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D.V. Ionescu
Institutul de Calcul
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D.V. Ionescu, The canonical form of a determinant and its applications. (Romanian) Acad. R. P. Romîne Fil. Cluj Stud. Cerc. Mat. 10 1959 33–44.
[Forma canonică a unui determinant și aplicațiile sale]
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Studii si Cercetari Matematice Cluj
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THE CANONICAL FORM OF A DETERMINANT AND ITS APPLICATIONS
OF
It is known that any matrix with elements from a given field can be reduced to a canonical form by conveniently repeated elementary transformations. The elementary transformations are the following:
T_(1): one line is replaced by another line;
T_(2): multiply the elements of one line by a factor and add them to the elements of another line.
T_(1): one line is replaced by another line;
T_(2): multiply the elements of one line by a factor and add them to the elements of another line.
By changing the word "line" in the previous definitions with the word "column", we obtain the transformationsT_(1)^('),T_(2)^(').
In particular, if the matrix is ​​square
A=||{:[a_(11),a_(12),dots,a_(1n)],[a_(21),a_(22),dots,a_(2n)],[*,*,dots,*],[a_(n1),a_(n2),dots,a_(nn)]:}||
its canonical form is
A^(**)=||[[a_(1),,,,],[,a_(2),,,],[,,ddots,,],[,,,a_(j),],[,,,0,],[0,,,ddots,],[,,,,0]]
with all elements, except the diagonal, equal to zero.
We will call the canonical form of the determinantD=|A|, the determinantD^(**)=|A^(**)|.
We will call the canonical form of the determinantD=|A|, the determinantD^(**)=|A^(**)|.
In general we haveD=D^(**)orD=-D^(**), from which it follows:
1^(@)IfD!=0, then the canonical form ofDis
1^(@)IfD!=0, then the canonical form ofDis
3 - Mathematics studies and research
D^(**)=|{:[a_(1),,,0],[,a_(2),,],[,,ddots,],[,0,,a_(n)]:}|
with all diagonal elements nonzero, and vice versa.
2^(@)IfD=0, then the canonical form ofDis
2^(@)IfD=0, then the canonical form ofDis
D^(**)=||[[a_(1),,,,,0],[,a_(2),,,,0],[,,ddots,,,],[,,,a_(j),,],[0,,,,0,],[,,,,0]]
with at least one element of the diagonal equal to zero, and vice versa.
In this paper we will make applications of the canonical form of a determinant. Although we have done several, we will limit ourselves to only two, namely:
1^(@). proving the well-known formula that relates the reciprocal determinant to a given order of the determinantD, by the determinantD;
2^(@). proving the well-known Sylvester identity.
The proofs we will make will be based on the following idea:
Let us assume that we have to calculate a determinantDeltawhich corresponds to a determinantDIf an invariant can be highlightedf(D,Delta)for any elementary transformationT_(1),T_(2),T_(1)^('),T_(2)^('), then we will have
In this paper we will make applications of the canonical form of a determinant. Although we have done several, we will limit ourselves to only two, namely:
1^(@). proving the well-known formula that relates the reciprocal determinant to a given order of the determinantD, by the determinantD;
2^(@). proving the well-known Sylvester identity.
The proofs we will make will be based on the following idea:
Let us assume that we have to calculate a determinantDeltawhich corresponds to a determinantDIf an invariant can be highlightedf(D,Delta)for any elementary transformationT_(1),T_(2),T_(1)^('),T_(2)^('), then we will have
f(D,Delta)=f(D^(**),Delta^(**))
whereDelta^(**)is the determinant that corresponds to the canonical formD^(**)of the determinant D. If the determinantDelta^(**)is calculated directly and easily fromD^(**), then the previous formula will give usDelta.
§ 1. The reciprocal determinant of the orderjof a determinant
Whether
{:(1)D=|{:[a_(11),a_(12),dots,a_(1n)],[a_(21),a_(22),dots,a_(2n)],[*,*,dots,*],[a_(n1),a_(n2),dots,a_(nn)]:}|:}
an arbitrary determinant and consider its minors of the orderj.
