Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 17

Universal property of the determinant

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.

1. ${}\triangle$ is multilinear.
2. ${}\triangle$ 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.

1. If a row of ${}M$ is multiplied with ${}s\in K$, then ${}\triangle$ is multiplied by ${}s$.
2. If ${}M$ contains a zero row, then ${}\triangle (M)=0$.
3. If in ${}M$ two rows are swapped, then ${}\triangle$ is multiplied with the factor ${}-1$.
4. If a multiple of a row is added to another row, then ${}\triangle$ does not change.
5. 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 Lemma 16.8 .
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

determinant.

Proof

The determinant fulfills, due to Theorem 16.9 , Theorem 16.10 and Lemma 16.4 , 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 Lemma 17 2. , 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.

$\Box$

The multiplication theorem for determinants

We discuss several important theorems about the determinant.

Theorem

Let ${}K$ denote a field, and ${}n\in \mathbb {N} _{+}$. Then for matrices ${}A,B\in \operatorname {Mat} _{n}(K)$, the relation

${}\det {\left(A\circ B\right)}=\det A\cdot \det B\,$

holds.

Proof

We fix the matrix ${}B$.

Suppose first that ${}\det B=0$. Then, due to Theorem 16.11 the matrix ${}B$ is not invertible and therefore, also ${}A\circ B$ is not invertible. Hence, ${}\det A\circ B=0$.

Suppose now that ${}B$ is invertible. In this case, we consider the well-defined mapping

$\delta \colon \operatorname {Mat} _{n}(K)\longrightarrow K,A\longmapsto (\det A\circ B)(\det B)^{-1}.$

We want to show that this mapping equals the mapping ${}A\mapsto \det A$, by showing that it fulfills all the properties which, according to Theorem 17.3 , characterize the determinant. If ${}z_{1},\ldots ,z_{n}$ denote the rows of ${}A$, then ${}\delta (A)$ is computed by applying the determinant to the rows ${}z_{1}B,\ldots ,z_{n}B$, and then by multiplying with ${}(\det B)^{-1}$. Hence the multilinearity and the alternating property follows from Exercise 16.29 . If we start with ${}A=E_{n}$, then ${}A\circ B=B$ and thus

${}\delta (E_{n})=(\det B)\cdot (\det B)^{-1}=1\,.$

$\Box$

Theorem

Let ${}K$ denote a field, and let ${}M$ denote an ${}n\times n$-matrix over ${}K$. Then

${}\det M=\det {M^{\text{tr}}}\,.$

Proof

If ${}M$ is not invertible, then, due to Theorem 16.11 , the determinant is ${}0$ and the rank is smaller than ${}n$. This does also hold for the transposed matrix, so that its determinant is again ${}0$. So suppose that ${}M$ is invertible. We reduce the statement in this case to the corresponding statement for the elementary matrices, which can be verified directly, see Exercise 16.13 . Because of Lemma 12.18 , there exist elementary matrices ${}E_{1},\ldots ,E_{s}$ such that

${}D=E_{s}\cdots E_{1}M\,$

is a diagonal matrix. Due to Exercise 4.20 , we have

${}{D^{\text{tr}}}={M^{\text{tr}}}{E_{1}^{\text{tr}}}\cdots {E_{s}^{\text{tr}}}\,$

and

${}{M^{\text{tr}}}={D^{\text{tr}}}({E_{s}^{\text{tr}}})^{-1}\cdots ({E_{1}^{\text{tr}}})^{-1}\,.$

The diagonal matrix ${}D$ is not changed under transposing it. Since the determinants of the elementary matrices are also not changed under transposition, we get, using Theorem 17.4 ,

${}{\begin{aligned}\det {M^{\text{tr}}}&=\det {\left({D^{\text{tr}}}({E_{s}^{\text{tr}}})^{-1}\cdots ({E_{1}^{\text{tr}}})^{-1}\right)}\\&=\det {D^{\text{tr}}}\cdot \det({E_{s}^{\text{tr}}})^{-1}\cdots \det({E_{1}^{\text{tr}}})^{-1}\\&=\det D\cdot \det {\left(E_{s}^{-1}\right)}\cdots \det {\left(E_{1}^{-1}\right)}\\&=\det {\left(E_{1}^{-1}\right)}\cdots \det {\left(E_{s}^{-1}\right)}\cdot \det D\cdot \\&=\det {\left(E_{1}^{-1}\cdots E_{s}^{-1}\cdot D\right)}\\&=\det M.\end{aligned}}$

$\Box$

Corollary

Let ${}K$ be a field, and let ${}M={\left(a_{ij}\right)}_{ij}$ be an ${}m\times n$-matrix over ${}K$. For ${}i,j\in \{1,\ldots ,n\}$, let ${}M_{ij}$ be the matrix which arises from ${}M$, by leaving out the ${}i$-th row and the ${}j$-th column. Then (for ${}n\geq 2$ and for every fixed ${}i$ and ${}j$)

${}\det M=\sum _{i=1}^{n}(-1)^{i+j}a_{ij}\det M_{ij}=\sum _{j=1}^{n}(-1)^{i+j}a_{ij}\det M_{ij}\,.$

Proof

For ${}j=1$, the first equation is the recursive definition of the determinant. From that statement, the case ${}i=1$ follows, due to Fact *****. By exchanging columns and rows, the statement follows in full generality, see Exercise 17.11 .

