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

$\det {\begin{pmatrix}a&b\\c&d\end{pmatrix}}=ad-cb.$

For a ${}3\times 3$-matrix, we can use the rule of Sarrus to compute the determinant. We repeat the first column as the fourth column and the second column as the fifth column. The products of diagonals from up to down enter with a positive sign, and the products of the other diagonals enter with a negative sign.

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

${}M={\begin{pmatrix}b_{1}&\ast &\cdots &\cdots &\ast \\0&b_{2}&\ast &\cdots &\ast \\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&b_{n-1}&\ast \\0&\cdots &\cdots &0&b_{n}\end{pmatrix}}\,,$

we have

${}\det M=b_{1}b_{2}\cdots b_{n-1}b_{n}\,.$

In particular, for the

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

${}K$-linear.

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

$\Phi {\left(\sum _{i=1}^{m_{1}}a_{i1}v_{i1},\ldots ,\sum _{i=1}^{m_{n}}a_{in}v_{in}\right)}=\sum _{(i_{1},\ldots ,i_{n})\in \{1,\ldots ,m_{1}\}\times \cdots \times \{1,\ldots ,m_{n}\}}a_{i_{1}1}\cdots a_{i_{n}n}\Phi {\left(v_{i_{1}1},\ldots ,v_{i_{n}n}\right)}\,.$

Proof

See Exercise 16.17 .

$\Box$

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

$\Phi (v_{1},\ldots ,v_{r-1},v_{r},v_{r+1},\ldots ,v_{s-1},v_{s},v_{s+1},\ldots ,v_{n})=-\Phi (v_{1},\ldots ,v_{r-1},v_{s},v_{r+1},\ldots ,v_{s-1},v_{r},v_{s+1},\ldots ,v_{n})\,.$

This means that if we swap two vectors, them the sign is changing.

Proof

Due to the definition of alternating and Lemma 16.6 , 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

${}\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\su\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}=s\det {\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,$

holds.

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 16.22 .

$\Box$

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

alternating.

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 Lemma 16.8 , 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

over ${}K$. Then the following statements are equivalent.

1. We have ${}\det M\neq 0$.
2. The rows of ${}M$ are linearly independent.
3. ${}M$ is invertible.
4. We have ${}\operatorname {rk} \,M=n$.

Proof

The relation between rank, invertibility and linear independence was proven in Corollary 12.16 . 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 Theorem 16.9 and Theorem 16.10 , 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 ${}1$, due to Lemma 16.4 , 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 (or rows).

For a proof of the relation between determinant and volume just mentioned, see Fact *****.

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