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Notice that the generated matrix is actually $(N+1)\times(N+1)$ as the indices run from 0 to $N$. You can see this in your images, which show an even-sized matrix for odd $N$ and vice-versa. … , {n + 1, n + 1}] If you need a non-sparse matrix, use Normal. I'm not sure why the bottom-right matrix element in your odd-$N$ image is zero. …

,{60,62}->2,{61,59}->2,{61,61}->3,{62,60}->2,{62,62}->1,{62,63}->2,{63,62}->2,{63,63}->3,{64,64}->5}]/10; we see that the commutator vanishes exactly, A.B - B.A // Norm 0 With the numerical matrix …

Assuming that you meant $$ \langle n\lvert M \rvert m \rangle = m^2 \delta_{m,n}+(1-m)(\delta_{m,n+2}+\delta_{m,n-2}) $$ and ranges of $n,m\in\{1,2,3,\ldots,a\}$, you can define this matrix with M[a … isn't Hermitian and that you are using Dirac-notation matrix elements, I suspect that there is something wrong in these definitions. …

an $n\times n$ symmetric matrix $M$: M[n_Integer?Positive] := Array[m @@ Sort[{##}] &, {n, n}] an $\vec{y}$-vector with only 1-elements: y[n_Integer? …

See this book on page 53 (section 3.4.1): SpinQ[S_] := IntegerQ[2S] && S>=0 op[S_?SpinQ, n_Integer, k_Integer, a_?MatrixQ] /; 1<=k<=n && Dimensions[a] == {2S+1,2S+1} := KroneckerPro …

M = {{1, 1, 0}, {1, 0, 0}}; ArrayPad[M, {{0, Length[Transpose[M]]}, {Length[M], 0}}] (* {{0, 0, 1, 1, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0 …

data = {{0, 0, 0, 0}, {1, 2, 92.355, 49.}, {1, 7, 86.7407, 46.5}, {6, 2, 92.355, 49.}, {6, 7, 87.237, 46.5}}; FirstCase[data, {1, 7, x_, ___} -> x] (* 86.7407 *)

I'm assuming that the matrix can be of any size, and that $a$ can stand for anything: check[u_] := DiagonalMatrixQ[u] && Apply[SameQ, Diagonal[u]*(-1)^Range[Length[u]]] check @ {{-a, 0, …

matrixExample = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 9, 9, 9}}; Mean[matrixExample] (* {5, 17/3, 19/3, 7} *) % // N (* {5., 5.66667, 6.33333, 7.} *)

Look up a Toeplitz matrix: m[n_] := a^(ToeplitzMatrix[n]-1) Alternatively, an explicit construction as a rule-based matrix: m[n_] := SparseArray[{i_, j_} -> a^Abs[i - j], {n, n}] and the same with the …

14, 11, 3, 8, 7}, {4, 12, 19, 13, 3}}; f[p_] := Total[MapThread[Part, {A, p}]] M = ResourceFunction["MaximizeOverPermutations"]; M[f, 5] (* {{{4, 5, 2, 1, 3}}, 81} *) Let's try it for a 20×20 matrix …

X = {TensorProduct[{1, 0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}]/Sqrt[2]}; Flatten[X, {{1, 3, 5}, {2, 4, 6}}] (* {{0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, …

/. a[0] -> x] (* (-x+a[1])(-x+a[2])(-a[1]+a[2])(-x+a[3])(-a[1]+a[3])(-a[2]+a[3])(-x+a[4])(-a[1]+a[4])(-a[2]+a[4])(-a[3]+a[4]) *) This method is exponentially faster than actually building the matrix …

x[t_] = {x1[t], x2[t], x3[t], x4[t]}; A = {{-K, K, 0, 0}, {K, -k - K, k, 0}, {0, k, -k - K, K}, {0, 0, K, -K}}; Thread[m*x''[t] == A.x[t]] {m x1"[t] == -K x1[t] + K x2[t], m x2"[t] == K x1[ …

Depending on your taste, d = MapThread[{#1[[1]] - #2[[1]], #1[[2]]} &, {X, Y}] or d = Transpose@{X[[All, 1]] - Y[[All, 1]], X[[All, 2]]} define the differences d, if you need them. The symbol D …