Determinant/Field/Recursively/Multiplication theorem/Section

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 fact 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 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

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 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 ${}E_{1},\ldots ,E_{s}$ such that

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

is a 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

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.

$\Box$