{:(2)a({:[i_(1)",",i_(2)",",dots","],[k_(1)",",k_(2)",",dots","],[k_(j)]:}):}
formed with the elements of the determinantD, common to the lines of ranki_(1),i_(2),dots dots,i_(j)and rank columnsk_(1),k_(2),dots,k_(j). The groups (i_(1),i_(2),dots,i_(j)),(k_(1),k_(2),dots,k_(j))are made withjindices among indices1,2,dots,nThe number of all
minors of the form (2) is((n)/(j))^(2)We will arrange these minors in a determinantDelta_(j)with((n)/(j))lines and((n)/(j))columns in the following way. We arrange all the groupings ofjINDICATORS(alpha_(1),alpha_(2),dots,alpha_(j))take from the clues1,2,dots,nin a rowS, so that(alpha_(1),alpha_(2),dots,alpha_(j)),(alpha_(1)^('),alpha_(2)^('),dots,alpha_(j)^('))being two consecutive terms of the sequence, let us havealpha_(h) <= alpha_(h)^(')forh=1,2,dots,j.
minors of the form (2) is((n)/(j))^(2)We will arrange these minors in a determinantDelta_(j)with((n)/(j))lines and((n)/(j))columns in the following way. We arrange all the groupings ofjINDICATORS(alpha_(1),alpha_(2),dots,alpha_(j))take from the clues1,2,dots,nin a rowS, so that(alpha_(1),alpha_(2),dots,alpha_(j)),(alpha_(1)^('),alpha_(2)^('),dots,alpha_(j)^('))being two consecutive terms of the sequence, let us havealpha_(h) <= alpha_(h)^(')forh=1,2,dots,j.
We will form the determinantDelta_(j), making the groups (i_(1),i_(2),dots,i_(j)) and(k_(1),k_(2),dots,k_(j))to go through all the terms of the sequenceS.
It is noted
{:(3)Delta_(j)=|a({:[i_(1)",",i_(2)",",dots","],[k_(1)",",k_(2)",",dots","],[k_(j),]:})|:}
reciprocal determinant of the orderj, hisDThese determinants were first considered by Cauchy. It was shown that
{:(4)Delta_(j)=D^(((n-1)/(j-1))):}
We will give the proof of this formula using the canonical form of the determinantD.
Paying attention to the first two lines of the determinantD, we will write the determinantDelta_(j)in the form of
Delta_(j)={:[ dots dots dots dots dots dots],[a((1,2,i_(3),dots,i_(j))/(k_(1),k_(2),k_(3),dots,k_(j)))],[ dots dots dots dots dots dots dots],[a((1,i_(2),i_(3),dots,i_(j))/(k_(1),k_(2),k_(3),dots,k_(j)))],[ dots dots dots dots dots dots","],[a((2,i_(2),i_(3),dots,)/(k_(1),k_(2),k_(3),dots,k_(j)))],[ dots dots dots dots dots dots dots],[a({:[i_(1)","i_(2)","i_(3)","dots","],[k_(1)","k_(2)","k_(3)","dots","k_(j)],[dots dots dots dots dots dots","dots]:})]:}
in which the elements of the column (k_(1),k_(2),dots,k_(j)).
In the first lines of the determinantDelta_(j)s-they highlighted the lines with minors of the orderjof the determinantDcontaining the first two lines;(i_(3),i_(4),dots,i_(j))being a group of some kind withj-2indices taken from indices3,4,dots,n, the number of these lines is((n-2)/(j-2)).
In the first lines of the determinantDelta_(j)s-they highlighted the lines with minors of the orderjof the determinantDcontaining the first two lines;(i_(3),i_(4),dots,i_(j))being a group of some kind withj-2indices taken from indices3,4,dots,n, the number of these lines is((n-2)/(j-2)).
