10
mat = {TensorProduct[{1, 0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}]/Sqrt[2]};
FixedPoint[ArrayFlatten, mat] // MatrixForm
$\left(
\begin{array}{cccccccc}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & \...
8
mat = {TensorProduct[{1, 0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}]/ Sqrt[2]};
ArrayFlatten[ArrayFlatten /@ mat] // MatrixForm
$\left(
\begin{array}{cccccccc}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 &...
8
First we run the original with an example (which should have been in the post).
Mmax = 400;
W = 1024;
deltaX = .1;
SeedRandom[1111];
lambdaMatrix = RandomReal[1, {W, W}];
lPoly = Developer`ToPackedArray[RandomReal[{-10, 10}, {Mmax + 1, W}]];
AbsoluteTiming[
res = MomentComputing[Mmax, 5, lambdaMatrix, lPoly, lPoly];]
(* Out[52]= {12.1522, Null} *)
What I ...
7
How to find the eigenvalue of this abstract matrix (reference answer:
-2,-2,1).
I am assuming $A^2$ means $A*A$? If so, then may be
e = IdentityMatrix[3];
a = {{a11, a21, a31}, {a21, a22, a32}, {a31, a32, a33}}(*abstract symmetric matrix*)
eqs = Thread[Flatten[a.a + a] == Flatten[2*e]];
sol = FindInstance[eqs, Flatten[a]];
Eigenvalues[a /. sol]
7
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, 0},
{0, 0, 0, 1/Sqrt[2], 0, 0, 0, 0}} *)
6
This suggests minimal polynomial of $A$ is $x^2+x-2$. This has roots -2,1 (the eigenvalues not counting multiplicities). Note $m(A)$ divides $p(A)$ (the characteristic polynomial, which by Cayley-Hamilton $p(A)=0$)
(-2,-2,1) or (-2,1,1) could be eigenvalues of $A$: e.g. using diagonal matrices.
This also case when examine other instances from @Nasser code.
...
5
Your specific example can be solve with a general $P$, see code below.
(*Data*)
A = {{-2, -2, 1}, {2, x, -2}, {0, 0, -2}};
B = {{2, 1, 0}, {0, -1, 0}, {0, 0, y}};
(*Search for x and y based on characteristic polynomial*)
n = Length@A;
Id = IdentityMatrix@n;
solxy = SolveAlways[Det[A - l*Id] == Det[B - l*Id], l]
(*Update data*)
A = A /. solxy[[1]];
B = B /. ...
5
One of many ways:
mat = Normal@
SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> -1/2,
Band[{2, 1}] -> -1/2}, {10, 10}];
mat // MatrixForm
The Normal@ is not really necessary.
5
Recommend that you look at Array
Clear["Global`*"]
genMat[n_Integer?Positive, f_] := Array[f, {n, n}]
genMat[4, f]
(* {{f[1, 1], f[1, 2], f[1, 3], f[1, 4]}, {f[2, 1], f[2, 2], f[2, 3],
f[2, 4]}, {f[3, 1], f[3, 2], f[3, 3], f[3, 4]}, {f[4, 1], f[4, 2], f[4, 3],
f[4, 4]}} *)
genMat[4, Times]
(* {{1, 2, 3, 4}, {2, 4, 6, 8}, {3, 6, 9, 12}, {...
5
Another option
(m = {TensorProduct[{1, 0}, {0, 1}, {0, 1}, {0, 1}, {0, 1}]}/Sqrt[2]) // MatrixForm
And now
m1 = ArrayFlatten[m[[1, 1]], 2]
m2 = ArrayFlatten[m[[1, 2]], 2]
Join[m1, m2, 2] // MatrixForm
4
After a second thought I notice it's not that hard to implement:
ClearAll[myD, δ]
myD[x_[k_, l_], x_[β_, α_]] := δ[k, β] δ[l, α] - δ[k, α] δ[l, β]
myD[a_ b_, c_] := a myD[b, c] + b myD[a, c]
SetAttributes[δ, Orderless]
δ /: δ[a_, b_] h_[former___, b_, latter___] := h[former, a, latter]
δ[a_, a_] = \[FormalCapitalD];
The following rule isn't necessary but ...
4
Clear["Global`*"]
A[n_Integer?Positive] :=
DiagonalMatrix[Table[1, n]] +
DiagonalMatrix[Table[-1/2, n - 1], 1, n] +
DiagonalMatrix[Table[-1/2, n - 1], -1, n];
A[5] // MatrixForm
4
There are only seven (six unique) solutions up to row/column sums of $10^8$ (of course there are 1152 equivalent squares by permutation/transposition for each one):
$$
\left(
\begin{array}{cccc}
2 & 16 & 108 & 180 \\
24 & 192 & 9 & 15 \\
144 & 18 & 150 & 90 \\
160 & 20 & 135 & 81 \\
\end{array}
\right),\...
3
Assuming you have definitions already for n, single argument t[x], and two-argument t[x,y], then this is fairly straightforward:
g[i_, n_] :=
Table[(t[i + 1] - t[j, i])/h[i] KroneckerDelta[k,
i] + (t[j, i] - t[i])/h[i] KroneckerDelta[k, i + 1] -
1/6 (t[j, i] - t[i]) (t[i + 1] -
t[j, i]) (1 + (t[i + 1] - t[j, i])/h[i]) KroneckerDelta[k,
...
3
It looks like RowReduce and MatrixRank implicitly assume that a does not have a particular value that reduces the rank of the matrix. They can detect that matrix A is degenerate for any value of a, but not that B is degenerate for a == 2.
I think you might be able to get around this with SingularValueList, which seems to handle the problem better:
Assuming[a ...
