6

You can divide both sides by T. This reduces the number of variables to 2: $$ j=\frac{J}{T},\quad b=\frac{B}{T}.$$ To visualize inequality, one can use ContourPlot. The inequality is fulfilled in the green area. ContourPlot[Sqrt[b^2 + j^2] Cosh[2 j] - j Sinh[2 Sqrt[b^2 + j^2]],{j,0,1},{b,0,1}, PlotRange->{-3,1}, Contours->20, ColorFunctionScaling->...


4

LeastSquare seems to use MKL lib internally and is already parallelized. Run the following code and observe cpu load: SeedRandom[1] ; size = 3000 ; m = RandomReal[{-1,1},{size,size}] ; b = RandomReal[{-1,1},size] ; LeastSquares[m,b] ; // AbsoluteTiming Check your parallel options and compare with observed cpu load: SystemOptions["ParallelOptions"] ...


4

Mathematica is correct. In general, there will only be the trivial solution. You are trying so solve an equation $ M x = b $ with $ b= 0$. This will have a nontrivial solution if and only if $\mathrm{det} M = 0$, because otherwise the matrix can be inverted, i.e. there exists a matrix $M^{-1}$ such that $M M^{-1} = M^{-1} M = I$, where $I$ is the identity ...


2

d = 3; Clear[A]; Clear[theta] Array[A, {d, d}]; Array[theta, {d, d}]; For[m = 0, m < d, m++, For[n = 0, n < d, n++, A[m, n] = Exp[Sum[theta[i, j], {i, 0, m}, {j, 0, n}]] ] ]; (mat = Table[A[i, j], {i, 0, d - 1}, {j, 0, d - 1}]) // MatrixForm Dimensions[mat] (* {3, 3} *) Det[mat] // Simplify


2

The equations are consistent, as can be shown as follows. First, note that the first six expressions, {g1, g2, g3, g4, g5, g6}, are linear in {px, py, pz, ptx, pty, ptz}. The coefficient matrix of the first six expressions with respect to these variables is m = Last@CoefficientArrays[{g1, g2, g3, g4, g5, g6}, {px, py, pz, ptx, pty, ptz}] // Normal; The ...


2

Combining equations we obtain a conventional linear equation in fully unique way without any constraints on the matrices: $$\left\{ \begin{array}{l}XA^\dagger C^\dagger=ABJD^\dagger-BJB^\dagger C^\dagger,\\ X C^\dagger + B J D^\dagger = 0;\end{array}\right.$$ or transposing $$\left\{ \begin{array}{l}CAX^\dagger=D J B^\dagger A^\dagger-C B JB^\dagger,\\ C X^\...


1

Instead of Labeled, use Text Clear["Global`*"]; eqns = {(1 - x)/2, -2 - 3 x}; pt = {x, eqns[[1]]} /. Solve[Equal @@ eqns, x, Reals][[1]] (* {-1, 1} *) plot = Plot[eqns, {x, -3, 1}, PlotRange -> {-1, 3}, PlotStyle -> Thick, PlotLabels -> {y == (1 - x)/2, y == -2 - 3 x}, AxesLabel -> {x, y}, ImageSize -> {Automatic, 200}];...


1

With Method->Reduce you get the result as ConditionalExpressions. Solve[Simplify[ a.X + X.a\[ConjugateTranspose] + b.ji.b\[ConjugateTranspose]] == ConstantArray[0, {n, n}] && Simplify[X.c\[ConjugateTranspose] + b.ji.d\[ConjugateTranspose]] == ConstantArray[0, {n, no}], Flatten[X], Method -> Reduce] Alternatively, you can ...


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