In the following lines, the minors in the determinant were highlighted.D, withjlines of which the first is the line1-aand then the line a2-afrom the determinantD;(i_(2),dots,i_(j))is a group withj-1indices taken from
indices(3,4,dots,n), the number of these linesepsilonyou are((n-2)/(j-1)). Finally, in the following lines (i_(1),i_(2)dots,i_(j)) is a grouping withjindices taken from indices3,4,dots,nThe number of these lines is((n-2)/(j))It is easy to verify that the number of all lines is
indices(3,4,dots,n), the number of these linesepsilonyou are((n-2)/(j-1)). Finally, in the following lines (i_(1),i_(2)dots,i_(j)) is a grouping withjindices taken from indices3,4,dots,nThe number of these lines is((n-2)/(j))It is easy to verify that the number of all lines is
((n-2)/(j-2))+((n-2)/(j-1))+((n-2)/(j-1))+((n-2)/(j))=((n)/(j)).
If in the determinantDchange the first line with the second, we obtain the determinant
bar(D)=|{:[a_(21),a_(22),dots,a_(2n)],[a_(11),a_(12),dots,a_(1n)],[dots,dots,dots,dots],[a_(n1),a_(n2),dots,a_(nn)]:}|=-D
and the reciprocal determinant of the orderjcorresponding tobar(D)is
Taking into account that
a({:[2",",1","i_(3)",",dots",",i_(j)],[k_(1)",",k_(2)",",k_(3)",",dots","],[k_(j)]:})=-a({:[1","],[k_(1)","],[k_(2)","],[k_(3)","]:},dots,i_(j),dots,k_(j))
and permuting the line marked bya((1,i_(2),dots,i_(j))/(k_(1),k_(2),dots,k_(j)))with the line marked bya({:[2",",i_(2)",",dots","],[k_(1)",",k_(2),dots","]:}), bar(Delta)_(j)changing the sign at each permutation, we deduce that
bar(Delta)_(j)=(-1)^(((n-1)/(j-1)))Delta_(j)
becausebar(Delta_(j))change the sign of((n-2)/(j-1))times, you can take out -1 factor on the firstn-2lines and
((n-2)/(j-2))+((n-2)/(j-1))=((n-1)/(j-1))
It is shown analogously that if in the determinantDtwo random rows or two random columns are swapped, we have the formulas
{:(5) bar(D)=-D","quad bar(Delta)_(j)=(-1)^(((n-1)/(j-1)))Delta_(j):}
urd was noted withbar(D)what becomesDthrough the change made and then throughbar(Delta), the reciprocal determinant of the orderjhis/herbar(D).
In the determinantD, let's add to the elements of the 1st line, the elements of the line a2-multiplied bylambda, that is, let us consider the determinant
bar(bar(D))=|{:[a_(11)+lambdaa_(21),a_(12)+lambdaa_(22),dots,a_(1n)+lambdaa_(2n)],[a_(21),a_(22),dots,a_(2n)],[a_(n1),a_(n2),dots,.],[.,dots,a_(nn)]:}|
The reciprocal determinant of the orderjhis/herbar(bar(D))is
When 1 gives the elements of any line ofD, the elements of another row multiplied by a certain number are added, or to the elements of a certain column the elements of another column multiplied by a certain number are added, we have the formulas
{:(6) bar(bar(D))=D","quad bar(bar(Delta))_(j)=Delta_(j):}
where was it noted withbar(bar(D))what becomesDthrough the transformation made and then throughbar(bar(Delta)), the reciprocal determinant of the orderjhis/herbar(bar(D)).
WhetherD^(**)the canonical determinant ofDandDelta_(j)^(**)reciprocal determinant of the orderjhis/herD^(**).
Because the determinantD^(**)is obtained fromDthrough transformationsT_(1),T_(2),T_(1)^('),T_(2)^(')conveniently repeated and we have formulas (4), (5), the determinantDelta_(j)^(**)will be equal toDelta_(j)or will it differ fromDelta_(j)by sign.
IfD=0, in canonical formD^(**)there will be at least one zero on the main diagonal, andDelta_(j^(⃛))^(j)𝒿will also have at least one zero element on the main diagonal, that is, we will haveDelta_(j)^(**)=0It follows therefore that ifD=0, then we will also haveDelta_(j)=0.
