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Skipping indices in a product



The Next CEO of Stack OverflowWhat's the best way to generate all the upper triangular matrix whose singular values are given?What is the fastest way to obtain the eigenvalues of a Wishart matrix?Evaluating the product of a matrix sequenceMapping over two indices with a conditionOuter product using the quantum mathematica packageWhy does Eigenvalues work for a matrix $M$ but not $M$?Conditions on a productHow to put conditions on indices of productProduct of matrices with symbolic boundsInverting a matrix when its elements are given by difficult expressions?










2












$begingroup$


I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_lambda_ine lambda_j fracA - lambda_i Ilambda_j-lambda_i$, where $lambda_i$ ($lambda_j$) denote the eigenvalues of $A$. How can this be implemented in Mathematica? Just gave a try here:



A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
Eigenvalues[A]



2, -1, 1, 1




Tj = Product[(A - Eigenvalues[A][[i]] IdentityMatrix[4])/(
Eigenvalues[A][[j]] - Eigenvalues[A][[i]]), i, 1, 4]









share|improve this question











$endgroup$











  • $begingroup$
    in which part exactly you want to exclude it in Tj !??
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    do you want it to be skipped put not Zero right !?
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    @Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
    $endgroup$
    – Tobias Fritzn
    5 hours ago










  • $begingroup$
    @Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
    $endgroup$
    – Tobias Fritzn
    5 hours ago






  • 2




    $begingroup$
    That product is presumably a matrix multiplication?
    $endgroup$
    – J. M. is slightly pensive
    4 hours ago















2












$begingroup$


I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_lambda_ine lambda_j fracA - lambda_i Ilambda_j-lambda_i$, where $lambda_i$ ($lambda_j$) denote the eigenvalues of $A$. How can this be implemented in Mathematica? Just gave a try here:



A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
Eigenvalues[A]



2, -1, 1, 1




Tj = Product[(A - Eigenvalues[A][[i]] IdentityMatrix[4])/(
Eigenvalues[A][[j]] - Eigenvalues[A][[i]]), i, 1, 4]









share|improve this question











$endgroup$











  • $begingroup$
    in which part exactly you want to exclude it in Tj !??
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    do you want it to be skipped put not Zero right !?
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    @Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
    $endgroup$
    – Tobias Fritzn
    5 hours ago










  • $begingroup$
    @Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
    $endgroup$
    – Tobias Fritzn
    5 hours ago






  • 2




    $begingroup$
    That product is presumably a matrix multiplication?
    $endgroup$
    – J. M. is slightly pensive
    4 hours ago













2












2








2





$begingroup$


I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_lambda_ine lambda_j fracA - lambda_i Ilambda_j-lambda_i$, where $lambda_i$ ($lambda_j$) denote the eigenvalues of $A$. How can this be implemented in Mathematica? Just gave a try here:



A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
Eigenvalues[A]



2, -1, 1, 1




Tj = Product[(A - Eigenvalues[A][[i]] IdentityMatrix[4])/(
Eigenvalues[A][[j]] - Eigenvalues[A][[i]]), i, 1, 4]









share|improve this question











$endgroup$




I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_lambda_ine lambda_j fracA - lambda_i Ilambda_j-lambda_i$, where $lambda_i$ ($lambda_j$) denote the eigenvalues of $A$. How can this be implemented in Mathematica? Just gave a try here:



A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
Eigenvalues[A]



2, -1, 1, 1




Tj = Product[(A - Eigenvalues[A][[i]] IdentityMatrix[4])/(
Eigenvalues[A][[j]] - Eigenvalues[A][[i]]), i, 1, 4]






matrix operators products






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









That Gravity Guy

2,1411615




2,1411615










asked 5 hours ago









Tobias FritznTobias Fritzn

1795




1795











  • $begingroup$
    in which part exactly you want to exclude it in Tj !??
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    do you want it to be skipped put not Zero right !?
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    @Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
    $endgroup$
    – Tobias Fritzn
    5 hours ago










  • $begingroup$
    @Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
    $endgroup$
    – Tobias Fritzn
    5 hours ago






  • 2




    $begingroup$
    That product is presumably a matrix multiplication?
    $endgroup$
    – J. M. is slightly pensive
    4 hours ago
















  • $begingroup$
    in which part exactly you want to exclude it in Tj !??
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    do you want it to be skipped put not Zero right !?
    $endgroup$
    – Alrubaie
    5 hours ago










  • $begingroup$
    @Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
    $endgroup$
    – Tobias Fritzn
    5 hours ago










  • $begingroup$
    @Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
    $endgroup$
    – Tobias Fritzn
    5 hours ago






  • 2




    $begingroup$
    That product is presumably a matrix multiplication?
    $endgroup$
    – J. M. is slightly pensive
    4 hours ago















$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
5 hours ago




$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
5 hours ago












$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
5 hours ago




$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
5 hours ago












$begingroup$
@Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
$endgroup$
– Tobias Fritzn
5 hours ago




