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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









      $endgroup$













        Your Answer





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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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                                What would you call this weird metallic apparatus that allows you to lift people? Is it possible to give , in economics, an example of a relation ( set of ordered pairs) that is not a function? Maximum summed subsequences with non-adjacent items How many time has Arya actually used Needle? Is there public access to the Meteor Crater in Arizona? Belief In God or Knowledge Of God. Which is better? What is the meaning of 'breadth' in breadth first search? Why is there Net Work Done on a Pressure/Volume Cycle? How much damage would a cupful of neutron star matter do to the Earth? Sum letters are not two different Why does it sometimes sound good to play a grace note as a lead in to a note in a melody? What initially awakened the Balrog? How to pronounce 伝統色 QGIS virtual layer functionality does not seem to support memory layers AppleTVs create a chatty alternate WiFi network How to write capital alpha? How often does castling occur in gra...
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                                Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye

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                                Contributing propertyKeeper of the RegisterHistoric districtHistory of the National Register of Historic PlacesNational Park ServiceProperty typesCharlestonColumbiaGreenvilleNorth CharlestonRock Hill Houses on the National Register of Historic Places in South CarolinaGeorgian architecture in South CarolinaHouses completed in 1786Plantation houses in South CarolinaNational Register of Historic Places in Charleston, South CarolinaHouses in Charleston, South Carolina plantation houseAshley RiverCharlestonNational Register of Historic PlacesRevolutionary WarWilliam LowndesU.S. CongressSouth Carolina Inter-State and West Indian ExpositionGeorgianDoricbalustradespedimentoculusentablaturemodilliondentillightship roofterra cottatilescorbelledcornicesRegency stylespiral staircaseskylight Lowndes Grove U.S. National Register of Historic Places Lowndes Grove in 1940 Show map of South Carolina Show map of the United States Location 260 St. Margaret St., Charleston, South Carolina Coordinates 32...
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