Usually, the definition of the determinant of a $n\times n$ matrix $A=(a_{ij})$ is as the following:
$$\det(A):=\sum_{\sigma\in S_n}\text{sgn}(\sigma)\prod_{i=1}^na_{i,\sigma(i)}.$$
In Gilbert Strang's Linear Algebra and Its Application, the author points out that it can be understood as the volume of a box in $n$-dimensional space. However, I am wondering if it can be defined in this way.
Here are my questions:
- [EDIT: Can the determinant be defined as the volume of a box in $n$-dimensional space?]
- Are there other definitions of the determinant for a square matrix?
- What is the "advantages" of the different definition? [EDIT: For instance, as Qiaochu said in the comment, how easy is it to prove that $\det(AB)=\det(A)\det(B)$ with that definition?]
Best Answer
Most properties of determinants follow more or less immediately when you use the following definition.
If $f:V\to V$ is an endomorphism of a vector space of finite dimension $n$ and $\Lambda^nV$ is the $n$th exterior power of $V$, then there is an induced morphism $\Lambda^n(f):\Lambda^nV\to\Lambda^nV$. Now $\Lambda^nV$ is a one dimensional vector space, so there is a canonical isomorphism of $k$-algebras $\operatorname{End}(\Lambda^nV)\cong k$. The image of the map $\Lambda^n(f)\in\operatorname{End}(\Lambda^nV)$ in $k$ under this isomorphism is the determinant of $f$.
If one wants to define the determinant of a matrix $A\in M_n(k)$, then one considers the corresponding map $k^n\to k^n$ determined by $A$, and proceeds as above.
Of course, in this approach one has to prove the properties of exterior powers and maps induced on them—but this is neither conceptually nor practically complicated.