A basic math identity often used in integrals I'm just wondering about why $y_i=A_{ij}x_j$ implies
$$d^Ny=|\det A|d^Nx.$$
I see that $\det A$ is the product of the eigenvalues of a diagonal matrix but still don't exactly see how. Please help.
 A: To add to the answer of @ACuriousMind, it may be remarked that the determinant of a square matrix is equal to the (signed) volume of the parallelepiped spanned by its columns.
Since this parallelepiped is exactly the image of the unit cube, the intuitive meaning of the substitution theorem (the name under which I know it) is that you can change your coordinates and still get the same value of the integral as long as you correct for the infinitesimal deformation of the unit cube at every point.
A: In German, this property is known as the Transformationssatz, but I do not know any appropriate translation for it.
This is, however, a special case of coordinate tranformations changing the measure by the determinant of their Jacobian, since obviously $\frac{\partial y_i}{\partial x_j} = A_{ij}$. That it is the determinant that plays a role in the transformation of the measure follows from some rather general algebraic considerations:
First off: $\mathrm{d}^Nx$, however widespread it may be, is a terrible notation (in my opinion, obscuring the underlying differential geometry). Properly, we should say that we are integrating the $n$-form $\mathrm{d}x_1 \wedge \dots \wedge \mathrm{d}x_N$ (with some prefactor $f(\vec x)$). Now, each $\mathrm{d}x_i$, as a $1$-form, i.e. a section of the cotangent bundle, transforms by the inverse of the Jacobian matrix, i.e. $\mathrm{d}x_i \mapsto \frac{\partial x_i}{\partial y_j}\mathrm{d}y_j = A^{-1}_{ij}\mathrm{d}y_j$ (summation over repeated indices implied).
Using the abstract exterior algebra definition of the determinant, it follows that 
$$\mathrm{d}x_1 \wedge \dots \wedge \mathrm{d}x_N \mapsto A^{-1}_{1i_1}\mathrm{d}y_{i_1} \wedge \dots \wedge A^{-1}_{Ni_N}\mathrm{d}y_{i_N} = \det(A^{-1})\mathrm{d}y_1 \wedge \dots \wedge \mathrm{d}y_N$$
Or, returning to the initial notation,
$$\mathrm{d}^N x \mapsto \det(A^{-1})\mathrm{d}^N y$$
Using $\det(A^{-1}) = \det(A)^{-1}$ now yields the desired result.
We should remark that it is also possible to forgo just applying the definition of the determinant in abstract terms, but that we may also just use the antisymmetry of the $\wedge$ to gain some antisymmetrization by the Levi-Civita $\epsilon$, and then just compare the obtained sum with the expression for the determinant using the Levi-Civita symbol.
