# Determinant/Alternating/Section

We want to show that the recursively defined determinant is a multilinear and alternating mapping. To make sense of this, we identify
    $\operatorname {Mat} _{n}(K)\cong (K^{n})^{n}\,,$
that is, we identify a matrix with the $n$-tuple of its rows. Thus, in the following, we consider a matrix as a column tuple
    ${\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}},$
where the entries $v_{i}$ are row vectors of length $n$.

## Theorem

Let $K$ be a field, and $n\in \mathbb {N} _{+}$. Then the determinant
    $\operatorname {Mat} _{n}(K)=(K^{n})^{n}\longrightarrow K,M\longmapsto \det M,$
is multilinear. This means that for every $k\in \{1,\ldots ,n\}$, and for every choice of $n-1$ vectors $v_{1},\ldots ,v_{k-1},v_{k+1},\ldots ,v_{n}\in K^{n}$, and for any $u,w\in K^{n}$, the identity
    $\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u+w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}=\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}+\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,$
holds, and for $s\in K$, the identity
    $\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\su\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}=s\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,$

### Proof

Let
    $M:={\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,,M':={\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}{\text{ and }}{\tilde {M}}:={\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u+w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}},$
where we denote the entries and the matrices arising from deleting a row in an analogous way. In particular, $u=\left(a_{k1},\,\ldots ,\,a_{kn}\right)$ and $w=\left(a_{k1}',\,\ldots ,\,a_{kn}'\right)$. We prove the statement by induction over $n$, For $i\neq k$, we have ${\tilde {a}}_{i1}=a_{i1}=a'_{i1}$ and
    $\det {\tilde {M}}_{i}=\det M_{i}+\det M'_{i}\,$
due to the induction hypothesis. For $i=k$, we have $M_{k}=M_{k}'={\tilde {M}}_{k}$ and ${\tilde {a}}_{k1}=a_{k1}+a'_{k1}$. Altogether, we get
    ${\begin{aligned}\det {\tilde {M}}&=\sum _{i=1}^{n}(-1)^{i+1}{\tilde {a}}_{i1}\det {\tilde {M}}_{i}\\&=\sum _{i=1,\,i\neq k}^{n}(-1)^{i+1}a_{i1}(\det {M}_{i}+\det {M}'_{i})+(-1)^{k+1}(a_{k1}+a'_{k1})(\det {\tilde {M}}_{k})\\&=\sum _{i=1,\,i\neq k}^{n}(-1)^{i+1}a_{i1}\det {M}_{i}+\sum _{i=1,\,i\neq k}^{n}(-1)^{i+1}a_{i1}\det {M}'_{i}+(-1)^{k+1}a_{k1}\det M_{k}+(-1)^{k+1}a'_{k1}\det M_{k}\\&=\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det {M}_{i}+\sum _{i=1,\,i\neq k,\,}^{n}(-1)^{i+1}a_{i1}\det {M}'_{i}+(-1)^{k+1}a'_{k1}\det M_{k}\\&=\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det {M}_{i}+\sum _{i=1}^{n}(-1)^{i+1}a'_{i1}\det {M}'_{i}\\&=\det M+\det M'.\end{aligned}}$
The compatibility with the scalar multiplication is proved in a similar way, see exercise.
$\Box$

## Theorem

Let $K$ be a field and $n\in \mathbb {N} _{+}$. Then the determinant
    $\operatorname {Mat} _{n}(K)=(K^{n})^{n}\longrightarrow K,M\longmapsto \det M,$
is

### Proof

We proof the statement by induction over $n$, So suppose that $n\geq 2$ and set $M={\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}={\left(a_{ij}\right)}_{ij}$. Let $v_{r}$ and $v_{s}$ with $r<s$ be the relevant row. By definition, we have $\det M=\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det M_{i}$. Due to the induction hypothesis, we have $\det M_{i}=0$ for $i\neq r,s$, because two rows coincide in these cases. Therefore,
    $\det M=(-1)^{r+1}a_{r1}\det M_{r}+(-1)^{s+1}a_{s1}\det M_{s}\,,$
where $a_{r1}=a_{s1}$. The matrices $M_{r}$ and $M_{s}$ consist in the same rows, however, the row $z=v_{r}=v_{s}$ is in $M_{r}$ the $(s-1)$-th row and in $M_{s}$ the $r$-th row. All other rows occur in both matrices in the same order. By swapping altogether $s-r-1$ times adjacent rows, we can transform $M_{r}$ into $M_{s}$. Due to the induction hypothesis and fact, their determinants are related by the factor $(-1)^{s-r-1}$, thus $\det M_{s}=(-1)^{s-r-1}\det M_{r}$. Using this, we obtain
    ${\begin{aligned}\det M&=(-1)^{r+1}a_{r1}\det M_{r}+(-1)^{s+1}a_{s1}\det M_{s}\\&=a_{r1}{\left((-1)^{r+1}\det M_{r}+(-1)^{s+1}(-1)^{s-r-1}\det M_{r}\right)}\\&=a_{r1}{\left((-1)^{r+1}+(-1)^{2s-r}\right)}\det M_{r}\\&=a_{r1}{\left((-1)^{r+1}+(-1)^{r}\right)}\det M_{r}\\&=0.\end{aligned}}$
$\Box$
The property of the determinant to be alternating simplifies its computation. In particular, it is clear how the determinat behaves under elementary row operations. If a row is multiplied with a number $s$, the determinant has to be multiplied with $s$ as well. If two rows are swapped, then the sign of the determinant changes. If a row (or a multiple of a row) is added to another row, then the determinant does not change.