$begingroup$
@Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
$endgroup$
– Tobias Fritzn
5 hours ago












$begingroup$
@Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
$endgroup$
– Tobias Fritzn
5 hours ago




$begingroup$
@Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
$endgroup$
– Tobias Fritzn
5 hours ago




2




2




$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive
4 hours ago




$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive
4 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

Here is my pedestrian implementation of your formula:



a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;

ClearAll[t]
t[amat_, j_] := Module[
evals, usable,
evals = Eigenvalues[amat];
usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
Dot @@
Table[
(amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
i, usable
]
]

t[a, 4]


Mathematica graphics



You do not provide an example of desired output, so I will let you check whether this is what you expect.






share|improve this answer









$endgroup$












  • $begingroup$
    Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
    $endgroup$
    – Tobias Fritzn
    4 hours ago



















3












$begingroup$

Something like this?



Clear[A, evals, T]
A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
T[A_?MatrixQ, j_Integer] := With[
evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
Dot @@ Table[
If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
i, Length@A
]
]

MatrixForm /@ Array[T[A, #] &, 4]


enter image description here






share|improve this answer









$endgroup$




















    1












    $begingroup$

    This



    A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
    e=Eigenvalues[A];
    Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
    DeleteCases[Tuples[Range[4],2],i_,i_]]


    generates your twelve matricies with i not equal to j.



    Put Dot@@ in front of that Map to form the dot product of the 12 matricies.



    That works by forming every possible distinct i,j pair and then using those in the Map



    If it might be easier to read you can also write it this way



    Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
    (A-ei*IdentityMatrix[4])/(ej-ei))&,
    DeleteCases[Tuples[Range[4],2],i_,i_]]





    share|improve this answer











    $endgroup$












    • $begingroup$
      Should e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?
      $endgroup$
      – That Gravity Guy
      4 hours ago










    • $begingroup$
      @ThatGravityGuy Yes! Good catch. Thank you! Corrected.
      $endgroup$
      – Bill
      4 hours ago











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Here is my pedestrian implementation of your formula:



    a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;

    ClearAll[t]
    t[amat_, j_] := Module[
    evals, usable,
    evals = Eigenvalues[amat];
    usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
    Dot @@
    Table[
    (amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
    i, usable
    ]
    ]

    t[a, 4]


    Mathematica graphics



    You do not provide an example of desired output, so I will let you check whether this is what you expect.






    share|improve this answer









    $endgroup$












    • $begingroup$
      Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
      $endgroup$
      – Tobias Fritzn
      4 hours ago
















    4












    $begingroup$

    Here is my pedestrian implementation of your formula:



    a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;

    ClearAll[t]
    t[amat_, j_] := Module[
    evals, usable,
    evals = Eigenvalues[amat];
    usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
    Dot @@
    Table[
    (amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
    i, usable
    ]
    ]

    t[a, 4]


    Mathematica graphics



    You do not provide an example of desired output, so I will let you check whether this is what you expect.






    share|improve this answer









    $endgroup$












    • $begingroup$
      Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
      $endgroup$
      – Tobias Fritzn
      4 hours ago














    4












    4








    4





    $begingroup$

    Here is my pedestrian implementation of your formula:



    a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;

    ClearAll[t]
    t[amat_, j_] := Module[
    evals, usable,
    evals = Eigenvalues[amat];
    usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
    Dot @@
    Table[
    (amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
    i, usable
    ]
    ]

    t[a, 4]


    Mathematica graphics



    You do not provide an example of desired output, so I will let you check whether this is what you expect.






    share|improve this answer









    $endgroup$



    Here is my pedestrian implementation of your formula:



    a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;

    ClearAll[t]
    t[amat_, j_] := Module[
    evals, usable,
    evals = Eigenvalues[amat];
    usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
    Dot @@
    Table[
    (amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
    i, usable
    ]
    ]

    t[a, 4]


    Mathematica graphics



    You do not provide an example of desired output, so I will let you check whether this is what you expect.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 4 hours ago









    MarcoBMarcoB

    38.1k556114




    38.1k556114











    • $begingroup$
      Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
      $endgroup$
      – Tobias Fritzn
      4 hours ago

















    • $begingroup$
      Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
      $endgroup$
      – Tobias Fritzn
      4 hours ago
















    $begingroup$
    Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
    $endgroup$
    – Tobias Fritzn
    4 hours ago





    $begingroup$
    Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
    $endgroup$
    – Tobias Fritzn
    4 hours ago












    3












    $begingroup$

    Something like this?



    Clear[A, evals, T]
    A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
    T[A_?MatrixQ, j_Integer] := With[
    evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
    Dot @@ Table[
    If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
    i, Length@A
    ]
    ]

    MatrixForm /@ Array[T[A, #] &, 4]


    enter image description here






    share|improve this answer









    $endgroup$

















      3












      $begingroup$

      Something like this?



