Let $\mM = (M_{ij})$ be an $n\times n$ matrix with entries in a field $K$ . The matrix $\mM$ is really the same thing as a list of $n$ column vectors of size $n$ . Consequently, the determinant operation may be regarded as a mapping $$\det:\overbrace{K^n\times\ldots\times K^n}^{n \mbox{ times}}\rightarrow K$$ The determinant of a matrix $\mM$ is then defined to be $\det(\mM_1,\ldots,\mM_n),$ where $\mM_j\in K^n$ denotes the $j\supth$ column of $\mM$ .

Starting with the definition \begin{equation} \label{eq:detdef} \det(\mM_1,\ldots,\mM_n) = \sum_{\pi\in S_n} \mathrm{sgn}(\pi) M_{1\pi_1} M_{2\pi_2}\cdots M_{n\pi_n} \end{equation}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 $1$ .

Let us prove this. We proceed by representing elements of $K^n$ as linear combinations of $$ \be_1=\begin{pmatrix} 1\\0\\0\\\vdots\\0\end{pmatrix},\quad \be_2=\begin{pmatrix} 0\\1\\0\\\vdots\\0\end{pmatrix},\quad \ldots\quad \be_n=\begin{pmatrix} 0\\0\\0\\ \vdots\\1\end{pmatrix}, $$ the standard basis of $K^n$ . Let $\mM$ be an $n\times n$ matrix. The $j\supth$ column is represented as $\sum_i M_{ij}\be_i$ ; whence using multilinearity