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Please help me understand the following notation



The 2019 Stack Overflow Developer Survey Results Are InWhy not use the identity matrix instead of the Kronecker delta?Matrix Transformation Onto?Need help understanding matrix norm notationDeterminants and the following questionLinear Algebra matrix notationUnderstanding an eigen decomposition notationHelp Determinant Binary MatrixStrange math notation, vertical bar with parentheses.Confused about diagonal matix notationInterpretation of Einstein notation for matrix multiplication










2












$begingroup$


Can someone kindly tell me the meaning of the following notation:



A book defined the following matrix $(a_ij)_3times 3$ :



$a_ij=begincases d_ij& ineq j\d_ii+sum_j=1^3 d_ij&i=jendcases$



where $d_ij$ are elements defined for $1le i,jle 3$.



I dont understand the case for $i=j$



Does it mean $d_11=d_11+d_12+d_13$?






linear-algebra matrices proof-verification proof-explanation







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    What is k in this problem? You haven't defined it.
    $endgroup$
    – Don Thousand
    Mar 23 at 3:57






  • 2




    $begingroup$
    Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
    $endgroup$
    – Theo Bendit
    Mar 23 at 3:58










  • $begingroup$
    @TheoBendit,i have edited the question,can you help now
    $endgroup$
    – Math_Freak
    Mar 23 at 4:09










  • $begingroup$
    @DonThousand,can you help now,i have edited it
    $endgroup$
    – Math_Freak
    Mar 23 at 4:09










  • $begingroup$
    Why the downvotes
    $endgroup$
    – Math_Freak
    Mar 23 at 4:22















2












$begingroup$


Can someone kindly tell me the meaning of the following notation:



A book defined the following matrix $(a_ij)_3times 3$ :



$a_ij=begincases d_ij& ineq j\d_ii+sum_j=1^3 d_ij&i=jendcases$



where $d_ij$ are elements defined for $1le i,jle 3$.



I dont understand the case for $i=j$



Does it mean $d_11=d_11+d_12+d_13$?






linear-algebra matrices proof-verification proof-explanation







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    What is k in this problem? You haven't defined it.
    $endgroup$
    – Don Thousand
    Mar 23 at 3:57






  • 2




    $begingroup$
    Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
    $endgroup$
    – Theo Bendit
    Mar 23 at 3:58










  • $begingroup$
    @TheoBendit,i have edited the question,can you help now
    $endgroup$
    – Math_Freak
    Mar 23 at 4:09










  • $begingroup$
    @DonThousand,can you help now,i have edited it
    $endgroup$
    – Math_Freak
    Mar 23 at 4:09










  • $begingroup$
    Why the downvotes
    $endgroup$
    – Math_Freak
    Mar 23 at 4:22













2












2








2


0



$begingroup$


Can someone kindly tell me the meaning of the following notation:



A book defined the following matrix $(a_ij)_3times 3$ :



$a_ij=begincases d_ij& ineq j\d_ii+sum_j=1^3 d_ij&i=jendcases$



where $d_ij$ are elements defined for $1le i,jle 3$.



I dont understand the case for $i=j$



Does it mean $d_11=d_11+d_12+d_13$?






linear-algebra matrices proof-verification proof-explanation







share|cite|improve this question











$endgroup$




Can someone kindly tell me the meaning of the following notation:



A book defined the following matrix $(a_ij)_3times 3$ :



$a_ij=begincases d_ij& ineq j\d_ii+sum_j=1^3 d_ij&i=jendcases$



where $d_ij$ are elements defined for $1le i,jle 3$.



I dont understand the case for $i=j$



Does it mean $d_11=d_11+d_12+d_13$?






linear-algebra matrices proof-verification proof-explanation




linear-algebra matrices proof-verification proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 5:25









David G. Stork

12.2k41836




12.2k41836










asked Mar 23 at 3:55









Math_FreakMath_Freak

535




535







  • 1




    $begingroup$
    What is k in this problem? You haven't defined it.
    $endgroup$
    – Don Thousand
    Mar 23 at 3:57






  • 2




    $begingroup$
    Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
    $endgroup$
    – Theo Bendit
    Mar 23 at 3:58










  • $begingroup$
    @TheoBendit,i have edited the question,can you help now
    $endgroup$
    – Math_Freak
    Mar 23 at 4:09










  • $begingroup$
    @DonThousand,can you help now,i have edited it
    $endgroup$
    – Math_Freak
    Mar 23 at 4:09










  • $begingroup$
    Why the downvotes
    $endgroup$
    – Math_Freak
    Mar 23 at 4:22












  • 1




    $begingroup$
    What is k in this problem? You haven't defined it.
    $endgroup$
    – Don Thousand
    Mar 23 at 3:57






  • 2




    $begingroup$
    Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
    $endgroup$
    – Theo Bendit
    Mar 23 at 3:58










  • $begingroup$
    @TheoBendit,i have edited the question,can you help now
    $endgroup$
    – Math_Freak
    Mar 23 at 4:09










