determinant as a multilinear mapping

Let $𝐌=(M_{ij})$ be an $n\times n$ matrix with entries in a field$K$. The matrix $𝐌$ is really the same thing as a list of $n$ columnvectors of size $n$. Consequently, the determinant operation may beregarded as a mapping

$$det:\stackrel{n\text{ times}}{\overbrace{K^n\times \mathrm{\dots }\times K^n}}\to K$$

The determinant of a matrix $𝐌$ is then defined to be$det({𝐌}_1,\mathrm{\dots },{𝐌}_n),$where ${𝐌}_j\in K^n$ denotes the $j^{\text{th}}$ column of $𝐌$.

Starting with the definition

$$det({𝐌}_1,\mathrm{\dots },{𝐌}_n)=\sum _{\pi \in S_n}\mathrm{sgn}(\pi )M_{1{\pi }_1}{M}_{2{\pi }_2}\mathrm{\cdots }{M}_{n{\pi }_n}$$(1)

the following properties are easily established:

1. the determinant is multilinear;

2. the determinant is anti-symmetric;

3. the determinant of the identity matrix is $1$.

These three properties uniquely characterize the determinant, andindeed can — some would say should — be used as the definition ofthe determinant operation.

Let us prove this. We proceed by representing elements of $K^n$as linear combinations of

$${𝐞}_1=\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill \mathrm{⋮}\hfill \\ \hfill 0\hfill \end{array}\right),{𝐞}_2=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 1\hfill \\ \hfill 0\hfill \\ \hfill \mathrm{⋮}\hfill \\ \hfill 0\hfill \end{array}\right),\mathrm{\dots }\mathit{\hspace{1em}}{𝐞}_n=\left(\begin{array}{c}\hfill 0\hfill \\ \hfill 0\hfill \\ \hfill 0\hfill \\ \hfill \mathrm{⋮}\hfill \\ \hfill 1\hfill \end{array}\right),$$

the standard basis of $K^n$. Let $𝐌$ be an $n\times n$ matrix.The $j^{\text{th}}$ column is represented as ${\sum }_i{M}_{ij}{𝐞}_i$; whenceusing multilinearity

$det(𝐌)$$=det({\displaystyle \sum _i}{M}_{i1}{𝐞}_i,{\displaystyle \sum _i}{M}_{i2}{𝐞}_i,\mathrm{\dots },{\displaystyle \sum _i}{M}_{in}{𝐞}_i)$

$={\displaystyle \sum _{i_1,\mathrm{\dots },i_n=1}^n}{M}_{i_{1}1}{M}_{i_{2}2}\mathrm{\cdots }{M}_{i_{n}n}det({𝐞}_{i_1},{𝐞}_{i_2},\mathrm{\dots },{𝐞}_{i_n})$

The anti-symmetry assumption implies that the expressions$det({𝐞}_{i_1},{𝐞}_{i_2},\mathrm{\dots },{𝐞}_{i_n})$ vanish if any two of theindices $i_1,\mathrm{\dots },i_n$ coincide. If all $n$ indices are distinct,

$$det({𝐞}_{i_1},{𝐞}_{i_2},\mathrm{\dots },{𝐞}_{i_n})=\pm det({𝐞}_1,\mathrm{\dots },{𝐞}_n),$$

the sign in the above expression beingdetermined by the number of transpositions required to rearrangethe list $(i_1,\mathrm{\dots },i_n)$ into the list $(1,\mathrm{\dots },n)$. The sign istherefore the parity of the permutation $(i_1,\mathrm{\dots },i_n)$. Since wealso assume that

$$det({𝐞}_1,\mathrm{\dots },{𝐞}_n)=1,$$

we now recover the original definition(1).