Determinant/Field/Recursively/Multilinearity/No proof/Section

We want to show that the recursively defined determinant is a "multilinear“ and "alternating“ mapping, where we identify

${}\operatorname {Mat} _{n}(K)\cong (K^{n})^{n}\,,$

so a matrix is identified with the ${}n$-tuple of the rows of the matrix. We consider a matrix as a tuple of columns

${\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}$

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

This proof was not presented in the lecture.

$\Box$

Theorem

Let ${}K$ be a field, and ${}n\in \mathbb {N} _{+}$. Then the determinant

$\operatorname {Mat} _{n}(K)=(K^{n})^{n}\longrightarrow K,M\longmapsto \det M,$

has the following properties.

1. If in ${}M$ two rows are identical, then ${}\det M=0$. This means that the determinant is alternating.
2. If we exchange two rows in ${}M$, then the determinant changes with factor ${}-1$.

Proof

This proof was not presented in the lecture.

$\Box$

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 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 ${}\neq 0$.

$\Box$