IfD!=0, its canonical form is
D^(**)=|{:[a_(1),,0],[,a_(2),],[0,ddots,],[0,,a_(n)]:}|=a_(1)a_(2)dotsa_(n)
determinantDelta_(3)^(j^(⃛)), corresponding to 1uiD^(**), it is
{:(7)Delta_(j)^(**)=|{:[a_(1)a_(2)dotsa_(j),,,],[a_(1)a_(3)dotsa_(j+1),,],[,ddots,,0],[0,ddots,],[,,ddots,a_(n-j+1)a_(n-j+2)dotsa_(n)]:}|=(a_(1)a_(2)dotsa_(n))^(((n-1)/(j-1)))=(D^(**))^(((n-1)/(j-1))):}
Formulas (5) and (6) show that
( bar(Delta)_(j))/((( bar(D)))^(((n-1)/(j-1))))=(Delta_(j))/(D^(((n-1)/(j-1))))quad;quad( bar(bar(Delta))_(j))/(( bar(bar(D)))^(((n-1)/(j-1))))=(Delta_(j))/(D^(((n-1)/(j-1))))
and therefore how much(Delta_(j))/(D^(((n-1)/(j-1))))is invariant for transformationsT_(1),T_(2),T_(1)^('),T_(2)^('). So we will have
{:(8)(Delta_(j))/(D^(((n-1)/(j-1))))=(Delta_(j)^(**))/((D^(**))^(((n-1)/(j-1)))):}
But formula (7) shows that the second member of formula (8) is 1, which results in formula (4), which is also valid forD=0, as shown above.
§ 2. Sylvester's Identity
Let's consider the determinants
{:(9)D=|{:[a_(11),a_(12),dots,a_(1n)],[a_(21),a_(22),dots,a_(2n)],[*,*,dots,],[a_(n1),a_(n2),dots,a_(nn)]:}|","quad B=|{:[b_(11),b_(12),dots,b_(1m)],[b_(21),b_(22),dots,b_(2m)],[*,*,dots,*],[b_(m1),b_(m2),dots,b_(mm)]:}|:}
and the elements
{:(10)C_(ik)=|{:[,,,{:[q_(1k)],[.],[.],[D]:}],[p_(i1),dots,p_(in),b_(ik)]:}|:}
wherei,k=1,2,dots,mWe want to calculate the determinant
{:(11)C=|{:[c_(11),c_(12),dots,c_(1m)],[c_(21),c_(22),dots,c_(2m)],[*,*,*,*],[c_(m1),c_(m2),*,*],[*,c_(mm)]:}|:}
and prove Sylvester's identity.
{:(12)C=D^(m-1)|{:[a_(11),dots,a_(1n),q_(11),dots,q_(1m)],[*,,*,*,,*],[*,,*,*,,*],[*,,*,*,,*],[a_(n1),dots,a_(nn),q_(n1),dots,q_(nm)],[p_(11),dots,p_(1n),b_(11),dots,b_(1m)],[*,,*,*,,*],[*,,*,*,,*],[p_(m_(1)),dots,p_(mn),b_(m_(1)),dots,b_(mm)]:}|:}
We will give a proof based on reducing a determinant to its canonical form.
Whetherbar(D)the determinant obtained fromDby swapping two lines between them, andbar(C)the determinant formed with the elementsbar(c_(ik))obtained by making in the determinantc_(ik)the same line changes as in the determinantDWe have
{:(13) bar(D)=-D","quad bar(c_(ik))=-c_(ik)","quad bar(C)=(-1)^(m)C:}
and these formulas are also valid if in the determinantDtwo columns are swapped with each other.
Whetherbar(bar(D))the determinant obtaineddin D, adding to the elements of one line the elements of another line multiplied by the factorlambdaWe denote bybar(bar(C))the determinant formed with the elementsbar(c_(ik))obtained by making in the determinantc_(ik)the same transformation as in the determinantDWe will have
{:(14) bar(bar(D))=D","quad bar(bar(c_(ik)))=c_(ik)","quad bar(bar(C))=C:}
these formulas being valid even if in the determinantDare added to the elements of one column, the elements of another column multiplied by a factormu.