$\Box$

The determinant of a linear mapping

Let

$\varphi \colon V\longrightarrow V$

be a linear mapping from a vector space of dimension ${}n$ into itself. This is described by a matrix ${}M\in \operatorname {Mat} _{n}(K)$ with respect to a given basis. We would like to define the determinant of the linear mapping, by the determinant of the matrix. However, we have here the problem whether this is well-defined, since a linear mapping is described by quite different matrices, with respect to different bases. But, because of Corollary 11.12 , when we have two describing matrices ${}M$ and ${}N$, and the matrix ${}B$ for the change of bases, we have the relation ${}N=BMB^{-1}$. The multiplication theorem for determinants yields then

${}\det N=\det {\left(BMB^{-1}\right)}=(\det B)(\det M){\left(\det B^{-1}\right)}=(\det B){\left(\det B^{-1}\right)}(\det M)=\det M\,,$

so that the following definition is in fact independent of the basis chosen.

Definition

Let ${}K$ denote a field, and let ${}V$ denote a ${}K$-vector space of finite dimension. Let

$\varphi \colon V\longrightarrow V$

be a linear mapping, which is described by the matrix ${}M$, with respect to a basis. Then

${}\det \varphi :=\det M\,$

is called the determinant of the linear mapping ${}\varphi$.

Cramer's rule

Definition

For a square matrix ${}M\in \operatorname {Mat} _{n}(K)$, we call

$M^{\operatorname {adj} }=(b_{ij}){\text{ with }}b_{ij}=(-1)^{i+j}\det M_{ji}$

the adjugate matrix of ${}M$, where ${}M_{ji}$ arises from ${}M$ by deleting the ${}j$-th row and the ${}i$-th column.

Note that in this definition, for the entries of the adjugate, the rows and the columns are swapped.

Theorem

Let ${}K$ be a field, and let ${}M$ denote an ${}n\times n$-matrix over ${}K$. Then

${}(M^{\operatorname {adj} })\cdot M=(\det M)E_{n}=M\cdot (M^{\operatorname {adj} })\,$

If ${}M$ is

invertible, then

${}M^{-1}={\frac {1}{\det M}}\cdot M^{\operatorname {adj} }\,.$

Proof

Let ${}M={\left(a_{ij}\right)}_{ij}$. Let the coefficients of the adjugate matrix be denoted by

${}b_{ik}=(-1)^{i+k}\det M_{ki}\,.$

The coefficients of the product ${}(M^{\operatorname {adj} })\cdot M$ are

${}c_{ij}=\sum _{k=1}^{n}b_{ik}a_{kj}=\sum _{k=1}^{n}(-1)^{i+k}a_{kj}\det M_{ki}\,.$

In case ${}j=i$, this is ${}\det M$, as this sum is the expansion of the determinant with respect to the ${}j$-th column. So let ${}j\neq i$, and let ${}N$ denote the matrix that arises from ${}M$ by replacing in ${}M$ the ${}i$-th column by the ${}j$-th column. If we expand ${}N$ with respect to the ${}i$-th column, then we get

${}0=\det N=\sum _{k=1}^{n}(-1)^{i+k}a_{kj}\det M_{ki}=c_{ij}\,.$

Therefore, these coefficients are ${}0$, and the first equation holds.
The second equation is proved similarly, where we use now the expansion of the determinant with respect to the rows.

$\Box$

The following statement is called Cramer's rule.

Theorem

Let ${}K$ be a field, and let

${\begin{matrix}a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}&=&c_{1}\\a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n}&=&c_{2}\\\vdots &\vdots &\vdots \\a_{n1}x_{1}+a_{n2}x_{2}+\cdots +a_{nn}x_{n}&=&c_{n}\end{matrix}}$

be an inhomogeneous linear system over ${}K$. Suppose that the describing matrix ${}M={\left(a_{ij}\right)}_{ij}$ is invertible. Then the unique solution for ${}x_{j}$ is given by

${}x_{j}={\frac {\det {\begin{pmatrix}a_{11}&\cdots &a_{1,j-1}&c_{1}&a_{1,j+1}&\cdots &a_{1n}\\\vdots &\ddots &\vdots &\vdots &\vdots &\ddots &\vdots \\a_{n1}&\cdots &a_{n,j-1}&c_{n}&a_{n,j+1}&\cdots &a_{nn}\end{pmatrix}}}{\det M}}$.

Proof

For an invertible matrix ${}M$, the solution of the linear system ${}Mx=c$ can be found by applying ${}M^{-1}$, that is, ${}x=M^{-1}c$. Using Theorem 17.9 , this means ${}x={\frac {1}{\det M}}(M^{\operatorname {adj} })c$. For the ${}j$-th component, this means

${}x_{j}={\frac {1}{\det M}}{\left(\sum _{k=1}^{n}(-1)^{k+j}(\det M_{kj})\cdot c_{k}\right)}\,.$

The right-hand factor is the expansion of the determinant of the matrix shown in the numerator with respect to the ${}j$-th column.

$\Box$

Example

We solve the linear system

${}{\begin{pmatrix}4&3\\-1&5\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}={\begin{pmatrix}2\\7\end{pmatrix}}\,$

using Cramer's rule. This yields

${}x_{1}={\frac {\det {\begin{pmatrix}2&3\\7&5\end{pmatrix}}}{\det {\begin{pmatrix}4&3\\-1&5\end{pmatrix}}}}=-{\frac {11}{23}}\,$

and

${}x_{2}={\frac {\det {\begin{pmatrix}4&2\\-1&7\end{pmatrix}}}{\det {\begin{pmatrix}4&3\\-1&5\end{pmatrix}}}}={\frac {30}{23}}\,.$

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