[Math] Is a correlation matrix with positive determinant PSD

correlationdeterminantmatricespositive-semidefinite

Please note: I'm not interested in the difference between positive definiteness and semi-definiteness for this question.

A correlation matrix is a symmetric positive semi-definite matrix with 1s down the diagonal and off-diagonal terms $ -1 \leq M_{ij} \leq 1$.

Since a correlation matrix must be positive semi-definite, it must have a positive (or zero) determinant, but does a positive determinant imply positive definiteness? In other words, if I have a matrix with 1s down the diagonal, off-diagonals satisfying $ -1 \leq M_{ij} \leq 1$ and positive determinant, is that enough to prove that the matrix is positive definite (and thus an acceptable correlation matrix)?

Thank you.

Best Answer

The answer to your question, as I now understand it, is no. In particular, we can construct a matrix of you particular pattern with a positive determinant that fails to be positive definite.

In particular, consider the matrix

$$ M = \pmatrix{ 1&-1&-1&0&0&0\\ -1&1&-1&0&0&0\\ -1&-1&1&0&0&0\\ 0&0&0&1&-1&-1\\ 0&0&0&-1&1&-1\\ 0&0&0&-1&-1&1\\ } $$ which has eigenvalues $-1-1,2,2,2,2$

Related Solutions

[Math] Necessary and sufficient conditions for a matrix to be a valid correlation matrix.

Yes, although the restriction that all entries are between $-1$ and $1$ is implied by the other properties (and so is not needed).

Let $\Sigma$ be a $n \times n$, symmetric, positive semidefinite matrix with $1$'s along the main diagonal.

First, $\Sigma$ is a covariance matrix. Since $\Sigma$ is symmetric and positive semidefinite, $\Sigma$ has a nonnegative symmetric square root $\Sigma^{1/2}$. Let $X$ be a $n$-vector of independent random variables, each with variance $1$. (For example, $X$ could be an $n$-vector of independent $N(0,1)$ random variables.) Construct the $n$-vector $\Sigma^{1/2} X$. Then, by properties of covariance matrices, $$\text{cov} (\Sigma^{1/2} X) = \Sigma^{1/2} \text{cov}(X) \Sigma^{1/2} = \Sigma^{1/2} I \Sigma^{1/2} = \Sigma.$$ Thus $\Sigma$ is the covariance matrix for the random vector $\Sigma^{1/2} X$. (This derivation is on the Wikipedia article for covariance matrices.)

Second, since $\Sigma$ has $1$'s on its diagonal, the standard deviation of each random variable in $\Sigma^{1/2} X$ is $1$. Thus the correlation matrix $R$ for the random vector $\Sigma^{1/2} X$ is $$R = I^{-1} \Sigma I^{-1} = \Sigma.$$ Thus $\Sigma$ is a correlation matrix as well.

[Math] Diagonal dominance versus positive semi-definiteness

Quick counter example

>>> a=2*ones(3,3)+eye(3)
a =

   3   2   2
   2   3   2
   2   2   3

>>> eig(a)
ans =

   1.00000
   1.00000
   7.00000

Best Answer

Related Solutions

[Math] Necessary and sufficient conditions for a matrix to be a valid correlation matrix.

[Math] Diagonal dominance versus positive semi-definiteness

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