WhetherD^(**)canonical form of the determinantD, that is,
and
and
{:[D^(**)=|{:[a_(1),,,,0],[,a_(2),,,,0],[,,ddots,,,],[,,,a_(j),,],[0,,,,ddots,],[,,,,,0]:}|],[c_(ik)^(**)=|{:[,,,,|{:[q_(1k)^(**)],[vdots],[]:}],[,D^(**),,],[,,,,q_(nk)^(**)],[p_(11)^(**),dots,.,p_(in)^(**),b_(ik)]:}|]:}
hatep_(ij^(¨))^(**)andq_(k^(¨))^(**)are the elements deduced from the elementsp_(i1),dots,p_(in)andq_(1k),dots,q_(k,k)the transformations that bringDto its canonical formD^(**)We have
C^(**)=|{:[c_(i1)^(**),cdots,c_(1,n)^(**)],[*,,*],[*,,*],[*,,*],[c_(m1)^(**),cdots,c_(mm)^(**)]:}|
Ifj <= n-2, then it is immediately seen - expanding the determinantc_(c_(k))^(**)by the elements of the penultimate column - thatc_(i_(k)^(**))^(**)=0, and thereforeC^(**)=0.
But ifj=n-1, butm > 1, then expanding the determinant sik by the elements of the penultimate column, we obtain
and therefore
and therefore
c_(ik)^(**)=-a_(1)a_(2)dotsa_(n-1)p_(in)^(**)q_(nk)^(**)
C^(**)=(-a_(1)a_(2)dotsa_(n-1))^(m)|{:[p_(1n)^(**)q_(n1)^(**),p_(1n)^(**)q_(n2)^(**),dots,p_(1n)^(**)q_(nm)^(**)],[p_(2n)^(**)q_(n1)^(**),p_(2n)^(**)q_(n2)^(**),dots,p_(n2)^(**)q_(nm)^(**)],[*,*,dots,*],[p_(mn)^(**)q_(n1)^(**),p_(mn)^(**)q_(n2)^(**),dots,p_(mn)^(**)q_(nm)^(**)]:}|=0.
So we have shown that ifD=0, we also haveC^(**)=0, provided that in the case whenj=n-1to havem > 1From formulas (13) and (14) it follows that the determinantC^(**)is equal toCor differs fromCby the sign. It follows that if we haveD=0, we also haveC=0(the case whenj=n-1andm=1, reserved for later).
Let's assumeD!=0, and letD^(**)its canonical form
D^(**)=|{:[a_(1),,,0],[,a_(2),,],[,,ddots,],[0,,a_(n)]:}|;
ELEMENTSc_(ik)^(**)are appropriate
c_(ik)^(**)=|{:[a_(1),,,0],[,a_(2),,],[,,,vdots],[,,ddots,],[0,,,a_(n)^(**)],[q_(ki)^(**),p_(ki)^(**),dots,q_(in)^(**)],[q_(kik)^(**)]:}|
Developing this determinant, we take the elements of its last column, and continuing ona_(1)a_(2)dotsa_(n)=D^(**)in the factor, we have
c_(ik)^(**)=D^(**)(b_(ik)-(p_(i1)^(**)q_(1k)^(**))/(a_(1))-(p_(i2)^(**))/(a_(2))(q_(2k)^(**))/(a_(2)^(**))-dots-(p_(in)^(**))/(a_(n))q_(nk)^(**))=D^(**)b_(ik)^(').
determinantC^(**)can be written in the form
C^(**)=(D^(**))^(m)|{:[1,0,dots,0,(q_(11)^(**))/(a_(1)),(q_(12)^(**))/(a_(1)),dots,(q_(1m)^(**))/(a_(1))],[0,1,dots,0,(q_(21)^(**))/(a_(2)),(q_(22)^(**))/(a_(2)),dots,(q_(2m)^(**))/(a_(2))],[.,.,dots,.,.,.,.,dots],[0,0,dots,1,(q_(n1)^(**))/(a_(n)),(q_(n2)^(**))/(a_(n)),dots,(q_(nm)^(**))/(a_(n))],[0,0,dots,0,0,b_(11)^('),b_(12)^('),dots],[0,0,dots,0,b_(21)^('),b_(22)^('),dots,b_(1m)^(')],[.,.,dots,.,.,b_(2m)^('),.,.]:}
Multiplying the elements of the 1st, 2nd lines,dots, ofn-a withp_(11)^(**),p_(12)^(**),dots,p_(1n)^(**)and adding to the elements of the line a(n+1)-a, these become
p_(11)^(**),p_(12)^(**),dots,p_(1n)^(**),b_(11),b_(12),dots,b_(1m).
