I am trying to perform some calculations with a matrix that has let's say size N but unknown, please How can I write this in mathematica?

This is the matrix in question


closed as unclear what you're asking by MarcoB, Yves Klett, dr.blochwave, Mr.Wizard Jun 29 '15 at 8:35

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


The following function, along the lines of the suggestion by Guesswhoitis, produces what you appear to want.

m[r_] := SparseArray[{Band[{1, 1}] -> n, Band[{2, 1}] -> -2, Band[{1, 2}] -> -2}, {r, r}]

It is unclear whether the n in the picture is the same as n, the dimension of the matrix. (Do not use N, which is a reserved term.) If not, change Band[{1, 1}] -> n to Band[{1, 1}] -> whateveryouwant. As a sample result,

m[5] // Normal
(* {{n, -2, 0, 0, 0}, {-2, n, -2, 0, 0}, {0, -2, n, -2, 0}, 
    {0, 0, -2, n, -2}, {0, 0, 0, -2, n}} *)

Edit: In light of the OP's comment, I have changed the dimension to r while leaving the matrix diagonal elements n, now unspecified.

  • $\begingroup$ Thank you for your answer. I misleaded you with the "N", the square matrix K is of size r x r ; with r arbitrary. I have a formulae that I want to compute: Formulae= V_transpose * K * V with V a vector of dimension "r" such that V=(1,0,0,..,0). $\endgroup$ – Zakariae Ben Slimane Jun 29 '15 at 10:39
  • 1
    $\begingroup$ @Zakariae, that's just the element in the first row and first column, no? $\endgroup$ – J. M. will be back soon Jun 29 '15 at 13:08
  • $\begingroup$ It was the inverse of K, I typed the code below where there is an error In[45]:= v[r_] := UnitVector[r, 1]; m[r_] := SparseArray[{Band[{1, 1}] -> n, Band[{2, 1}] -> -2, Band[{1, 2}] -> -2}, {r, r}]; Formulae[r_] := v[r].Inverse[m[r]].v[r]; Formulae[2] Out[48]= n/(n^2-4) In[49]:= Formulae[Q] During evaluation of In[49]:= SparseArray::adims: Array dimension specification {Q,Q} should be Automatic, a non-negative machine integer, or a list of non-negative machine integers. >> Out[49]= UnitVector[Q,1].SparseArray[{Band[{1,1}]->n,Band[{2,1}]->-2,Band[{1,2}]->-2},{Q,Q}]^-1.UnitVector[Q,1] $\endgroup$ – Zakariae Ben Slimane Jun 30 '15 at 1:12

Assuming bbgodfrey's interpretation, I'd also expect this to be faster if dimension is large:

ToeplitzMatrix[PadRight[{#, -2}, #]] &

Though more memory hungry, it produces a packed array, so depending on what you're doing, it may have some performance benefits in use compared to a sparse realization (but the reverse could also be true, again, depends on what you're doing with it after creation).


Another approach is to use DiagonalMatrix and its optional third argument

n = 5; 
mat[n_]:= DiagonalMatrix[ConstantArray[n, n]] + 
 DiagonalMatrix[ConstantArray[-2, n - 1], 1] + 
 DiagonalMatrix[ConstantArray[-2, n - 1], -1];

Not the answer you're looking for? Browse other questions tagged or ask your own question.