# Tag Info

## New answers tagged linear-algebra

0

This seems to work. kmat = {{3 - 3 w, -4/3 + a + 2 w, 5/12 + b - w/2, 0}, {-4/3 + a + 2 w, -1 - 6 a - 3 x, 3/4 + 3 a - 3 b + 2 x - y/2, -2/3 - a - x/2 + z/2}, {5/12 + b - w/2, 3/4 + 3 a - 3 b + 2 x - y/2, -1/2 + 6 b, 19/12 - b + y/2 - 2 z}, {0, -2/3 - a - x/2 + z/2, 19/12 - b + y/2 - 2 z, -3 + 3 z}}; vec = Array[t, Length[kmat]]; ...

1

I think LUDecomposition should be able to do this. Using code from help in Mathematica docs: a = {{1, 2, 3}, {2, 3, 4}, {-1, 0, 2}}; {lu, p, c} = LUDecomposition[a] (u = lu SparseArray[{i_, j_} /; j >= i -> 1, {3, 3}]) // MatrixForm If you want all pivots to be +1, this can now be easily done. d = Position[Diagonal[u], -1] (u[[First@#]] = ...

2

For square matrices: You can try this: RandomMatrix[rank_, m_] := Sum[TensorProduct @@ RandomReal[{-1, 1}, {2, m}], {i, rank}]; It returns a pseudorandom m-by-m matrix with rank rank. Example usage: MatrixRank@RandomMatrix[5, 10] (*5*) Rectangular matrices For rectangular matrices, try this: RandomMatrix[rank_, m_, n_] := Sum[RandomReal[{-1, ...

1

You can do it as Assuming[ {s0 > 0, s1 > 0, s2 > 0, s3 > 0, s4 > 0, s5 > 0, s6 > 0, s7 > 0, s8 > 0}, Inverse[{{s0, s1, s2}, {s3, s4, s5}, {s6, s7, s8}}]]  For instance, If you get $s0=-1$, then you will get the error $Assumptions::fas: Warning: one or more assumptions evaluated to False.. Also you can determine the condition ... 4 In addition to using Solve one can augment the matrix by a row containing the modulus in each position and use HermiteDecomposition. Any zero row (modulo the modulus) in the resulting HNF corresponds to a null vector in the conversion matrix. i1 = IntervalDifferenceMatrix[{{0}, {1}, {4}}]; i2 = Append[i1, ConstantArray[12, 3]]; {uu, hnf} = ... 0 One rather obvious way, useful for the situation you gave, is as follows: You can evaluate the expression at e=0: (a*k + (a^2)*b*c + b*e)/.e->0 gives a*k+a^2*b*c, and (a*k + (a^2)*b*c + b*e)/.a->0 gives b*e. 9 The Susyno and LieART Mathematica packages can do this. I know best the first one (I wrote it), so let me use it as an example in this answer. Assuming that you have installed and loaded the package in a Mathematica session, your example ($\mathbf{10}\times\mathbf{27}$in$SU(3)\$) is computed as follows: ReduceRepProduct[SU3, {{3, 0}, {2, 2}}] This ...

0

I haven't found out why the method presented in the question doesn't produce the right formula. But if you want the formula to present an arbitrary vector that's in 3d cartesian coordinate system as a linear combination of vectors forming a basis, or in other words, want to transform the vector from 3d cartesian basis to an arbitrary one, you can do this: ...

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