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}}
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1$\begingroup$ BTW, column vectors are not a necessity. $\endgroup$– Αλέξανδρος ΖεγγCommented Jun 4, 2020 at 12:54
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$\begingroup$ @ΑλέξανδροςΖεγγ You're right: I realized this from the corresponding answers. $\endgroup$– BekasoCommented Jun 4, 2020 at 12:58
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2 Answers
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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}
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$\begingroup$ 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? $\endgroup$ Commented Jun 5, 2020 at 3:07
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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}
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$\begingroup$ 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. $\endgroup$– chrisCommented Jun 5, 2020 at 7:45