Determinant/Multiplication theorem/Fact/Proof

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

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

To fact