Determinant/Field/Recursively/Introduction/Section

Definition

Let ${}K$ be a field, and let ${}M$ denote an ${}n\times n$-matrix over ${}K$ with entries ${}a_{ij}$. For ${}i\in \{1,\ldots ,n\}$, let ${}M_{i}$ denote the ${}(n-1)\times (n-1)$-matrix, which arises from ${}M$, when we remove the first column and the ${}i$-th row. Then one defines recursively the determinant of ${}M$ by

${}\det M={\begin{cases}a_{11}\,,&{\text{ for }}n=1\,,\\\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det M_{i}&{\text{ for }}n\geq 2\,.\end{cases}}\,$

The determinant is only defined for square matrices. For small ${}n$, the determinant can be computed easily.

Example

For a ${}2\times 2$-matrix

${}M={\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,,$

we have

$\det {\begin{pmatrix}a&b\\c&d\end{pmatrix}}=ad-cb.$

For a ${}3\times 3$-matrix, we can use the rule of Sarrus to compute the determinant. We repeat the first column as the fourth column and the second column as the fifth column. The products of diagonals from up to down enter with a positive sign, and the products of the other diagonals enter with a negative sign.

Example

For a ${}3\times 3$-matrix ${}M={\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}$, we have

$\det {\begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{pmatrix}}=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}.$

This is called the rule of Sarrus.

Lemma

For an upper triangular matrix

${}M={\begin{pmatrix}b_{1}&\ast &\cdots &\cdots &\ast \\0&b_{2}&\ast &\cdots &\ast \\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&b_{n-1}&\ast \\0&\cdots &\cdots &0&b_{n}\end{pmatrix}}\,,$

we have

${}\det M=b_{1}b_{2}\cdots b_{n-1}b_{n}\,.$

In particular, for the

identity matrix we get ${}\det E_{n}=1$.

Proof

This follows with a simple induction directly from the recursive definition of the determinant.

$\Box$