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Your question seems to miss some points in What topics can I ask about here?. This answer is for helping you to write the code by yourself and if you encounter any problem, feel free to post another question.
In Mathematica use:
(* comment *) syntax for commenting your code
LowerTriangularize function for accessing the lower triangular part of a matrix
...
3
Try FindGeometricTransform to describe the transformation {x,y}<->{u,v}.
Therefore it's necessary to know the three points A,B,C (I added Buv ).
{Axy, Bxy, Cxy} = {{0.2, 0.8}, {0.1, 0.15}, {0.8, 0.25}}
{Auv, Buv, Cuv} = {{0, 75}, {0, -45.378 }, {100, 0}, {0, -45.378 }}
trafo = FindGeometricTransform[ {Auv, Buv, Cuv}, {Axy,
Bxy, Cxy}] [[2]]
Last ...
3
If you have got all your matrices in matrices then use ResourceFunction["LinearlyIndependent"] for example:
LinearlyIndependent = ResourceFunction["LinearlyIndependent"];
matrices = RandomInteger[20, {16, 4, 4}];
LinearlyIndependent[Flatten /@ matrices]
To extract your matrices you can do this:
onebigmatrix = RandomInteger[20, {4, 64}];
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2
As suggested by @bill s, you can write
x = Array[a, 3];
deriv = D[x . x, {x}]
(* {2 a[1], 2 a[2], 2 a[3]} *)
Note that the you need to put {x} rather than x (otherwise it will attempt to interpret the 2nd item in the list as the order of the derivative - see the help for D - the syntax is rather over loaded).
The easiest way to substitute values is perhaps
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2
ψ = {a, b};
ψ1 = Conjugate[ψ];
R = KroneckerProduct[ψ, ψ1]
(* {{a*Conjugate[a], a*Conjugate[b]},
{b*Conjugate[a], b*Conjugate[b]}} *)
R // TeXForm
$$
\left(
\begin{array}{cc}
a a^* & a b^* \\
b a^* & b b^* \\
\end{array}
\right)
$$
In Mathematica, there is no concept of row vectors vs. column vectors, so there is no need to ...
2
B = ArrayFlatten[Transpose[A, {3, 1, 4, 2}]];
Ai = Transpose[ArrayReshape[Inverse[B], {n, n, n, n}], {2, 1, 4, 3}];
1
Presumably you are in an affine space; given a point {p1,p2,p3} on your line and a vector {u1,u2,u3} on the line, any point on the line has coordinates {p1,p2,p3}+x*{u1,u2,u3}, where $x$ is real. You still have to determine what distribution you want for $x$ (saying "a random point" is not enough). If you choose a normal distribution you could ...
1
ComplexExpand assumes that all variables are real except those mentioned as second argument.
Try:
ψ = {α Cos[B t/2] +
I β Sin[B t/2] E^(-I ϕ), α I Sin[
B t/2] E^(I ϕ) + β Cos[B t/2]};
ComplexExpand[Conjugate[ψ], {α, b }] // Simplify
This gives:
1
Here is a solution that uses AffineTransform and Solves for coefficients. You can specify a scale.
{a, b, c, p} = {{0.2, 0.8}, {0.1, 0.15}, {0.8, 0.25}, {0.6, 0.7}};
(* Specify scale here; for example, u and v's lengths are 1/2 *)
scale = 1/2;
(* Define the parameters *)
ClearAll[u0, ux, uy, v0, vx, vy, sol, UV];
m := {{ux, uy}, {vx, vy}};
V := {u0, v0};
UV ...
1
Assume a,b,c are given as:
{a, b, c} = {{0.2, 0.8}, {0.1, 0.15}, {0.8, 0.25}};
The origin of the uv system is on a-b: orig= b+ x(a-b), where we need to determine x. This is done by using the fact that (c-orig) is perpendicular to (a-b):
orig = b + x (a - b) /. Solve[(b + x (a - b) - c).(a - b) == 0, x][[1]];
With the origin, it is easy to determine unit ...
1
To clear up ambiguities, the following code finds the transformation from the points in xy space $(0,1),(1,0)$ and $(0,0)$ to the points in uv space $a,b$ and $c$ projected onto $\overline{ab}$ respectively.
With[{a = {ax, ay}, b = {bx, by}, c = {cx, cy},
m = {{mx, mxy}, {myx, myy}}},
With[{v = Projection[c - b, a - b, Dot] + b},
m /. Simplify@
First@...
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