Definition:Minor of Determinant/Notation

From ProofWiki
Jump to navigation Jump to search

Definition

Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$.

Let $D := \map \det {\mathbf A}$ denote the determinant of $\mathbf A$.

Let:

$\set {a_1, a_2, \ldots, a_k}$ be the indices of the $k$ selected rows of $\mathbf A$
$\set {b_1, b_2, \ldots, b_k}$ be the indices of the $k$ selected columns of $\mathbf A$

where all of $a_1, \ldots, a_k$ and all of $b_1, \ldots, b_k$ are between $1$ and $n$.


Let:

$\mathbf B := \mathbf A \sqbrk {a_1, a_2, \ldots, a_k; b_1, b_2, \ldots, b_k}$

be the submatrix formed from rows $\set {a_1, a_2, \ldots, a_k}$ and columns $\set {b_1, b_2, \ldots, b_k}$.


The order-$k$ minor of $D$ formed from rows $r_1, r_2, \ldots, r_k$ and columns $s_1, s_2, \ldots, s_k$ can be denoted:

$\map D {a_1, a_2, \ldots, a_k \mid b_1, b_2, \ldots, b_k}$.


Each element of $D$ is an order $1$ minor of $D$, and can be denoted:

$\map D {a_i \mid b_j}$


Minor of Order $n - 1$

Let a submatrix $\mathbf B$ of $\mathbf A$ be of order $n - 1$.

Let:

$j$ be the row of $\mathbf A$ which is not included in $\mathbf B$
$k$ be the column of $\mathbf A$ which is not included in $\mathbf B$.

Thus, let $\mathbf B := \map {\mathbf A} {j; k}$.

Then $\map \det {\mathbf B}$ can be denoted:

$D_{i j}$


That is, $D_{i j}$ is the minor of order $n - 1$ obtained from $D$ by deleting all the elements of row $i$ and column $j$.