A question for the statisticians and other math lovers: Are all symmetric matrices with diagonal elements 1 and other values between $-1$ and 1 correlation matrices?

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6No, it must also be positive definite. – hard2fathom Jan 06 '19 at 15:01
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@hard2fathom, thank you for your answer! What is this? – Math123 Jan 06 '19 at 15:06
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Related: https://stats.stackexchange.com/questions/72790/bound-for-the-correlation-of-three-random-variables – Julius Vainora Jan 06 '19 at 17:47
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I thought this must be asked & answered before, but cannot find it, so here it goes ... Let $S$ be a covariance matrix (for the algebra that follows it does not matter if it is theoretical or empirical). Let $D$ be a diagonal matrix with the diagonal of $S$. Then the correlation matrix $R$ is given by $$ R= D^{-1/2} S D^{-1/2} $$ (how you can see this directly is explained here.)
To see that $S$ must be positive (semi)-definite (abbreviated posdef), let $X$ be a random variable with covariance amtrix $S$, and $c$ a vector. Then $$ \DeclareMathOperator{\var}{\mathbb{V}ar} \var(c^T X)= c^T S c \ge 0 $$ since variance is always nonnegative. Then this transfers to the correlation matrix: $$ c^T R c = c^T D^{-1/2} S D^{-1/2} c = (D^{-1/2} c)^T S (D^{-1/2} c) \ge 0 $$ Armed with this it is easy to make an counterexample, the following is not a correlation matrix: $$ \begin{pmatrix} 1 & -0.9 & -0.9 \\ -0.9& 1 & -0.9 \\ -0.9 & -0.9 & 1 \end{pmatrix} $$

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