0
$\begingroup$

How can I reduce this covariance matrix? And also plot its eigenvalues? and how to compute its eigenvectors? Can we compute its eigenvalues and eigenvectors without reducing in compact form?

{{x + Cosh[2*t], -I, (x + Sinh[2*t])/Sqrt[2], 0, -((-3*x + Sinh[2*t])/Sqrt[2]), 0}
,{I, x + Cosh[2*t], 0, (x + Sinh[2*t])/Sqrt[2], 0, (-3*x + Sinh[2*t])/Sqrt[2]}
,{(x + Sinh[2*t])/Sqrt[2], 0, (1 + x + Cosh[2*t])/2, -I, (1 + 3*x - Cosh[2*t])/2, 0}
,{0, (x + Sinh[2*t])/Sqrt[2], I, (1 + x + Cosh[2*t])/2, 0, (-1 - 3*x + Cosh[2*t])/2}
,{-((-3*x + Sinh[2*t])/Sqrt[2]), 0, (1 + 3*x - Cosh[2*t])/2, 0, (1 + 9*x + Cosh[2*t])/2,-I}
,{0, (-3*x + Sinh[2*t])/Sqrt[2], 0, (-1 - 3*x + Cosh[2*t])/2, I, (1 + 9*x + Cosh[2*t])/2}
}
$\endgroup$
1
  • 1
    $\begingroup$ Eigenvalues, Eigenvectors, or both combined, Eigensystem. $\endgroup$ Jun 19, 2018 at 11:42

1 Answer 1

1
$\begingroup$

Here's a shot at plotting the eigenvalues:

evs = Eigenvalues[m]

{1/2 Root[-64 x-80 x Cosh[2 t]-48 x Sinh[2 t]+(4+44 x+12 Cosh[2 t]+40 x Cosh[2 t]+8 x Sinh[2 t]) #1+(-4-12 x-4 Cosh[2 t]) #1^2+#1^3&,1]

,1/2 Root[-64 x-80 x Cosh[2 t]-48 x Sinh[2 t]+(4+44 x+12 Cosh[2 t]+40 x Cosh[2 t]+8 x Sinh[2 t]) #1+(-4-12 x-4 Cosh[2 t]) #1^2+#1^3&,2]

,1/2 Root[-64 x-80 x Cosh[2 t]-48 x Sinh[2 t]+(4+44 x+12 Cosh[2 t]+40 x Cosh[2 t]+8 x Sinh[2 t]) #1+(-4-12 x-4 Cosh[2 t]) #1^2+#1^3&,3]

,1/2 Root[-16-48 x+16 Cosh[2 t]+48 x Cosh[2 t]+16 x Sinh[2 t]+(-4-28 x+4 Cosh[2 t]+40 x Cosh[2 t]+8 x Sinh[2 t]) #1+(-12 x-4 Cosh[2 t]) #1^2+#1^3&,1]

,1/2 Root[-16-48 x+16 Cosh[2 t]+48 x Cosh[2 t]+16 x Sinh[2 t]+(-4-28 x+4 Cosh[2 t]+40 x Cosh[2 t]+8 x Sinh[2 t]) #1+(-12 x-4 Cosh[2 t]) #1^2+#1^3&,2]

,1/2 Root[-16-48 x+16 Cosh[2 t]+48 x Cosh[2 t]+16 x Sinh[2 t]+(-4-28 x+4 Cosh[2 t]+40 x Cosh[2 t]+8 x Sinh[2 t]) #1+(-12 x-4 Cosh[2 t]) #1^2+#1^3&,3]}

Plot3D[Evaluate@evs, {x, -2, 2}, {t, 0, 2},
  PlotRange -> {-10, 20},  AxesLabel -> {"x", "t"},
  PlotStyle -> Opacity[0.3], Mesh -> None
]

Eigenvalues

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.