668: For Principal Integral Domain, Rectangle Matrix over Domain, and Square Matrix Over Domain, Sum of Principal Ideals by Specified-Dimensional Subdeterminants of Product Is Contained in Sum of Principal Ideals by Same-Dimensional Subdeterminants of Rectangle Matrix|T.B.P.

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description/proof of that for principal integral domain, rectangle matrix over domain, and square matrix over domain, sum of principal ideals by specified-dimensional subdeterminants of product is contained in sum of principal ideals by same-dimensional subdeterminants of rectangle matrix

Topics

About: ring

Starting Context

The reader knows a definition of principal integral domain.

Target Context

The reader will have a description and a proof of the proposition that for any principal integral domain, any rectangle matrix over the domain, and any rectangular-matrix-columns-dimensional or rectangular-matrix-rows-dimensional square matrix over the domain, the sum of the principal ideals by the specified-dimensional subdeterminants of the product is contained in the sum of the principal ideals by the same-dimensional subdeterminants of the rectangle matrix.

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

Main Body

1: Structured Description

Here is the rules of Structured Description.

Entities:

R

\in {the principal integral domains}

M

\in {the m x n matrices over R}

N

\in {the n x n matrices over R}

O

\in {the m x m matrices over R}

k

\in N

1 \leq k \leq m i n (m, n)

I_{k} (M)

= the sum of the principal ideals by the k -dimensional subdeterminants of M

I_{k} (M N)

= the sum of the principal ideals by the k -dimensional subdeterminants of M N

I_{k} (O M)

= the sum of the principal ideals by the k -dimensional subdeterminants of O M

//

Statements:

I_{k} (M N) \subseteq I_{k} (M)

\land

I_{k} (O M) \subseteq I_{k} (M)

//

As an immediate corollary,

I_{k} (O M N) \subseteq I_{k} (M)

I_{k} (O M N) \subseteq I_{k} (M N) \subseteq I_{k} (M)

2: Natural Language Description

For any principal integral domain,

R

, any

m x n

matrix over

R

M

, any

n x n

matrix over

R

N

, any

m x m

matrix over

R

O

, any natural number,

k

, such that

1 \leq k \leq m i n (m, n)

, the sum of the principal ideals by the

k

-dimensional subdeterminants of

M

I_{k} (M)

, the sum of the principal ideals by the

k

-dimensional subdeterminants of

M N

I_{k} (M N)

, and the sum of the principal ideals by the

k

-dimensional subdeterminants of

O M

I_{k} (O M)

I_{k} (M N) \subseteq I_{k} (M)

and

I_{k} (O M) \subseteq I_{k} (M)

3: Proof

Generally, for any matrix,

A

I_{k} (A)

denotes the sum of the principal ideals by the

k

-dimensional subdeterminants of

A

.

Hereafter, let us denote the

(j, l)

component of any matrix,

A

, as

A_{l}^{j}

; the

l

-th column of

A

A_{l}

; the submatrix of

A

with the

B = {r_{1}, . . ., r_{k}}

rows as

A (B)

M N = (\begin{matrix} \sum_{j_{1} \in {1, . . ., n}} M_{j_{1}} N_{1}^{j_{1}}, . . ., \sum_{j_{m} \in {1, . . ., n}} M_{j_{m}} N_{m}^{j_{m}} \end{matrix})

.

Let us take the

k x k

submatrix of

M N

by the

B := {r_{1}, . . ., r_{k}}

-th rows and the

{c_{1}, . . ., c_{k}}

-columns: it is

(\begin{matrix} \sum_{j_{1} \in {1, . . ., n}} M (B)_{j_{1}} N_{c_{1}}^{j_{1}}, . . ., \sum_{j_{k} \in {1, . . ., n}} M (B)_{j_{k}} N_{c_{k}}^{j_{k}} \end{matrix})

.

Its determinant is

| (\begin{matrix} \sum_{j_{1} \in {1, . . ., n}} M (B)_{j_{1}} N_{c_{1}}^{j_{1}}, . . ., \sum_{j_{k} \in {1, . . ., n}} M (B)_{j_{k}} N_{c_{k}}^{j_{k}} \end{matrix}) | = \sum_{j_{1} \in {1, . . ., n}} N_{c_{1}}^{j_{1}} | (\begin{matrix} M (B)_{j_{1}}, . . ., \sum_{j_{k} \in {1, . . ., n}} M (B)_{j_{k}} N_{c_{k}}^{j_{k}} \end{matrix}) |

(because determinant is linear with respect to each column)

= . . . = \sum_{j_{1} \in {1, . . ., n}} . . . \sum_{j_{k} \in {1, . . ., n}} N_{c_{1}}^{j_{1}} . . . N_{c_{k}}^{j_{k}} | (\begin{matrix} M (B)_{j_{1}}, . . ., M (B)_{j_{k}} \end{matrix}) |

.

When

{j_{1}, . . ., j_{k}}

has a duplication, the term is

0

; otherwise, the term is a multiple of the

k x k

subdeterminant of

M

.

So, each

k x k

subdeterminant of

M N

is a linear combination of some

k x k

subdeterminants of

M

.

So, while each element of

I_{k} (M N)

is a linear combination of the

k x k

subdeterminants of

M N

, it is a linear combination of the

k x k

subdeterminants of

M

, which means that it is an element of

I_{k} (M)

.

The set of the

k x k

subdeterminants of

O M

equals the set of the

k x k

subdeterminants of

(O M)^{t}

, where

(O M)^{t}

is the transposition of

O M

, because the transposition of any square matrix has the same determinant with the original matrix.

(O M)^{t} = M^{t} O^{t}

I_{k} (O M) = I_{k} (M^{t} O^{t}) \subseteq I_{k} (M^{t}) = I_{k} (M)

References

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2024-07-07