3
The simplest way to do that is to use:
EV = FullSimplify[Table[Eigenvalues[J /. f], {f, FP}]]
3
This will do it:
Reverse /@ Partition[Range[0, 999], 10, 1]
{{9, 8, 7, 6, 5, 4, 3, 2, 1, 0}, ... {999, 998, 997, 996, 995, 994, 993, 992, 991, 990}}
Replace Range[0, 999] by your x.
2
First, don't use N as symbol, it's predfined in Mathematica
Try Reduce to solve the problem
expr = (n*
k*(\[Beta]1*(1 - q) + \[Beta]2*
q)) - ((k + \[Mu])*(\[Gamma]1 + \[Mu])*(\[Gamma]2 + \[Mu]) -\[Beta]2*n*k*q*(\[Gamma]1 + \[Mu]) - \[Beta]1*n*k*(1 - q)*(\[Gamma]2 + \[Mu])) ;
var = Variables[expr];
Reduce[Join[{expr > 0}, Map[# > 0 &, var]]] //...
2
You do not need to iterate manually. First use ReplaceAll to substitute all eight solutions contained in FP, which will give you a list of 8 J expressions. Then Map the Eigenvalues function over the list: this will apply Eigenvalues to each element of the list of Js in turn, to give you the eight results, one for each solution from FP:
Eigenvalues /@ (J /. ...
2
All credit to @Roman who deserves the accept for discovering 7095816 was the target number. With this number pinned down I tried to solve this myself with Mathematica, but it was still too slow:
variables = Array[x, 16];
matrix = ArrayReshape[variables, {4, 4}];
target = 7095816;
uniqueConstraint = (And @@ (Unequal @@@ Subsets[variables, {2}]));
...
2
Is this a formula with the Einstein sum convention. If so then the summation over two same indices has to be carried through.
g[i_, j_, k_, n_] :=
Sum[(t[i + 1] - t[j, i])/h[ii] KroneckerDelta[k,
ii] + (t[j, i] - t[ii])/h[ii] KroneckerDelta[k, ii + 1] -
1/6 (t[j, i] - t[ii]) (t[ii + 1] -
t[j, i]) (1 + (t[ii + 1] - t[j, i])/h[ii]) ...
2
This seems to be a hasty start into Mathematica. Your inputs are right but used without insight.
Use
B = {{k1 + h, -k1, 0}, {-k1,h + k1´ + k2, -k2´}, {0, -k2, h + k2´}};
C0 = {{CA0}, {CB0}, {CC0}};
C1[t_] = h.(IdentityMatrix - Exp[-Bt])(B^-1)*C0;
C2[t_] = {CA2[t], CB2[t], CC2[t]};
B// MatrixForm
C0 // MatrixForm
C1[t] // MatrixForm
C2[t] // MatrixForm
This ...
2
Works fine for me:
a = {{1, -1, 0, 0}, {-1, 1, -1, 0}, {0, -1, 1, -1}, {0, 0, -1, -1}};
{Da, Sa} = Eigensystem[a];
Inverse[Transpose[Sa]].a.Transpose[Sa] == DiagonalMatrix[Da] // FullSimplify
(* True *)
Transpose[Sa].DiagonalMatrix[Da].Inverse[Transpose[Sa]] == a // FullSimplify
(* True *)
1
I am not sure if I have got your attention properly, but you can see if code below works for you or not,
matM = Module[{block, n = 4, jup = 10},
block[i_, j_] := Which[i == j, Cos[i],
Abs[i - j] == 1, Sin[Min[i, j]],
True, 0
] IdentityMatrix[n];
SparseArray @ ArrayFlatten[Table[block[i, j], {i, 0, jup}, {j, 0, jup}]]
]
1
The special linear group is defined by
In words: SL(2,ℤ) is the group of 2 × 2 real matrices with determinant one. SL(2,ℤ) = {A ∈ Z2×2 : Det(A) = 1}
In representation theory:
SL(2,ℤ) is a real, non-compact simple Lie group, with the Lie-Algebra of the Tr-free {\displaystyle 2\times 2}2\times 2-Matrices.
A vector space bass of this 3-dimensional vector is ...
1
Not too hard to do, if you recall that one can use a rotation matrix to perform the required orthogonal similarity transformation:
{{{a, 2}, {2, b}}, TrigExpand[RotationMatrix[2 ArcTan[u]]]} /.
Solve[With[{rot = TrigExpand[RotationMatrix[2 ArcTan[u]]]},
Flatten[Thread /@ Thread[rot.{{a, 2}, {2, b}}.Transpose[rot] ==
...
answered 2 days ago
1
This script can be applied when the same elements are in the same row.
A = {{1, 2, 3, 5}, {4, 5, 6, 8}, {7, 8, 9, 3}, {3, 56, 8, 2}, {4, 5, 6, 8}};
B = {{6, 7, 9, 1}, {2, 5, 0, 8}, {1, 2, 3, 7}, {34, 56, 78, 56}};
l = Length /@ {A, B} // Min
(* 4 *)
truefalse = MapThread[Equal, {A[[;; l, 2]], B[[;; l, 2]]}]
(* {False, True, False, True} *)
index = Pick[...
1
Another possibility using FeynCalc 9.3 or above (could be interesting for particle physicists)
CLC[i, j, k] CSI[j].CSI[k] // PauliSimplify[#, PauliReduce -> True] & // FCE
2 I CSI[i]
CSI[i].CSI[i] // PauliSimplify // FCE
3
CSI is a shortcut for a Cartesian Pauli sigma matrix, while CLC denotes a Cartesian Levi-Civita tensor. PauliSimplify handles ...
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