Doing analogous operations for rank linesn+2,n+3,dots,n+mget
C^(**)=(D^(**))^(m)|{:[1,0,dots,0,(q_(11)^(**))/(a_(1)),(q_(12)^(**))/(a_(1)),dots,(q_(1m)^(**))/(a_(1))],[0,1,dots,0,(q_(21)^(**))/(a_(2)),(q_(22)^(**))/(a_(2)),dots,(q_(2m)^(**))/(a_(2))],[.,.,dots,.,.,.,.,dots],[0,0,dots,1,(q_(n1)^(**))/(a_(n)),(q_(n2)^(**))/(a_(n)),dots,.],[p_(11)^(**),p_(12)^(**),dots,p_(1n)^(**),b_(11),b_(12),dots,dots],[p_(21)^(**),p_(22)^(**),dots,p_(2n)^(**),b_(21)^(**),b_(22),dots,b_(1m)],[.,.,dots,.,.,.,dots,b_(2m)],[p_(m1)^(**),p_(m2)^(**),dots,p_(mn)^(**),b_(m1),b_(m2),dots,b_(mm)]:}|
or
{:(15)C^(**)=(D^(**))^(m-1)|{:[a_(1),0,dots,0,q_(11)^(**),q_(12)^(**),dots,q_(1m)^(**)],[0,a_(2),dots,0,q_(21)^(**),q_(22)^(**),dots,q_(2m)^(**)],[.,.,dots,.,.,.,dots,.],[0,0,dots,a,q_(n1)^(**),q_(n2)^(**),dots,q_(nm)^(**)],[p_(11)^(**),p_(12)^(**),dots,p_(1n)^(**),b_(11),b_(12),dots,b_(1m)],[p_(21)^(**),p_(22)^(**),dots,p_(2n)^(**),b_(21),b_(22),dots,b_(2m)],[.,.,dots,.,.,.,dots,.],[p_(m1)^(**),p_(m2)^(**),dots,p_(mn)^(**),b_(m1),b_(m2),dots,b_(mm)]:}|:}
From formulas (13) and (14) it follows that
(( bar(C)))/((( bar(D)))^(m))=(C)/(D^(m))quad"şi"quad( bar(bar(C)))/(( bar(bar(D)))^(m))=(C)/(D^(m))ş
MEAN(C)/(D^(m))is an invariant for the transformationsT_(1),T_(2),T_(1)^('),T_(2)^(')We will then have
(C)/(D^(m))=(C^(**))/((D^(**))^(m))
and taking into account formula (15), we deduce that
{:(16)(C)/(D^(m))=(Delta^(**))/(D^(**)):}
whereDelta^(**)is the determinant in the second member of formula (15).
Let us now consider the determinant
Delta=|{:[a_(11),a_(12),dots,a_(1n),q_(11),q_(12),dots,q_(1m)],[a_(21),a_(22),dots,a_(2n),q_(21),q_(22),dots,q_(2m)],[.,.,dots,.,.,.,dots,.],[a_(n1),a_(n2),dots,a_(nn),q_(n1),q_(n2),dots,q_(nm)],[p_(11),p_(12),dots,p_(1n),b_(11),b_(12),dots,b_(1m)],[p_(21),p_(22),dots,p_(2n),b_(21),b_(22),dots,b_(2m)],[.,.,dots,.,.,.,dots,.],[p_(m1),p_(m2),dots,p_(mn),b_(m1),b_(m2),dots,b_(mm)]:}|
Whetherbar(D)the determinant which is obtained by changing toDtwo lines or two columns between them. We denote bybar(Delta)the determinant which is obtained by making the same transformation as inDon rows or columns. We will have
bar(D)=-D,quad bar(Delta)=-Delta
Either nowbar(bar(D))the determinant obtained by adding to the elements of a line ofD, the elements of another line multiplied by a factorlambda, or which is obtained by adding to the elements of a column ofD, the elements of another column multiplied by a factormuWe notecu bar(bar(Delta))the determinant that is obtained fromDeltamaking the same transformation as inD, on rows or on columns. We will have
bar(D)=D,quad bar(bar(Delta))=Delta
From these formulas it follows that
(( bar(Delta)))/(( bar(D)))=(Delta )/(D),quad(( bar(Delta)))/(( bar(D)))=(Delta )/(D)
MEAN(Delta )/(D)is an invariant for the transformationsT_(1),T_(2),T_(1)^('),T_(2)^(')It
follows that
follows that
{:(17)(Delta )/(D)=(Delta^(**))/(D^(**)):}
whereD^(**)is the canonical form ofD, andDelta^(**)is the determinant in the second member of formula (15).
