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

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 MDLD/vector space over a field MDLD/field ${}K$ of dimension MDLD/dimension (fgvs) ${}n$ . A mapping MDLD/mapping

\triangle \colon V^{n}\longrightarrow K

is called a determinant function if the following two conditions are fulfilled.

${}\triangle$ is multilinear.MDLD/multilinear
${}\triangle$ is alternating.MDLD/alternating

Lemma Create referencenumber

Let ${}K$ be a field MDLD/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.MDLD/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,MDLD/upper triangular matrix we have ${}\triangle (M)=a_{11}a_{22}\cdots a_{nn}$ .

Proof

(1) and (2) follow directly from multilinearity.MDLD/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 Create referencenumber

Let ${}K$ be a field MDLD/field and ${}n\in \mathbb {N}$ . Then there exists exactly one determinant function MDLD/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,MDLD/standard vectors namely the

determinant.MDLD/determinant

Proof

The determinant MDLD/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

The multiplication theorem for determinants

We discuss several important theorems about the determinant.

Theorem Create referencenumber

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 fact the matrix ${}B$ is not invertible MDLD/invertible (matrix) 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 fact, 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. 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 Create referencenumber

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

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

Proof

If ${}M$ is not invertible, then, due to fact, 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. Because of fact, there exist elementary matrices MDLD/elementary matrices ${}E_{1},\ldots ,E_{s}$ such that

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

is a diagonal matrix.MDLD/diagonal matrix Due to exercise, 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 fact,

{}{\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 Create referencenumber

Let ${}K$ be a field,MDLD/field and let ${}M={\left(a_{ij}\right)}_{ij}$ be an ${}m\times n$ -matrix MDLD/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.MDLD/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.

\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 fact, 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