# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 16

    The determinant
Suppose an $n\times n$-matrix is given. Can we see "at a glance“ whether it is invertible? Does there exist an expression in the $n^{2}$ entries of the matrix to decide this? This question has a positive answer in terms of the determinant.

## Definition

Let $K$ be a field, and let $M$ denote an $n\times n$-matrix over $K$ with entries $a_{ij}$. For $i\in \{1,\ldots ,n\}$, let $M_{i}$ denote the $(n-1)\times (n-1)$-matrix, which arises from $M$, when we remove the first column and the $i$-th row. Then one defines recursively the determinant of $M$ by
    $\det M={\begin{cases}a_{11}\,,&{\text{ for }}n=1\,,\\\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det M_{i}&{\text{ for }}n\geq 2\,.\end{cases}}\,$
The determinant is only defined for square matrices. For small $n$, the determinant can be computed easily.

## Example

For a $2\times 2$-matrix
    $M={\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,,$
we have
    $\det {\begin{pmatrix}a&b\\c&d\end{pmatrix}}=ad-cb.$

## Example

For a $3\times 3$-matrix $M={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}$, we have
    $\det {\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}.$
This is called the rule of Sarrus.

## Lemma

For an upper triangular matrix
    $M={\begin{pmatrix}b_{1}&\ast &\cdots &\cdots &\ast \\0&b_{2}&\ast &\cdots &\ast \\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&b_{n-1}&\ast \\0&\cdots &\cdots &0&b_{n}\end{pmatrix}}\,,$
    $\det M=b_{1}b_{2}\cdots b_{n-1}b_{n}\,.$
identity matrix we get
$\det E_{n}=1$

### Proof

This follows with a simple induction directly from the recursive definition of the determinant.
$\Box$
    Multilinear and alternatings mappings
Wir führen zwei Begriffe ein, the we im Moment hauptsächlich zum weiteren Verständnis the Determinante brauchen.

## Definition

Let $K$ be a field, and let $V_{1},\ldots ,V_{n}$ and $W$ be vector spaces over $K$. A mapping
    $\Phi \colon V_{1}\times \cdots \times V_{n}\longrightarrow W$
is called multilinear if, for every $i\in \{1,\ldots ,n\}$ and every $(n-1)$-tuple $(v_{1},\ldots ,v_{i-1},v_{i+1},\ldots ,v_{n})$ with $v_{j}\in V_{j}$, the induced mapping
    $V_{i}\longrightarrow W,v_{i}\longmapsto \Phi (v_{1},\ldots ,v_{i-1},v_{i},v_{i+1},\ldots ,v_{n}),$
is
$K$
For $n=2$, this property is called bilinear. For example, the multiplikation in a field $K$, that is, the mapping
    $K\times K\longrightarrow K,(x,y)\longmapsto xy,$
is bilinear. Also, for a $K$-vector space $V$ and its dual space ${V}^{*}$, the evaluation mapping
    $V\times {V}^{*}\longrightarrow K,(v,f)\longmapsto f(v),$
is bilinear.

## Lemma

Let $K$ be a field, and let $V_{1},\ldots ,V_{n}$ and $W$ be vector spaces over $K$. Let
    $\Phi \colon V_{1}\times \cdots \times V_{n}\longrightarrow W$
be a multilinear mapping, and let $v_{1j},\ldots ,v_{m_{j}j}\in V_{j}$ and $a_{ij}\in K$. Then
    $\Phi {\left(\sum _{i=1}^{m_{1}}a_{i1}v_{i1},\ldots ,\sum _{i=1}^{m_{n}}a_{in}v_{in}\right)}=\sum _{(i_{1},\ldots ,i_{n})\in \{1,\ldots ,m_{1}\}\times \cdots \times \{1,\ldots ,m_{n}\}}a_{i_{1}}\cdots a_{i_{n}}\Phi {\left(v_{i_{1}1},\ldots ,v_{i_{n}n}\right)}\,.$

### Proof

$\Box$

## Definition

Let $K$ be a field, let $V$ and $W$ denote $K$-vector spaces, and let $n\in \mathbb {N}$. A multilinear mapping
    $\Phi \colon V^{n}=\underbrace {V\times \cdots \times V} _{n{\text{-mal}}}\longrightarrow W$
is called alternating if the following holds: Whenever in $v=(v_{1},\ldots ,v_{n})$, two entries are identical, that is $v_{i}=v_{j}$ for a pair $i\neq j$, then
    $\Phi (v)=0\,.$
For an alternating mapping, there is only one vector space occurring several times in the product on the left.

