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 11.8.1Let u_{1},
⋅⋅⋅
,u_{p}be vectors in ℝ^{M},M ≥ p. The parallelepipeddetermined bythese vectors will be denoted by P
(u1,⋅⋅⋅,up)
and it is defined as
( p )
{∑ }
P (u1,⋅⋅⋅,up) ≡ (j=1sjuj : sj ∈ [0,1]) .
The volume of this parallelepiped is defined as
1∕2
volume of P (u1,⋅⋅⋅,up) ≡ v (P (u1,⋅⋅⋅,up)) ≡ (det(G)) .
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.
A parallelepiped is a sort of a squashed box. Here is a picture which shows
PICT
the relationship between P
(u1,⋅⋅⋅,up−1)
and P
(u1,⋅⋅⋅,up)
. In a sense, we can define the volume
any way we want but if it is to be reasonable, the following relationship must hold. 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 )) (11.10)
p 1 p−1
(11.10)
In the case where p = 1, the parallelepiped P
(v)
consists of the single vector and the one
dimensional volume should be
|v|
=
(vT v)
^{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 11.8.2Let
{u1,⋅⋅⋅,up}
be vectors in ℝ^{M}. Then letting U =
( )
u1 u2 ⋅⋅⋅ up
,
( T )1∕2 1∕2
v(P (u1,⋅⋅⋅,up )) ≡ det U U = det(G ) , 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 11.10 holds.
Theorem 11.8.3With the above definition of volume, 11.10holds.
Proof: To check whether this is so, it is necessary to find
|cos(θ)|
. This involves finding a
vector perpendicular to P
(u1,⋅⋅⋅,up−1)
. Let
{w1, ⋅⋅⋅,wp}
be an orthonormal basis for
span
(u1,⋅⋅⋅,up)
such that span
(w1,⋅⋅⋅,wk)
= span
(u1,⋅⋅⋅,uk )
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 (u ,⋅⋅⋅,u ))
1 p
. Thus
|N ⋅u |
p
= v
(P (u ,⋅⋅⋅,u ))
1 p
. 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.