Definition:Minor of Determinant

From ProofWiki
Jump to navigation Jump to search

Definition

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

Consider the order $k$ square submatrix $\mathbf B$ obtained by deleting $n - k$ rows and $n - k$ columns from $\mathbf A$.


Let $\map \det {\mathbf B}$ denote the determinant of $\mathbf B$.

Then $\map \det {\mathbf B}$ is an order-$k$ minor of $\map \det {\mathbf A}$.


Thus a minor is a determinant formed from the elements (in the same relative order) of $k$ specified rows and columns.


Notation

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 $j$.


Example

Let $D$ be the determinant defined as:

$D = \begin {vmatrix} a_{1 1} & a_{1 2} & a_{1 3} \\ a_{2 1} & a_{2 2} & a_{2 3} \\ a_{3 1} & a_{3 2} & a_{3 3} \end{vmatrix}$


Then:

$\map D {1, 2 \mid 1, 3} = \begin {vmatrix} a_{1 1} & a_{1 3} \\ a_{2 1} & a_{2 3} \end {vmatrix}$


Note that $\map D {1, 2 \mid 1, 3}$ can also be denoted as $D_{3 2}$.


Also see

The equivalent term in the context of a matrix is a submatrix.


Sources