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I'm reading this paper, which contains the following covariance matrix:

Covariance Matrix

In the example there are six forecasters who estimate some quantity, and then we look at the covariances of those estimates.

I understand from this question that the rank of a covariance matrix is at most $n-1$, and I also understand that the rank of a matrix is also equal to its number of nonzero eigenvalues.

I entered this covariance matrix into MATLAB and it says the rank of the matrix is 6. MATLAB says this matrix also has 6 nonzero eigenvalues. What's going on?

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user1205901 - Reinstate Monica
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1 Answers1

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Don't confuse the sample size with the number of vectors for which you calculate the covariance.

You have $k=6$ forecasters, each of which provides some number $n$ of forecasts (which you don't give in your question). So you get a $k\times k$ covariance matrix, which can have a rank of at most $n-1$. But if $n>k$, then the matrix can certainly have full rank $k$, as it does here (it's block diagonal, and both blocks have nonzero determinant, so the matrix is nonsingular and has full rank).

Stephan Kolassa
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    There might not be any $n$ involved as this might not be a sample covariance matrix – Juho Kokkala Nov 05 '15 at 11:19
  • It is a sample covariance matrix, though. Actually the example in the paper is hypothetical but they're referring to a situation which would in the regular course of events involve a sample covariance matrix. – user1205901 - Reinstate Monica Nov 05 '15 at 11:52
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    Why is the rank tied to the number of forecasts $n$ rather than the number of forecasters $k$? I realise that it is true, but I don't understand why the number of forecasts should be relevant, since the size of the covariance matrix is completely determined by the number of forecasters. – user1205901 - Reinstate Monica Nov 05 '15 at 12:17
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    @user1205901: that's exactly [the question you link to](http://stats.stackexchange.com/q/178587/1352). See also [this question](http://stats.stackexchange.com/q/60622/1352), which the other one links to. – Stephan Kolassa Nov 05 '15 at 12:21
  • Thanks. Because the question I linked didn't have any answer I wasn't sure what to think of it, but the second link looks like it will resolve things. – user1205901 - Reinstate Monica Nov 05 '15 at 12:26
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    @user1205901: I have now provided an answer to the question you linked to; you might be interested to take a look. +1 to Stephan's answer, by the way. – amoeba Nov 05 '15 at 17:18