# Determinant/Field/Universal property/Section

The determinant fulfills the characteristic properties that it is multilinear and alternating. This, together with the property that the determinant of the identity matrix is $1$, determines already the determinant in a unique way.

## Definition

Let $V$ be a vector space over a field $K$ of dimension $n$. A mapping
    $\triangle \colon V^{n}\longrightarrow K$
is called a determinant function if the following two conditions are fulfilled.
* $\triangle$ is multilinear.
* $\triangle$ is alternating.

## Lemma

Let $K$ be a field and $n\in \mathbb {N} _{+}$. Let
    $\triangle \colon \operatorname {Mat} _{n}(K)\cong {\left(K^{n}\right)}^{n}\longrightarrow K$
be a determinant function. Then $\triangle$ fulfills the following properties.
* If a row of $M$ is multiplied with $s\in K$, then $\triangle$ is multiplied by $s$.
* If $M$ contains a zero row, then $\triangle (M)=0$.
* If in $M$ two rows are swapped, then $\triangle$ is multiplied with the factor $-1$.
* If a multiple of a row is added to another row, then $\triangle$ does not change.
* If $\triangle (E_{n})=1$, then, for an upper triangular matrix, we have $\triangle (M)=a_{11}a_{22}\cdots a_{nn}$.

### Proof

(1) and (2) follow directly from multilinearity. (3) follows from fact. To prove (4), we consider the situation where we add to the $s$-th row the $a$-multiple of the $r$-th row, $r<s$. Due to the parts already proven, we have
    $\triangle {\begin{pmatrix}\vdots \\v_{r}\\\vdots \\v_{s}+av_{r}\\\vdots \end{pmatrix}}=\triangle {\begin{pmatrix}\vdots \\v_{r}\\\vdots \\v_{s}\\\vdots \end{pmatrix}}+\triangle {\begin{pmatrix}\vdots \\v_{r}\\\vdots \\av_{r}\\\vdots \end{pmatrix}}=\triangle {\begin{pmatrix}\vdots \\v_{r}\\\vdots \\v_{s}\\\vdots \end{pmatrix}}+a\triangle {\begin{pmatrix}\vdots \\v_{r}\\\vdots \\v_{r}\\\vdots \end{pmatrix}}=\triangle {\begin{pmatrix}\vdots \\v_{r}\\\vdots \\v_{s}\\\vdots \end{pmatrix}}\,.$
(5). If a diagonal element is $0$, then set $r=\max {\left\{i\mid a_{ii}=0\right\}}$. We can add to the $r$-th row suitable multiples of the $i$-th rows, $i>r$, in order to achieve that the new $r$-th row is a zero row, without changing the value of the determinant function. Due to (2), this value is $0$.
In case no diagonal element is $0$, we may obtain, by several scalings, that all diagonal element are $1$. By adding rows, we obtain furthermore the identity matrix. Therefore,
    $\triangle (M)=a_{11}a_{22}\cdots a_{nn}\triangle (E_{n})=a_{11}a_{22}\cdots a_{nn}\,.$
$\Box$

## Theorem

Let $K$ be a field and $n\in \mathbb {N}$. Then there exists exactly one determinant function
    $\triangle \colon \operatorname {Mat} _{n}(K)=(K^{n})^{n}\longrightarrow K$
fulfilling
    $\triangle {\left(e_{1},\,e_{2},\ldots ,e_{n}\right)}=1\,,$
where $e_{i}$ denote the standard vectors, namely the

### Proof

The Determinant fulfills, due to fact, fact and fact, all the given properties. Uniqueness. For every matrix $M$, there exists a sequence of elementary row operations such that, in the end, we get an upper triangular matrix. Hence, due to fact, the value of the determinant function is determined by the values on the upper triangular matrices. Therefore, after scaling and row addition, it is even determined by its value on the identity matrix.
$\Box$