New answers tagged linear-algebra
0
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@...
8
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
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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}];
...
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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1
A simple way to solve your equations is:
e1 == W1 /. {x -> 0};
e2 == (D[W1, {x, 1}]) /. {x -> 1};
e3 = (W1 /. {x -> z}) - (W2 /. {x -> z});
e4 = ((D[W1, {x}]) /. {x -> z}) - ((D[W2, {x}]) /. {x -> z});
Solve[{e1, e2, e3, e4} == {0, 0, 0, 1}, {A1, A2, B1, B2}]
However, look at your equations:
TableForm[Transpose[Append[Transpose[R], {0, 0, ...
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