Most usual notation for extracted matrix

Question

Let $$A = \pmatrix{1&2&3\\4&5&6\\7&8&9}.$$

What is the most usual notation and english-math vocabulary term for "the extracted matrix obtained by removing the 3rd line and 2nd column", i.e.

$$\pmatrix{1&3\\4&6}$$

Would you note it $A_{3,2}$, ${A^{(3,2)}}$, etc. ? I know there's probably no consensus, but what's the most usual notation you find in textbooks?

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Example: This is useful when computing a determinant:

$$\det(A) = 7 \det A_{3,1} - 8 \det A_{3,2} + 9 \det A_{3,3}. $$

Answer 1

One calls it "the submatrix of $A$ obtained by removing row $3$ and column $2$." Alternatively, you may write "the submatrix of $A$ obtained by deleting the third row and second column".

For the notation, there's no consensus:

• Wikipedia adopts the exclusive notation $M_{ij}$ which denotes the determinant of the submatrix of $A$ obtained by deleting the $i$-th row and $j$-th entry. $$M_{ij} = \det((A_{p,q})_{p \ne i, q \ne j})$$ Wiki claims this notation to be "standard everywhere in the litterature".
• Some authors may prefer surrounding $M_{ij}$ by a pair of $|$'s to denote the determinant, so that $M_{ij}$ means the submatrix of $A$ obtained by deleting the $i$-th row and $j$-th entry. $$M_{ij} =(A_{p,q})_{p \ne i, q \ne j}$$ One example from Shayle R. Searle's Matrix Algebra Useful for Statistics.

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(Edit in response to OP's comment)

The author prefers to use $A_{ij}$ for the $(i,j)$-th partitioned matrix of $A$. I extract the bottom of page 72 of the book to illustrate this.

When matrices $\bf A$ and $\bf B$ are partitioned so that their submatrices are appropriately conformable for multiplication, the product $\bf AB$ can be expressed in partitioned form having submatrices that are functions of the submatrices $\bf A$ and $\bf B$. For example, if $$A = \begin{bmatrix} {\bf A}_{11} & {\bf A}_{12} \\ {\bf A}_{21} & {\bf A}_{22} \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} {\bf B}_{11} \\ {\bf B}_{21} \end{bmatrix},$$ then $${\bf AB} = \begin{bmatrix} {\bf A}_{11} & {\bf A}_{12} \\ {\bf A}_{21} & {\bf A}_{22} \end{bmatrix} \begin{bmatrix} {\bf B}_{11} \\ {\bf B}_{21} \end{bmatrix} = \begin{bmatrix} {\bf A}_{11} {\bf B}_{11} + {\bf A}_{12} {\bf B}_{21} \\ {\bf A}_{21} {\bf B}_{11} + {\bf A}_{22} {\bf B}_{21} \end{bmatrix}.$$

Answer 2

Let $A=[a_{ij}]$ be an $m$-by-$n$ matrix with complex entries. If $\emptyset \subset \alpha = \{\alpha_1,\dots,\alpha_{m'} \} \subseteq\{1,\dots,m\}$ and $\emptyset \subset \beta = \{\beta_1,\dots,\beta_{n'}\} \subseteq\{1,\dots,n\}$, then the submatrix of $A$ with respect to $\alpha$ and $\beta$, denoted $A[\alpha,\beta]$, is the $m'$-by-$n'$ matrix whose $(i,j)$-entry is $$a_{\alpha_i\beta_j}.$$

In the example above, $\alpha = \{1,2\}$, $\beta=\{1,3\}$, and $$A[\alpha,\beta] = \begin{bmatrix} a_{\alpha_1\beta_1} & a_{\alpha_1\beta_2} \\ a_{\alpha_2\beta_1} & a_{\alpha_2\beta_2} \end{bmatrix} = \begin{bmatrix} a_{11} & a_{13} \\ a_{21} & a_{23} \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 4 & 6 \end{bmatrix}.$$

If $\alpha = \beta$, then $A[\alpha,\alpha]$ is abbreviated to $A[\alpha]$.