified treatment of the
concept of p dimensional volume of a parallelepiped in ℝ^{M}. Here is the definition of what is meant by such
a thing.
Definition 13.5.1Let u_{1},
⋅⋅⋅
,u_{p}be vectors in ℝ^{M},M ≥ p. The parallelepipeddetermined by these vectorswill be denoted by P
(u1,⋅⋅⋅,up )
and it is defined as
({ p )}
P (u ,⋅⋅⋅,u ) ≡ ∑ su : s ∈ [0,1] .
1 p ( j=1 j j j )
The volume of this parallelepiped is defined as
volume of P (u ,⋅⋅⋅,u ) ≡ v(P (u ,⋅⋅⋅,u )) ≡ (det(G ))1∕2.
1 p 1 p
where G_{ij} = u_{i}⋅ u_{j}.
If the vectors are dependent, this definition will give the volume to be 0.
First lets observe the last assertion is true. Say u_{i} = ∑_{j≠i}α_{j}u_{j}. Then the i^{th} row is a linear
combination of the other rows and so from the properties of the determinant, the determinant of this
matrix is indeed zero as it should be. Indeed,
∑ ∑
Gik = αjuj ⋅uk = αjGjk
j⁄=i j⁄=i
showing that the k^{th} entry of the i^{th} row in G is a fixed linear combination of the k^{th} entries of the other
rows.
A parallelepiped is a sort of a squashed box. Here is a picture which shows P
(u1,⋅⋅⋅,up−1)
and
P
(u1,⋅⋅⋅,up)
.
PICT
In a sense, we can define the volume any way we want but if it is to be reasonable, the following
relationship must hold for N ⋅ u_{i} = 0 each i ≤ p − 1. The appropriate definition of the volume of
P
(u1,⋅⋅⋅,up)
in terms of P
(u1,⋅⋅⋅,up−1)
is v
(P (u1,⋅⋅⋅,up))
=
|u ||cos (θ)|v(P (u ,⋅⋅⋅,u )) (13.4)
p 1 p−1
(13.4)
In the case where p = 1, the parallelepiped P
(v)
consists of the single vector and the one dimensional
volume should be
|v|
=
(vTv)
^{1∕2} =
(v ⋅v)
^{1∕2}. Now having made this definition, I will show that this is
the appropriate definition of p dimensional volume for every p.
Definition 13.5.2Let
{u1,⋅⋅⋅,up}
be vectors in ℝ^{M}. Then letting U =
( )
u1 u2 ⋅⋅⋅ up
,
( )
v(P (u1,⋅⋅⋅,up)) ≡ det UT U 1∕2 = det(G)1∕2, Gij = ui ⋅uj
As just pointed out, this is the only reasonable definition of volume in the case of one vector. The next
theorem shows that it is the only reasonable definition of volume of a parallelepiped in the case of p vectors
because 13.4 holds.
Theorem 13.5.3With the above definition of volume, 13.4holds.
Proof: To check whether this is so, it is necessary to find
|cos (θ)|
. This involves finding a vector
perpendicular to P
(u ,⋅⋅⋅,u )
1 p− 1
. Let
{w ,⋅⋅⋅,w }
1 p
be an orthonormal basis for span
(u ,⋅⋅⋅,u )
1 p
such
that span
(w ,⋅⋅⋅,w )
1 k
= span
(u ,⋅⋅⋅,u )
1 k
for each k ≤ p. Such an orthonormal basis exists because of
the Gram Schmidt procedure. First note that since
which results from formally expanding along the top row. Note that from what was just discussed,
v (P (u1,⋅⋅⋅,up−1)) = ±A1p
where A_{1k} is the 1k^{th} cofactor of the above matrix, equal to N ⋅ w_{k}. Now it follows from the formula for
expansion of a determinant along the top row that for each j ≤ p,
because the matrix has two equal rows. Therefore, N points in the direction of the normal vector in the
above picture or else it points in the opposite direction to this vector. If j = p,N ⋅ u_{p} equals
±v
(P (u1,⋅⋅⋅,up))
. Thus
|N ⋅up|
= v
(P (u1,⋅⋅⋅,up))
. From the geometric description of the dot
product,
v(P (u1,⋅⋅⋅,up)) = |N ||up||cos(θ)|
Now N is perpendicular to each w_{k} and u_{k} for k ≤ p − 1 and so
The theorem shows that the only reasonable definition of p dimensional volume of a parallelepiped is
the one given in the above definition. Recall that these vectors are in ℝ^{M}. What is the role of ℝ^{M}? It is
just to provide an inner product. That is its only function.