Determinant/Recursively/Multilinear/Fact

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, Write another proof