# Why does this covariance matrix have a rank of $n$?

I'm reading this paper, which contains the following 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?

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).
• There might not be any $n$ involved as this might not be a sample covariance matrix – Juho Kokkala Nov 5 '15 at 11:19
• 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 5 '15 at 12:17