  • $begingroup$
    @DonThousand,can you help now,i have edited it
    $endgroup$
    – Math_Freak
    Mar 23 at 4:09










  • $begingroup$
    Why the downvotes
    $endgroup$
    – Math_Freak
    Mar 23 at 4:22







1




1




$begingroup$
What is k in this problem? You haven't defined it.
$endgroup$
– Don Thousand
Mar 23 at 3:57




$begingroup$
What is k in this problem? You haven't defined it.
$endgroup$
– Don Thousand
Mar 23 at 3:57




2




2




$begingroup$
Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
$endgroup$
– Theo Bendit
Mar 23 at 3:58




$begingroup$
Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
$endgroup$
– Theo Bendit
Mar 23 at 3:58












$begingroup$
@TheoBendit,i have edited the question,can you help now
$endgroup$
– Math_Freak
Mar 23 at 4:09




$begingroup$
@TheoBendit,i have edited the question,can you help now
$endgroup$
– Math_Freak
Mar 23 at 4:09












$begingroup$
@DonThousand,can you help now,i have edited it
$endgroup$
– Math_Freak
Mar 23 at 4:09




$begingroup$
@DonThousand,can you help now,i have edited it
$endgroup$
– Math_Freak
Mar 23 at 4:09












$begingroup$
Why the downvotes
$endgroup$
– Math_Freak
Mar 23 at 4:22




$begingroup$
Why the downvotes
$endgroup$
– Math_Freak
Mar 23 at 4:22










3 Answers
3






active

oldest

votes


















1












$begingroup$

I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.






share|cite|improve this answer









$endgroup$




















    0












    $begingroup$

    beginaligned
    a_11 & = d_11 + d _11 + d_12 + d _13 \
    a_12 & = d_12 \
    a_13 & = d_13 \
    a_21 & = d_21 \
    a_22 & = d_22 + d _21 + d_22 + d _23 \
    a_23 & = d_23 \
    a_31 & = d_31 \
    a_32 & = d_32 \
    a_33 & = d_33 + d _31 + d_32 + d _33
    endaligned






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$


      We have for $1leq i,jleq 3$



      beginalign*
      a_ij&=
      begincases
      d_ijqquadqquadqquadqquad ineq j\
      d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
      endcasestag1\
      &=
      begincases
      d_ijqquadqquadqquadqquad ineq j\
      d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
      endcasestag2\
      &=
      begincases
      d_ijqquadqquadqquadqquad ineq j\
      d_ii+sum_k=1^3 d_ikquadqquad i=j
      endcases\
      endalign*



      These variants are all valid and represent the same.



      • In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.


      • The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.







      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.






        share|cite|improve this answer









        $endgroup$

















          1












          $begingroup$

          I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.






          share|cite|improve this answer









          $endgroup$















            1












            1








            1





            $begingroup$

            I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.






            share|cite|improve this answer









            $endgroup$



            I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 23 at 5:18









            John OmielanJohn Omielan

            4,8962216




            4,8962216





















                0












                $begingroup$

                beginaligned
                a_11 & = d_11 + d _11 + d_12 + d _13 \
                a_12 & = d_12 \
                a_13 & = d_13 \
                a_21 & = d_21 \
                a_22 & = d_22 + d _21 + d_22 + d _23 \
                a_23 & = d_23 \
                a_31 & = d_31 \
                a_32 & = d_32 \
                a_33 & = d_33 + d _31 + d_32 + d _33
                endaligned






                share|cite|improve this answer









                $endgroup$

















                  0












                  $begingroup$

                  beginaligned
                  a_11 & = d_11 + d _11 + d_12 + d _13 \
                  a_12 & = d_12 \
                  a_13 & = d_13 \
                  a_21 & = d_21 \
                  a_22 & = d_22 + d _21 + d_22 + d _23 \
                  a_23 & = d_23 \
                  a_31 & = d_31 \
                  a_32 & = d_32 \
                  a_33 & = d_33 + d _31 + d_32 + d _33
                  endaligned






                  share|cite|improve this answer









                  $endgroup$















                    0












                    0








                    0





                    $begingroup$

                    beginaligned
                    a_11 & = d_11 + d _11 + d_12 + d _13 \
                    a_12 & = d_12 \
                    a_13 & = d_13 \
                    a_21 & = d_21 \
                    a_22 & = d_22 + d _21 + d_22 + d _23 \
                    a_23 & = d_23 \
                    a_31 & = d_31 \
                    a_32 & = d_32 \
                    a_33 & = d_33 + d _31 + d_32 + d _33
                    endaligned






                    share|cite|improve this answer









                    $endgroup$



                    beginaligned
                    a_11 & = d_11 + d _11 + d_12 + d _13 \
                    a_12 & = d_12 \
                    a_13 & = d_13 \
                    a_21 & = d_21 \
                    a_22 & = d_22 + d _21 + d_22 + d _23 \
                    a_23 & = d_23 \
                    a_31 & = d_31 \
                    a_32 & = d_32 \
                    a_33 & = d_33 + d _31 + d_32 + d _33
                    endaligned







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 23 at 5:44









                    MarianDMarianD

                    2,2611618




                    2,2611618





















                        0












                        $begingroup$


                        We have for $1leq i,jleq 3$



                        beginalign*
                        a_ij&=
                        begincases
                        d_ijqquadqquadqquadqquad ineq j\
                        d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
                        endcasestag1\
                        &=
                        begincases
                        d_ijqquadqquadqquadqquad ineq j\
                        d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
                        endcasestag2\
                        &=
                        begincases
                        d_ijqquadqquadqquadqquad ineq j\
                        d_ii+sum_k=1^3 d_ikquadqquad i=j
                        endcases\
                        endalign*



                        These variants are all valid and represent the same.