Formulas (16) and (17) then show us that
(C)/(D^(m))=(Delta )/(D)
from which it follows that
C=D^(m-1)Delta
and with this, Sylvester's identity (12) is proven.
The casej=n-1,m=1is trivial, the determinantCis reduced to a single elementc_(11)Sylvester's formula (12) is in this case a trivial identity, the factorD^(m-1)from the second member which appears as0^(@)should be considered equal to 1.
The casej=n-1,m=1is trivial, the determinantCis reduced to a single elementc_(11)Sylvester's formula (12) is in this case a trivial identity, the factorD^(m-1)from the second member which appears as0^(@)should be considered equal to 1.
THE CANONICAL FORM OF ONE DETERMINER AND ITS APPLICATIONS
(Brief summary)
It is known that elementary transformationsT_(1),T_(2)andT_(1)^('),T_(2)^('), produced over lines and columns, matrixA=||a_(i)^(k)||_(1)^(n)it is reduced to the canonical formA^(**). We call it the canonical form of the determinantD=|A|determinerD^(**)=|A^(**)|.
In this work, the application of the canonical form of the determinant is given, and the formula (4) for the mutual determinant is provedDelta(order)j) determinantD. Доказываться также тождество (12) Sylvester.
Proofs are based on the following idea:
Let it be required to calculate the determinantDelta, corresponding to some determinantD. If the invariant is foundf(D,Delta)transformedT_(1),T_(2),T_(1)^('),T_(2)^('), then from equality
Let it be required to calculate the determinantDelta, corresponding to some determinantD. If the invariant is foundf(D,Delta)transformedT_(1),T_(2),T_(1)^('),T_(2)^('), then from equality
f(D,Delta)=f(D^(**),Delta^(**))
is displayedDelta, becauseDelta^(**)it's easy to calculate.
LA FORME CANONIQUE D'UN DÉTERMINANT FIT SES APPLICATIONS
(Résumé)
On sait que par des transformations élémentairesT_(1),T_(2)andT_(1)^('),T_(2)^('), effected on the lines and columns of a matrixA=||a_(ik)||_(1)^(n), on ramène celle-ci à la forme canoniqueA^(**). On appelle forme canonique du determinantD=|A|, determining themD^(**)=|A^(**)|.
(Résumé)
On sait que par des transformations élémentairesT_(1),T_(2)andT_(1)^('),T_(2)^('), effected on the lines and columns of a matrixA=||a_(ik)||_(1)^(n), on ramène celle-ci à la forme canoniqueA^(**). On appelle forme canonique du determinantD=|A|, determining themD^(**)=|A^(**)|.
Dans ce travail, on fait des applications de la forme canonique d'un determinant, en démontrant la formulae (4) pour le réciproque determinantDelta, by orderj, of a determinantD, et en démontrant l'identité (12) by Sylvester.
The demonstrations given are based on the following idea:
Supposons que nous ayons à calculer un determinantDeltaqui correspond à un determinantDany And l'on peut mettre en évidence an invariantf(D,Delta)for the transformationsT_(1),T_(2),T^(')_(1),T^(')_(2), then equality
Supposons que nous ayons à calculer un determinantDeltaqui correspond à un determinantDany And l'on peut mettre en évidence an invariantf(D,Delta)for the transformationsT_(1),T_(2),T^(')_(1),T^(')_(2), then equality
f(D,Delta)=f(D^(**),Delta^(**))
we can shootDelta, easily calculatingDelta^(**).