## Lemma

Let $K$ be a field, let $V$ and $W$ denote $K$-vector spaces, and let $n\in \mathbb {N}$. Suppose that
    $\Phi \colon V^{n}=\underbrace {V\times \cdots \times V} _{n{\text{-mal}}}\longrightarrow W$
is an alternating mapping. Then
    $\Phi (v_{1},\ldots ,v_{r-1},v_{r},v_{r+1},\ldots ,v_{s-1},v_{s},v_{s+1},\ldots ,v_{n})=-\Phi (v_{1},\ldots ,v_{r-1},v_{s},v_{r+1},\ldots ,v_{s-1},v_{r},v_{s+1},\ldots ,v_{n})\,.$

### Proof

Due to the definition of alternating and fact, we have
    ${\begin{aligned}0&=\Phi (v_{1},\ldots ,v_{r-1},v_{r}+v_{s},v_{r+1},\ldots ,v_{s-1},v_{r}+v_{s},v_{s+1},\ldots ,v_{n})\\&=\Phi (v_{1},\ldots ,v_{r-1},v_{r},v_{r+1},\ldots ,v_{s-1},v_{s},v_{s+1},\ldots ,v_{n})+\Phi (v_{1},\ldots ,v_{r-1},v_{s},v_{r+1},\ldots ,v_{s-1},v_{r},v_{s+1},\ldots ,v_{n}).\end{aligned}}$
$\Box$
    The determinant is an alternating mapping
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.

## Theorem

Let $K$ be a field, and let $M$ denote an $n\times n$-matrix
$K$
* We have $\det M\neq 0$.
* The rows of $M$ are linearly independent.
* $M$ is invertible.
* We have $\operatorname {rk} \,M=n$.

### Proof

The relation between rank, invertibility and linear independence was proven in fact. Suppose now that the rows are linearly dependent. After exchanging rows, we may assume that $v_{n}=\sum _{i=1}^{n-1}s_{i}v_{i}$. Then, due to fact and fact, we get
    $\det M=\det {\begin{pmatrix}v_{1}\\\vdots \\v_{n-1}\\\sum _{i=1}^{n-1}s_{i}v_{i}\end{pmatrix}}=\sum _{i=1}^{n-1}s_{i}\det {\begin{pmatrix}v_{1}\\\vdots \\v_{n-1}\\v_{i}\end{pmatrix}}=0\,.$
Now suppose that the rows are linearly independent. Then, by exchanging of rows, scaling and addition of a row to another row, we can transform the matrix successively into the identity matrix. During these manipulations, the determinant is multiplied with some factor $\neq 0$. Since the determinant of the identity matrix is $1$, the determinant of the initial matrix is $\neq 0$.
$\Box$

## Remark

In case $K=\mathbb {R}$, the determinant is in tight relation to volumes of geometric objects. If we consider in $\mathbb {R} ^{n}$ vectors $v_{1},\ldots ,v_{n}$, then they span a parallelotope. This is defined by
    $P:={\left\{s_{1}v_{1}+\cdots +s_{n}v_{n}\mid s_{i}\in [0,1]\right\}}\,.$
It consists of all linear combinations of these vectors, where all the scalars belong to the unit interval. If the vectors are linearly independent, then this is a "voluminous“ body, otherwise it is an object of smaller dimension. Now the relation
    $\operatorname {vol} \,P=\vert {\det {\left(v_{1},\ldots ,v_{n}\right)}}\vert \,$
holds, saying that the volume of the parallelotope is the modulus of the determinant of the matrix, consisting of the spanning vectors as columns.
For a proof of the relation between determinant and volume just mentioned, see fact.