"Geometrically, it can be viewed as the scaling factor of the linear transformation described by the matrix."

Question

The sentence in the title can be found on Wikipedia with no further explanation. Could somebody tell me where I can find more about this? Does this mean that the determinant function is a linear transformation?

Answer 1

It's very well explained in ODE by Arnol'd V.I. in the chapter called "The Determinant of an Operator", but anyway it's true, you can think of determinant of a $n \times n$ matrix as of linear (with respect to each one of its arguments) function with signature $$R^{n^2} = R^n \times \ldots \times R^n \to R,$$ where you take column vectors of a matrix as inputs of this function. Such functions are called tensors (actually, skew-symmetric tensors), you can read about it here: Tensors - Wiki. As I remember, Calculus on Manifolds also has an amazing chapter about tensors.

Answer 2

The absolute value of the determinant is the factor by which the volume (Lebesgue measure, more generally) changes when mapping sets under a given transformation. For example, a determinant $3$ transformation on $\mathbb{R}^2$ will map shapes to other shapes with triple the area.

The determinant is not linear; finding the volume of a shape mapped under $T_1 + T_2$ is not simply a matter of smooshing the image under $T_1$ with the image under $T_2$! It's not difficult to see, in fact, that $I$ has determinant $1$ (it doesn't change the measure of shapes), but $I + I$ has determinant $2^n$, where $n$ is the dimension of the space.