# How to find eigenvalues and eigenvectors of the cokurtosis matrix?

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.

• 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 $K$. – Eric Perkerson Sep 9 '20 at 2:05
• the singular values and singular vectors lead to the matrix's eigenvalues and eigenvectors though? – develarist Sep 9 '20 at 2:09
• 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. – Eric Perkerson Sep 9 '20 at 3:36