行列式 与 n维平行多面体体积 公式的证明 determinant volume proof
Q: why determinant is volume of parallelepiped in any dimensions？
A: http://math.stackexchange.com/a/547522/227287
Suppose v1,v2,...vn are linear independent.
In 2d,
The signed area (2-volume) of (v1,v2) is vol(v1,v2).
v1=(x,0) v2'=(0,y)
v2' = v2 - C*v1.
vol(v1,v2) = vol(v1,v2') = x*y = det(v1,v2')= det(v1,v2 - C*v1)= det(v1,v2)
In 3d,
The signed volume (3-volume) of (v1,v2,v3) is vol(v1,v2,v3).
v1=(x,0,0) v2'=(0,y,0) v3'=(0,0,z)
v2' = v2 - C21*v1.
v3' = v3 - C31*v1 - C32*v2.
vol(v1,v2,v3) = vol(v1,v2',v3') = x*y*z = det(v1,v2',v3')
= det(v1, v2-C21*v1, v3-C31*v1-C32*v2)= det(v1, v2, v3-C31*v1-C32*v2)
= det(v1,v2,v3)
...
In n-d:
(signed) n-volume of (v1,v2,...,vn) is vol(v1,v2,...,vn).
v1=(x1,0,...,0) v2'=(0,x2,...,0) ... vn'=(0,...,0,xn)
v2' = v2 - C21*v1.
v3' = v3 - C31*v1 - C32*v2.
...vn' = vn - Cn1*v1 - Cn2*v2 -...- Cnn-1*vn-1
vol(v1,v2,...,vn) = vol(v1,v2',...,vn') = x1*x2*...*xn = det(v1,v2',...,vn')
= det(v1, v2 - C21*v1, ..., vn - Cn1*v1 - Cn2*v2 -...- Cnn-1*vn-1)
= det(v1,v2,..., vn-1, vn - Cn1*v1 - Cn2*v2 -...- Cnn-1*vn-1)
= det(v1,v2,..., vn)
if v1,v2 are linear dependent, det(v1,v2)=0 and vol(v1,v2)(which is signed area)=0.
if v1,v2,v3 are linear dependent, det(v1,v2,v3)=0 and vol(v1,v2,v3)(which is signed volume)=0.
if v1,v2, ..., vn are linear dependent, det(v1,v2,...,vn)=0 and vol(v1,v2,...,vn)=0.
So we always have vol(v1,v2,...,vn) = det(v1,v2,..., vn).
Determinant : Scaling factor of n-volume