Determinant/Alternating/Section

We want to show that the recursively defined determinant is a multilinear and alternating mapping. To make sense of this, we identify

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

that is, we identify a matrix with the ${}n$-tuple of its rows. Thus, in the following, we consider a matrix as a column tuple

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

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

Let

$M:={\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,,M':={\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}{\text{ and }}{\tilde {M}}:={\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u+w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}},$

where we denote the entries and the matrices arising from deleting a row in an analogous way. In particular, ${}u=\left(a_{k1},\,\ldots ,\,a_{kn}\right)$ and ${}w=\left(a_{k1}',\,\ldots ,\,a_{kn}'\right)$. We prove the statement by induction over ${}n$,  For ${}i\neq k$, we have ${}{\tilde {a}}_{i1}=a_{i1}=a'_{i1}$ and

${}\det {\tilde {M}}_{i}=\det M_{i}+\det M'_{i}\,$

due to the induction hypothesis. For ${}i=k$, we have ${}M_{k}=M_{k}'={\tilde {M}}_{k}$ and ${}{\tilde {a}}_{k1}=a_{k1}+a'_{k1}$. Altogether, we get

${}{\begin{aligned}\det {\tilde {M}}&=\sum _{i=1}^{n}(-1)^{i+1}{\tilde {a}}_{i1}\det {\tilde {M}}_{i}\\&=\sum _{i=1,\,i\neq k}^{n}(-1)^{i+1}a_{i1}(\det {M}_{i}+\det {M}'_{i})+(-1)^{k+1}(a_{k1}+a'_{k1})(\det {\tilde {M}}_{k})\\&=\sum _{i=1,\,i\neq k}^{n}(-1)^{i+1}a_{i1}\det {M}_{i}+\sum _{i=1,\,i\neq k}^{n}(-1)^{i+1}a_{i1}\det {M}'_{i}+(-1)^{k+1}a_{k1}\det M_{k}+(-1)^{k+1}a'_{k1}\det M_{k}\\&=\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det {M}_{i}+\sum _{i=1,\,i\neq k,\,}^{n}(-1)^{i+1}a_{i1}\det {M}'_{i}+(-1)^{k+1}a'_{k1}\det M_{k}\\&=\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det {M}_{i}+\sum _{i=1}^{n}(-1)^{i+1}a'_{i1}\det {M}'_{i}\\&=\det M+\det M'.\end{aligned}}$

The compatibility with the scalar multiplication is proved in a similar way, see exercise.

$\Box$

Theorem

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

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

is

alternating.

Proof

We proof the statement by induction over ${}n$,  So suppose that ${}n\geq 2$ and set ${}M={\begin{pmatrix}v_{1}\\\vdots \\v_{n}\end{pmatrix}}={\left(a_{ij}\right)}_{ij}$. Let ${}v_{r}$ and ${}v_{s}$ with ${}r<s$ be the relevant row. By definition, we have ${}\det M=\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det M_{i}$. Due to the induction hypothesis, we have ${}\det M_{i}=0$ for ${}i\neq r,s$, because two rows coincide in these cases. Therefore,

${}\det M=(-1)^{r+1}a_{r1}\det M_{r}+(-1)^{s+1}a_{s1}\det M_{s}\,,$

where ${}a_{r1}=a_{s1}$. The matrices ${}M_{r}$ and ${}M_{s}$ consist in the same rows, however, the row ${}z=v_{r}=v_{s}$ is in ${}M_{r}$ the ${}(s-1)$-th row and in ${}M_{s}$ the ${}r$-th row. All other rows occur in both matrices in the same order. By swapping altogether ${}s-r-1$ times adjacent rows, we can transform ${}M_{r}$ into ${}M_{s}$. Due to the induction hypothesis and fact, their determinants are related by the factor ${}(-1)^{s-r-1}$, thus ${}\det M_{s}=(-1)^{s-r-1}\det M_{r}$. Using this, we obtain

${}{\begin{aligned}\det M&=(-1)^{r+1}a_{r1}\det M_{r}+(-1)^{s+1}a_{s1}\det M_{s}\\&=a_{r1}{\left((-1)^{r+1}\det M_{r}+(-1)^{s+1}(-1)^{s-r-1}\det M_{r}\right)}\\&=a_{r1}{\left((-1)^{r+1}+(-1)^{2s-r}\right)}\det M_{r}\\&=a_{r1}{\left((-1)^{r+1}+(-1)^{r}\right)}\det M_{r}\\&=0.\end{aligned}}$

$\Box$

The property of the determinant to be alternating simplifies its computation. In particular, it is clear how the determinat behaves under elementary row operations. If a row is multiplied with a number ${}s$, the determinant has to be multiplied with ${}s$ as well. If two rows are swapped, then the sign of the determinant changes. If a row (or a multiple of a row) is added to another row, then the determinant does not change.