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Kurtosis is the fourth statistical moment of a random variable's distribution. Unlike the variance-covariance matrix $\Sigma$, which had a shape of $p\times p$, the kurtosis-cokurtosis matrix is shaped $p\times p^3$.

For bivariate data, $p=2$, the cokurtosis matrix is shaped $2\times 2^3 = 2\times 8$, $$K = \begin{pmatrix} k_{1,1,1,1} & k_{1,1,1,2} & k_{1,1,2,1} & k_{1,1,2,2} & k_{1,2,1,1} & k_{1,2,1,2} & k_{1,2,2,1} & k_{1,2,2,2} \\ k_{2,1,1,1} & k_{2,1,1,2} & s_{2,1,2,1} & k_{2,1,2,2} & k_{2,2,1,1} & k_{2,2,1,2} & k_{2,2,2,1} & k_{2,2,2,2} \\ \end{pmatrix} $$ with only $5$ of the $16$ elements being distinct. which of the elements are kurtosis values rather than cokurtosis values?

Does this matrix have a characteristic polynomial for getting its eigenvalues and eigenvectors? If so, how to find them.

This post suggests I should be looking for singular values and singular vectors instead of eigenvalues and eigenvectors since the matrix is non-square.

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  • $\begingroup$ Non-square matrices don't have characteristic polynomials, nor do they have eigenvectors and eigenvalues. You should change the question to: "How to find the singular values and singular vectors" instead, and you can do this numerically by taking the SVD of $K$. $\endgroup$ – Eric Perkerson Sep 9 '20 at 2:05
  • $\begingroup$ the singular values and singular vectors lead to the matrix's eigenvalues and eigenvectors though? $\endgroup$ – develarist Sep 9 '20 at 2:09
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    $\begingroup$ Not quite. The matrix doesn't have eigenvalues/vectors, so it doesn't make sense to talk about the singular values/vectors as "leading you" to something that doesn't exist. However, you can think of singular values as being a generalization of eigenvalues. What are you actually trying to do with this matrix $K$? It might be that you can use the singular values/vectors for your purpose. $\endgroup$ – Eric Perkerson Sep 9 '20 at 3:36

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