It is well-known that a covariance matrix must be semi-positive definite, however, is the converse true?

That is, does every semi-positive definite matrix correspond to a covariance matrix?


1 Answer 1


Going by the definitions of PD and PSD here, yes, I think so, since we can do this by construction. I'll assume for a slightly simpler argument that you mean for matrices with real elements, but with appropriate changes it would extend to complex matrices.

Let $A$ be some real PSD matrix; from the definition I linked to, it will be symmetric. Any real symmetric positive definite matrix $A$ can be written as $A = LL^T$. This can be done by $L=Q\sqrt{D}Q^T$ if $A=QDQ^T$ with orthogonal $Q$ and diagonal $D$ and $\sqrt{D}$ as matrix of component wise square roots of $D$. Thus, it needn't be full rank.

Let $Z$ be some vector random variable, of the appropriate dimension, with covariance matrix $I$ (which is easy to create).

Then $LZ$ has covariance matrix $A$.

[At least that's in theory. In practice there'd be various numerical issues to deal with if you wanted good results, and - because of the usual problems with floating point calculation - you'd only approximately get what you need; that is, the population variance of a computed $LZ$ usually wouldn't be exactly $A$. But this sort of thing is always an issue when we come to actually calculate things]

  • $\begingroup$ While it is true that a decomposition $A=LL'$ is possible without full rank, the Cholesky algorithm only works with regular $A$. So without full rank, it cannot be a Cholesky decomposition. Computationally, one could do this decomposition in the singular case by diagonalization. (Although this is far more expensive) $\endgroup$ Aug 13, 2014 at 12:14
  • $\begingroup$ @Horst: Why would $L=Q\sqrt{D}Q^T$ be lower triangular? $\endgroup$
    – amoeba
    Aug 13, 2014 at 12:41
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    $\begingroup$ @amoeba While one could organize it so it is, it doesn't have to be lower triangular for the argument to work - it's a feature of the Cholesky but it's not required for the result to work. $\endgroup$
    – Glen_b
    Aug 13, 2014 at 13:02
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    $\begingroup$ @Glen Is being symmetric a necessary condition for being PSD or is that definition one of many? $\endgroup$
    – 114
    Aug 13, 2014 at 13:42
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    $\begingroup$ @114 for the relation between symmetric and PSD see math.stackexchange.com/questions/516533/… $\endgroup$
    – Frank
    Aug 13, 2014 at 22:13

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