Is there any way to define the matrix M that achieves the following equation?
the two defined matrices are
{{(ω1 + ω4) / Sqrt[2]}, {(μ ω2 + ω3) / Sqrt[2]}, {(-μ ω2 + ω3) / Sqrt[2]}, {(-ω1 + ω4) / Sqrt[2]}} and
{{ω1}, {ω2}, {ω3}, {ω4}} $\begingroup$ $\endgroup$
2 
- 1$\begingroup$ BTW, column vectors are not a necessity. $\endgroup$Αλέξανδρος Ζεγγ– Αλέξανδρος Ζεγγ2020-06-04 12:54:00 +00:00Commented Jun 4, 2020 at 12:54
- $\begingroup$ @ΑλέξανδροςΖεγγ You're right: I realized this from the corresponding answers. $\endgroup$Bekaso– Bekaso2020-06-04 12:58:56 +00:00Commented Jun 4, 2020 at 12:58
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2 Answers
$\begingroup$ $\endgroup$
2 tt = {{(ω1 + ω4)/Sqrt[2]}, {(μ ω2 + ω3)/Sqrt[2]}, {(-μ ω2 + ω3)/Sqrt[2]}, {(-ω1 + ω4)/Sqrt[2]}} // Flatten mat =Last@ CoefficientArrays[tt, {ω1, ω2, ω3, ω4}] // Normal 
mat. {ω1, ω2, ω3, ω4} - tt // Simplify {0,0,0,0}
- $\begingroup$ Huh. Maybe I’m mistake but is this not just putting the initial vector/set of coupled equations in a matrix form? Seems magical but I know it’s simpler than it seems. Or maybe not? $\endgroup$CA Trevillian– CA Trevillian2020-06-05 03:07:48 +00:00Commented Jun 5, 2020 at 3:07
- $\begingroup$ @CATrevillian yes you are right. $\endgroup$chris– chris2020-06-05 07:43:38 +00:00Commented Jun 5, 2020 at 7:43
$\begingroup$ $\endgroup$
1 CoefficientArrays[] works very well for this:
m1 = Normal[Last[CoefficientArrays[{(ω1 + ω4)/Sqrt[2], (μ ω2 + ω3)/Sqrt[2], (-μ ω2 + ω3)/Sqrt[2], (-ω1 + ω4)/Sqrt[2]}, {ω1, ω2, ω3, ω4}]]] {{1/Sqrt[2], 0, 0, 1/Sqrt[2]}, {0, μ/Sqrt[2], 1/Sqrt[2], 0}, {0, -(μ/Sqrt[2]), 1/Sqrt[2], 0}, {-(1/Sqrt[2]), 0, 0, 1/Sqrt[2]}} Note how I used vectors instead of column matrices.
m1.{ω1, ω2, ω3, ω4} - {(ω1 + ω4)/Sqrt[2], (μ ω2 + ω3)/Sqrt[2], (-μ ω2 + ω3)/Sqrt[2], (-ω1 + ω4)/Sqrt[2]} {0, 0, 0, 0} - $\begingroup$ oops... I simplified my silly answer and I now notice its exactly the same as yours. Oh well. Don't know why the OP changed his mind about giving you the credit. $\endgroup$chris– chris2020-06-05 07:45:13 +00:00Commented Jun 5, 2020 at 7:45