Geometric Interpretation of Determinant of Transpose

Solution 1:

Write $A$ as a product of the form $E_1E_2\cdots E_kDF_1F_2\cdots F_\ell$, where each $E_i$ (resp. $F_j$) is an elementary matrix corresponding to row (resp. column) switching or row (resp. column) addition, and $D$ is a diagonal matrix. You are done if you can explain why

1. $\det A=\det(E_1)\cdots\det(E_k)\det(D)\det(F_1)\cdots\det(F_\ell)$ and
2. the determinant of each elementary matrix is equal to the determinant of its transpose.

But these are essentially what motivate the definition of determinant: we want to define the determinant as the volume of the parallelepiped generated by the columns of a matrix. Shears do not change the volume (thus the determinant of an elementary matrix for row/column addition is $1$) and scaling in one direction changes the volume in the same proportion (thus $\det D$ is the product of $D$'s diagonal entries). Since we want to define the volume as zero in the degenerate case, it follows that together with the previous two requirements, the volume must be signed and by flipping two columns, the sign flips too (thus the determinant of an elementary matrix for row/column switching is $-1$).

Solution 2:

To prove that the two (generalized) volumes are equal is going to require some algebra -- I don't see a way around that (because the transpose is defined algebraically). This proof doesn't require the use of matrices or characteristic equations or anything, though. I just use a geometric definition of the determinant and then an algebraic formula relating a linear transformation to its adjoint (transpose).

Consider this as the geometric definition of the determinant. If $I$ is the unit $n$-cube in $\Bbb R^n$, then we can define the determinant of a linear transformation $T$ as $$\underline T(I)=\det(T)I$$ This formula tells us that the transformation $\underline T$ just scales the unit $n$-cube by its determinant.

For vectors, you might recognize that the adjoint (transpose), $\overline T$, of a linear transformation, $T$, is given by $$T(v)\cdot w = v\cdot \overline T(w)$$ However, $I$ is not really a vector, it's an object called an $n$-blade or $n$-vector. So we have to use the modified formula $$\langle\underline T(A)B\rangle = \langle A\overline T(B)\rangle$$ where $\langle M\rangle$ returns the scalar part of the multivector $M$.

So for us, it becomes $\langle\underline T(I)I\rangle = \langle I\overline T(I)\rangle$. On the LHS, we get $$\langle\underline T(I)I\rangle = \langle \det(T)I^2\rangle = \det(T)I^2$$ And on the RHS, we get $$\langle I\overline T(I)\rangle = \langle \det(\overline T)I^2\rangle = \det(\overline T)I^2$$ Therefore $$\det(T) = \det(\overline T)\ \ \ \ \square$$