# Determinant/Field/Recursively/Multiplication theorem/No proof/Section

We discuss without proofs further important theorems about the determinant. The proofs rely on a systematic account of the properties which are characteristic for the determinant, namely the properties multilinear and alternating. By these properties, together with the condition that the determinant of the identity matrix is ${\displaystyle {}1}$, the determinant is already determined.

## Theorem

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

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

holds.

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

## Definition

Let ${\displaystyle {}K}$ be a field, and let ${\displaystyle {}M=(a_{ij})_{ij}}$ be an ${\displaystyle {}m\times n}$-matrix over ${\displaystyle {}K}$. Then the ${\displaystyle {}n\times m}$-matrix

${\displaystyle {M^{\text{tr}}}={\left(b_{ij}\right)}_{ij}{\text{ with }}b_{ij}:=a_{ji}}$
is called the transposed matrix for ${\displaystyle {}M}$.

The transposed matrix arises by interchanging the role of the rows and the columns. For example, we have

${\displaystyle {}{{\begin{pmatrix}t&n&o&d\\r&s&s&x\\a&p&e&y\end{pmatrix}}^{\text{tr}}}={\begin{pmatrix}t&r&a\\n&s&p\\o&s&e\\d&x&y\end{pmatrix}}\,.}$

## Theorem

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

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

### Proof

This proof was not presented in the lecture.
${\displaystyle \Box }$

This implies that we can compute the determinant also by expanding with respect to the rows, as the following statement shows.

## Corollary

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

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

### Proof

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

${\displaystyle \Box }$