2025-06-08

1155: For Commutative Ring, Some Expansions of Determinant of Product of Rectangular Matrices

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description/proof of for commutative ring, some expansions of determinant of product of rectangular matrices

Topics


About: matrices space

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that for any commutative ring, some expansions of the determinant of the product of any rectangular matrices whose product is square hold.

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: { the commutative rings }
M: { the m×nR matrices }
N: { the n×mR matrices }
//

Statements:
det(MN)=l1{1,...,n}...lm{1,...,n}Ml11...Mlmmdet(N1l1...Nml1...N1lm...Nmlm)

det(MN)=l1{1,...,n}...lm{1,...,n}N1l1...Nmlmdet(Ml11...Mlm1...Ml1m...Mlmm)
//


2: Note


The ring needs to be commutative, because "well-known properties of determinant" used in Proof requires it.

The expansions claim that when n<m, the determinant is inevitably 0, because each det(N1l1...Nml1...N1lm...Nmlm) cannot be nonzero because some rows are inevitably duplicated.

For example, when m=2 and n=1, M=(M11M12) and N=(N11N21), and det(MN)=det(M11N11M11N21M12N11M12N21)=M11N11M12N21M12N11M11N21=0: the commutativity required.

As any square matrix is a rectangular matrix, M and N can be m×m matrices.

In that case, also det(MN)=detMdetN holds.

For the mn case, "det(MN)=detMdetN" does not make sense, because each of detM and detN is not defined.


3: Proof


Whole Strategy: Step 1: see the components of MN; Step 2: expand det(MN) with respect to the 1st row, then with respect to the 2nd row, ...; Step 3: use det(MN)=det((MN)t)=det(NtMt) and apply the result of the Step 2.

Step 1:

MN=(l{1,...,n}MljNol)=(l{1,...,n}Ml1N1l...l{1,...,n}Ml1Nml...l{1,...,n}MlmN1l...l{1,...,n}MlmNml).

Step 2:

The 1st row of MN can be written as (l1{1,...,n}Ml11N1l1,...,l1{1,...,n}Ml11Nml1)=l1{1,...,n}(Ml11N1l1,...,Ml11Nml1).

By a well-known property of determinant, det(MN)=l1{1,...,n}det(Ml11N1l1...Ml11Nml1l{1,...,n}Ml2N1l...l{1,...,n}Ml2Nml...l{1,...,n}MlmN1l...l{1,...,n}MlmNml).

By a well-known property of determinant, det(MN)=l1{1,...,n}Ml11det(N1l1...Nml1l{1,...,n}Ml2N1l...l{1,...,n}Ml2Nml...l{1,...,n}MlmN1l...l{1,...,n}MlmNml).

For each of the new determinants, the 2nd row can be written as (l2{1,...,n}Ml22N1l2...l2{1,...,n}Ml22Nml2)=l2{1,...,n}(Ml22N1l2,...,Ml22Nml2).

By a well-known property of determinant, det(N1l1...Nml1l{1,...,n}Ml2N1l...l{1,...,n}Ml2Nml...l{1,...,n}MlmN1l...l{1,...,n}MlmNml)=l2{1,...,n}det(N1l1...Nml1Ml22N1l2...Ml22Nml2l{1,...,n}Ml3N1l...l{1,...,n}Ml3Nml...l{1,...,n}MlmN1l...l{1,...,n}MlmNml).

By a well-known property of determinant, =l2{1,...,n}Ml22det(N1l1...Nml1N1l2...Nml2l{1,...,n}Ml3N1l...l{1,...,n}Ml3Nml...l{1,...,n}MlmN1l...l{1,...,n}MlmNml).

So, det(MN)=l1{1,...,n}Ml11l2{1,...,n}Ml22det(N1l1...Nml1N1l2...Nml2l{1,...,n}Ml3N1l...l{1,...,n}Ml3Nml...l{1,...,n}MlmN1l...l{1,...,n}MlmNml)=l1{1,...,n}l2{1,...,n}Ml11Ml22det(N1l1...Nml1N1l2...Nml2l{1,...,n}Ml3N1l...l{1,...,n}Ml3Nml...l{1,...,n}MlmN1l...l{1,...,n}MlmNml).

And so on, after all, det(MN)=l1{1,...,n}...lm{1,...,n}Ml11...Mlmmdet(N1l1...Nml1...N1lm...Nmlm).

Step 3:

det(MN)=det((MN)t)=det(NtMt).

By the result of Step 2, det(NtMt)=l1{1,...,n}...lm{1,...,n}N1l1...Nmlmdet(Ml11...Ml1m...Mlm1...Mlmm)=l1{1,...,n}...lm{1,...,n}N1l1...Nmlmdet(Ml11...Mlm1...Ml1m...Mlmm).


References


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