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Could somebody explain to me in simple terms what an isotropic covariance matrix is? I can't find anything online.

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A covariance matrix $\mathbf C$ is called isotropic, or spherical, if it is proportionate to the identity matrix: $$\mathbf C = \lambda \mathbf I,$$ i.e. it is diagonal and all elements on the diagonal are equal.


This definition does not depend on the coordinate system; if we rotate coordinate system with an orthogonal rotation matrix $\mathbf V$, then the covariance matrix will transform into $$\mathbf V^\top \mathbf C \mathbf V = \mathbf V^\top \cdot \lambda \mathbf I \cdot\mathbf V = \mathbf V^\top \mathbf V \cdot \lambda \mathbf I = \lambda \mathbf I,$$ i.e. will stay the same.

Intuitively, isotropic covariance matrix corresponds to a "spherical" data cloud. A sphere remains a sphere after rotation.

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  • $\begingroup$ What if the variables can be rotated to get to $\lambda \mathbf I$ covariance matrix? $\endgroup$ – Aksakal Mar 30 '16 at 18:27
  • $\begingroup$ @Aksakal See update. $\endgroup$ – amoeba says Reinstate Monica Mar 30 '16 at 18:49
  • $\begingroup$ +1. But curiously, a completely different definition of "isotropic" also applies to $C$ because--as is usual with covariance matrices--it represents a quadratic form on a real vector space. But in this other sense, the only isotropic covariance matrix is the zero matrix! $\endgroup$ – whuber Mar 30 '16 at 20:11
  • $\begingroup$ @whuber Interesting! I did not remember that there exists a notion of "isotropic" quadratic forms. But reading the definition now, wouldn't any covariance matrix with at least one zero eigenvalue be "isotropic" in that sense? $\endgroup$ – amoeba says Reinstate Monica Mar 30 '16 at 20:15
  • $\begingroup$ You're right--I mis-specified the quantifier. By definition, an isotropic quadratic form has at least one nonzero isotropic vector (rather than all vectors being isotropic). $\endgroup$ – whuber Mar 30 '16 at 20:23
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The covariance is only a function of $|x - x'|$. You can find a definition there.

Edit: sorry I misread, for matrix, the right answer is amoeba's one.

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    $\begingroup$ The questions asks about covariance matrix. Of course a matrix can be seen as a function, but I guess this requires some elaboration for the OP. $\endgroup$ – amoeba says Reinstate Monica Mar 30 '16 at 18:24

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