      Clear[A, evals, T]
      A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
      T[A_?MatrixQ, j_Integer] := With[
      evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
      Dot @@ Table[
      If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
      i, Length@A
      ]
      ]

      MatrixForm /@ Array[T[A, #] &, 4]


      enter image description here






      share|improve this answer









      $endgroup$















        3












        3








        3





        $begingroup$

        Something like this?



        Clear[A, evals, T]
        A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
        T[A_?MatrixQ, j_Integer] := With[
        evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
        Dot @@ Table[
        If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
        i, Length@A
        ]
        ]

        MatrixForm /@ Array[T[A, #] &, 4]


        enter image description here






        share|improve this answer









        $endgroup$



        Something like this?



        Clear[A, evals, T]
        A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
        T[A_?MatrixQ, j_Integer] := With[
        evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
        Dot @@ Table[
        If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
        i, Length@A
        ]
        ]

        MatrixForm /@ Array[T[A, #] &, 4]


        enter image description here







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 4 hours ago









        That Gravity GuyThat Gravity Guy

        2,1411615




        2,1411615





















            1












            $begingroup$

            This



            A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
            e=Eigenvalues[A];
            Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
            DeleteCases[Tuples[Range[4],2],i_,i_]]


            generates your twelve matricies with i not equal to j.



            Put Dot@@ in front of that Map to form the dot product of the 12 matricies.



            That works by forming every possible distinct i,j pair and then using those in the Map



            If it might be easier to read you can also write it this way



            Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
            (A-ei*IdentityMatrix[4])/(ej-ei))&,
            DeleteCases[Tuples[Range[4],2],i_,i_]]





            share|improve this answer











            $endgroup$












            • $begingroup$
              Should e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?
              $endgroup$
              – That Gravity Guy
              4 hours ago










            • $begingroup$
              @ThatGravityGuy Yes! Good catch. Thank you! Corrected.
              $endgroup$
              – Bill
              4 hours ago















            1












            $begingroup$

            This



            A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
            e=Eigenvalues[A];
            Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
            DeleteCases[Tuples[Range[4],2],i_,i_]]


            generates your twelve matricies with i not equal to j.



            Put Dot@@ in front of that Map to form the dot product of the 12 matricies.



            That works by forming every possible distinct i,j pair and then using those in the Map



            If it might be easier to read you can also write it this way



            Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
            (A-ei*IdentityMatrix[4])/(ej-ei))&,
            DeleteCases[Tuples[Range[4],2],i_,i_]]





            share|improve this answer











            $endgroup$












            • $begingroup$
              Should e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?
              $endgroup$
              – That Gravity Guy
              4 hours ago










            • $begingroup$
              @ThatGravityGuy Yes! Good catch. Thank you! Corrected.
              $endgroup$
              – Bill
              4 hours ago













            1












            1








            1





            $begingroup$

            This



            A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
            e=Eigenvalues[A];
            Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
            DeleteCases[Tuples[Range[4],2],i_,i_]]


            generates your twelve matricies with i not equal to j.



            Put Dot@@ in front of that Map to form the dot product of the 12 matricies.



            That works by forming every possible distinct i,j pair and then using those in the Map



            If it might be easier to read you can also write it this way



            Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
            (A-ei*IdentityMatrix[4])/(ej-ei))&,
            DeleteCases[Tuples[Range[4],2],i_,i_]]





            share|improve this answer











            $endgroup$



            This



            A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
            e=Eigenvalues[A];
            Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
            DeleteCases[Tuples[Range[4],2],i_,i_]]


            generates your twelve matricies with i not equal to j.



            Put Dot@@ in front of that Map to form the dot product of the 12 matricies.



            That works by forming every possible distinct i,j pair and then using those in the Map



            If it might be easier to read you can also write it this way



            Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
            (A-ei*IdentityMatrix[4])/(ej-ei))&,
            DeleteCases[Tuples[Range[4],2],i_,i_]]






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 4 hours ago

























            answered 4 hours ago









            BillBill

            5,88569




            5,88569











            • $begingroup$
              Should e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?
              $endgroup$
              – That Gravity Guy
              4 hours ago










            • $begingroup$
              @ThatGravityGuy Yes! Good catch. Thank you! Corrected.
              $endgroup$
              – Bill
              4 hours ago
















            • $begingroup$
              Should e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?
              $endgroup$
              – That Gravity Guy
              4 hours ago










            • $begingroup$
              @ThatGravityGuy Yes! Good catch. Thank you! Corrected.
              $endgroup$
              – Bill
              4 hours ago















            $begingroup$
            Should e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?
            $endgroup$
            – That Gravity Guy
            4 hours ago




            $begingroup$
            Should e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?
            $endgroup$
            – That Gravity Guy
            4 hours ago












            $begingroup$
            @ThatGravityGuy Yes! Good catch. Thank you! Corrected.
            $endgroup$
            – Bill
            4 hours ago




            $begingroup$
            @ThatGravityGuy Yes! Good catch. Thank you! Corrected.
            $endgroup$
            – Bill
            4 hours ago

















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