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Skipping indices in a product



The Next CEO of Stack OverflowTensorContract of inverse matrixWhat'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 sequenceNearest Kronecker ProductMapping over two indices with a conditionBilinear Dot FunctionWhy does Eigenvalues work for a matrix $M$ but not $M$?Numerically computing the eigenvalues of an infinite-dimensional tridiagonal matrixInverting 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
    2 days ago










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






  • 2




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










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






  • 2




    $begingroup$
    That product is presumably a matrix multiplication?
    $endgroup$
    – J. M. is slightly pensive
    2 days 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 linear-algebra operators






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









Michael E2

150k12203482




150k12203482










asked 2 days ago









Tobias FritznTobias Fritzn

1795




1795











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










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






  • 2




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
















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










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






  • 2




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















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




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












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




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




2




2




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




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










4 Answers
4






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
    2 days 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









$endgroup$




















    2












    $begingroup$

    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
      2 days ago










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


















    0












    $begingroup$

    Another way:



    ClearAll[t];
    t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
    t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
    Fold[
    #1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
    IdentityMatrix[Length@A],
    Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
    ];


    Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With[] to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold[]). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.






    share|improve this answer









    $endgroup$













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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      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
        2 days 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
        2 days 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 2 days ago









      MarcoBMarcoB

      38.2k556114




      38.2k556114











      • $begingroup$
        Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
        $endgroup$
        – Tobias Fritzn
        2 days 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
        2 days 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
      2 days 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
      2 days 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 2 days ago









          That Gravity GuyThat Gravity Guy

          2,1411615




          2,1411615





















              2












              $begingroup$

              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
                2 days ago










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















              2












              $begingroup$

              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
                2 days ago










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













              2












              2








              2





              $begingroup$

              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 2 days ago

























              answered 2 days ago









              BillBill

              5,89569




              5,89569











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










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
















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










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















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




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












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




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











              0












              $begingroup$

              Another way:



              ClearAll[t];
              t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
              t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
              Fold[
              #1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
              IdentityMatrix[Length@A],
              Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
              ];


              Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With[] to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold[]). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.






              share|improve this answer









              $endgroup$

















                0












                $begingroup$

                Another way:



                ClearAll[t];
                t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
                t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
                Fold[
                #1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
                IdentityMatrix[Length@A],
                Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
                ];


                Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With[] to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold[]). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.






                share|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  Another way:



                  ClearAll[t];
                  t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
                  t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
                  Fold[
                  #1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
                  IdentityMatrix[Length@A],
                  Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
                  ];


                  Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With[] to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold[]). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.






                  share|improve this answer









                  $endgroup$



                  Another way:



                  ClearAll[t];
                  t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
                  t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
                  Fold[
                  #1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
                  IdentityMatrix[Length@A],
                  Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
                  ];


                  Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With[] to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold[]). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered yesterday









                  Michael E2Michael E2

                  150k12203482




                  150k12203482



























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