Question

I know that the determinant of a matrix is the volume of a parallelepiped (three dimensions or more) formed by columns of the matrix.

Is there a way to prove it by using Gram-Schmidt procedure?

Answer 1

Yes there is a way to prove it by gram schmidt procedure.

For any vectors v,w1,...,wm in Rk, we may find vectors B and C so that v = B+C, where B is perpendicular to all wi, i = 1,...,m and C is in the span of wi, i = 1,...,m.
Proof. Apply the Gram-Schmidt process to w1, . . . , wm to retrieve vectors a1, . . . , am′ that are
orthonormal and span the same space as wi, i = 1,...,m. Let B = v− (Summation) (i=1 to m’) v·aiai. Then B is perpendicular to all the ai and hence to the wi, and similarly v − B is in the span of the wi.
We can now define the volume of P by induction on k. The volume is the product of a certain “base” and “altitude” of P . The base of P is the area of the (k − 1)-dimensional parallelepiped with edges x2,...,xk. The Lemma gives x1 = B + C so that B is orthogonal to all of the xi, i≥2 and C is in the span of the xi, i≥2. The altitude is the length of B.