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bear with me as I'm very inexperienced with Mathematica, although I think my problem is reasonably straight forward:

I'm currently trying to find a matrix from a simple expression. I've found a way to extract the matrix, however there still remains a constraint that I must somehow impose in order for the matrix to be correct. Here is what I have:

nn = 10;
row[n_] := Sum[(p + 1 + R + 3 n^2) a[p], {p, 0, nn - 1}] + \[Alpha] a[n - 1]
vector = Table[a[i], {i, 0, nn - 1}];
matrix = Table[D[row[i], vector[[j]]], {i, 1, nn}, {j, 1, nn}]

This produces a matrix which is fantastic. However, I still have two "boundary conditions" left to impose which are (written in Mathematica code):

Sum[a[n],{n,0,nn-1}]=0
Sum[n^2a[n],{n,0,nn-1}]=0

Does anyone have any ideas of how I can impose these constraints upon the above set of code, such that the end matrix takes them into account?

Thank you :)

Edit:

nn = 10;
row[n_] := Sum[(p + 1 + R + 3 n^2) a[p], {p, 0, nn - 1}] + \[Alpha] a[n - 1]
vector = Table[a[i], {i, 0, nn - 1}];
matrix = Table[D[row[i], vector[[j]]], {i, 1, nn}, {j, 1, nn}]
matrix /. { a[0] -> -Sum[i^2 a[i],{i,2,nn-1}] - Sum[a[i],{i,1,nn-1}], a[1] -> -Sum[i^2 a[i],{i,2,nn-1}]}
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You may use the 2 equations to eliminate 2 of the a[n]. E.g. to eliminate a[0]and a[1]:

From Sum[a[n],{n,0,nn-1}]=0 and Sum[n^2a[n],{n,0,nn-1}]=0 we get:

a[0]== -Sum[a[i],{i,1,nn-1}]
a[1]== -Sum[i^2 a[i],{i,2,nn-1}]

Or:

a[0]==  -Sum[i^2 a[i],{i,2,nn-1}] - Sum[a[i],{i,1,nn-1}]
a[1]== -Sum[i^2 a[i],{i,2,nn-1}]

To insert this into matrix, you can use ReplaceAll (/.) and a Rule (.. -> ..), look this up in the manual:

matrix /. {
  a[0] -> -Sum[i^2 a[i],{i,2,nn-1}] - Sum[a[i],{i,1,nn-1}],
  a[1] -> -Sum[i^2 a[i],{i,2,nn-1}]}
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  • $\begingroup$ Thank you very much for your answer. I see, so I can impose these conditions upon the matrix simply by rearranging the constraining sums. How would I go about inserting this into the code? $\endgroup$
    – James
    Mar 23 at 11:02
  • $\begingroup$ I added this to my answer. $\endgroup$ Mar 23 at 11:07
  • $\begingroup$ Ah yes, amazing stuff! I've added an edit to my original post which is what I think my code should now look like. Does it look okay? Thank you again for you time, it is greatly appreciated. $\endgroup$
    – James
    Mar 23 at 11:26

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