# What is an isotropic (spherical) covariance matrix?

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

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.

• What if the variables can be rotated to get to $\lambda \mathbf I$ covariance matrix? Mar 30, 2016 at 18:27
• @Aksakal See update. Mar 30, 2016 at 18:49
• +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!
– whuber
Mar 30, 2016 at 20:11
• @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? Mar 30, 2016 at 20:15
• 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).
– whuber
Mar 30, 2016 at 20:23

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.

• 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. Mar 30, 2016 at 18:24