                        • In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.


                        • The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.







                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$


                          We have for $1leq i,jleq 3$



                          beginalign*
                          a_ij&=
                          begincases
                          d_ijqquadqquadqquadqquad ineq j\
                          d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
                          endcasestag1\
                          &=
                          begincases
                          d_ijqquadqquadqquadqquad ineq j\
                          d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
                          endcasestag2\
                          &=
                          begincases
                          d_ijqquadqquadqquadqquad ineq j\
                          d_ii+sum_k=1^3 d_ikquadqquad i=j
                          endcases\
                          endalign*



                          These variants are all valid and represent the same.



                          • In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.


                          • The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.







                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$


                            We have for $1leq i,jleq 3$



                            beginalign*
                            a_ij&=
                            begincases
                            d_ijqquadqquadqquadqquad ineq j\
                            d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
                            endcasestag1\
                            &=
                            begincases
                            d_ijqquadqquadqquadqquad ineq j\
                            d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
                            endcasestag2\
                            &=
                            begincases
                            d_ijqquadqquadqquadqquad ineq j\
                            d_ii+sum_k=1^3 d_ikquadqquad i=j
                            endcases\
                            endalign*



                            These variants are all valid and represent the same.



                            • In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.


                            • The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.







                            share|cite|improve this answer









                            $endgroup$




                            We have for $1leq i,jleq 3$



                            beginalign*
                            a_ij&=
                            begincases
                            d_ijqquadqquadqquadqquad ineq j\
                            d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
                            endcasestag1\
                            &=
                            begincases
                            d_ijqquadqquadqquadqquad ineq j\
                            d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
                            endcasestag2\
                            &=
                            begincases
                            d_ijqquadqquadqquadqquad ineq j\
                            d_ii+sum_k=1^3 d_ikquadqquad i=j
                            endcases\
                            endalign*



                            These variants are all valid and represent the same.



                            • In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.


                            • The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.








                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 23 at 18:01









                            Markus ScheuerMarkus Scheuer

                            64.1k460152




                            64.1k460152



























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                                People from County GalwayJacobite military personnel of the Williamite War in IrelandHouse of Burke1642 births1722 deathsMembers of the Irish House of LordsEarls of ClanricardePeers of Ireland created by James II IrishpeerWilliam Burke, 7th Earl of ClanricardeRichardisle of InishbofinWilliamite War in IrelandBattle of AughrimUlick Burke, 1st Viscount GalwayProtestantsprivate act of the English ParliamentCatholicWild GeeseSpainBolognaFontenoyHonora BurkePatrick Sarsfield, 1st Earl of Lucan John Burke, 9th Earl of Clanricarde (1642–1722) was an Irish peer. Burke was a younger son of William Burke, 7th Earl of Clanricarde and succeeded his brother Richard. He was created Baron Bophin (over the isle of Inishbofin where Burke is still a common surname amongst the islanders [1] ) in 1689 and commanded a foot regiment as its colonel during the Williamite War in Ireland. He was taken prisoner at the Battle of Aughrim in 1691 and outlawed. His younger brother Ulick Burke, 1st Viscount G...
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                                Football at the 1986 Brunei Merdeka Games Contents Teams Group stage Knockout stage References Navigation menu"Brunei Merdeka Games 1986".

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                                1986 in Asian footballFootball at the Brunei Merdeka Games1986 in Brunei1986 in Singaporean football1986 in Malaysian football1986 in Thai sport Bandar Seri BegawanBruneiMalaysia Cup Football at the 1986 Brunei Merdeka Games Tournament details Host country Brunei Dates 19 July – 27 July Teams 10 Final positions Champions Persib Bandung (1st title) Runners-up Malaysia B Third place   Brunei Fourth place Singapore XI Tournament statistics Matches played 10 Goals scored 25 (2.5 per match) ← 1985 1987 → The football tournament at the 1986 Brunei Merdeka Games was held from July 19 to 27 in Bandar Seri Begawan, Brunei. The tournament was also referred to as the Sultan Hassanal Bolkiah Cup, or the second Brunei Pesta Sukan Cup. Contents 1 Teams 2 Group stage 2.1 Group A 2.2 Group B 3 Knockout stage 3.1 Semi-finals 3.2 Bronze medal match 3.3 Gold medal match 4 References Teams It is likely that the Philippines team was the full national team, bu...
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