Are all symmetric matrices with diagonal elements 1 and other values between -1 and 1 correlation matrices?












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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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  • 5




    No, it must also be positive definite.
    – hard2fathom
    yesterday










  • @hard2fathom, thank you for your answer! What is this?
    – Math123
    yesterday










  • Related: stats.stackexchange.com/questions/72790/…
    – Julius
    yesterday
















6














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?










share|cite|improve this question









New contributor




Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 5




    No, it must also be positive definite.
    – hard2fathom
    yesterday










  • @hard2fathom, thank you for your answer! What is this?
    – Math123
    yesterday










  • Related: stats.stackexchange.com/questions/72790/…
    – Julius
    yesterday














6












6








6


1





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?










share|cite|improve this question









New contributor




Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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?







correlation mathematical-statistics multivariate-analysis covariance covariance-matrix






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Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question









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edited yesterday









kjetil b halvorsen

28.9k980208




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Math123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 5




    No, it must also be positive definite.
    – hard2fathom
    yesterday










  • @hard2fathom, thank you for your answer! What is this?
    – Math123
    yesterday










  • Related: stats.stackexchange.com/questions/72790/…
    – Julius
    yesterday














  • 5




    No, it must also be positive definite.
    – hard2fathom
    yesterday










  • @hard2fathom, thank you for your answer! What is this?
    – Math123
    yesterday










  • Related: stats.stackexchange.com/questions/72790/…
    – Julius
    yesterday








5




5




No, it must also be positive definite.
– hard2fathom
yesterday




No, it must also be positive definite.
– hard2fathom
yesterday












@hard2fathom, thank you for your answer! What is this?
– Math123
yesterday




@hard2fathom, thank you for your answer! What is this?
– Math123
yesterday












Related: stats.stackexchange.com/questions/72790/…
– Julius
yesterday




Related: stats.stackexchange.com/questions/72790/…
– Julius
yesterday










1 Answer
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6














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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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

    oldest

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    6














    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:
    $$
    begin{pmatrix} 1 & -0.9 & -0.9 \
    -0.9& 1 & -0.9 \
    -0.9 & -0.9 & 1 end{pmatrix}
    $$






    share|cite|improve this answer


























      6














      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:
      $$
      begin{pmatrix} 1 & -0.9 & -0.9 \
      -0.9& 1 & -0.9 \
      -0.9 & -0.9 & 1 end{pmatrix}
      $$






      share|cite|improve this answer
























        6












        6








        6






        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:
        $$
        begin{pmatrix} 1 & -0.9 & -0.9 \
        -0.9& 1 & -0.9 \
        -0.9 & -0.9 & 1 end{pmatrix}
        $$






        share|cite|improve this answer












        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:
        $$
        begin{pmatrix} 1 & -0.9 & -0.9 \
        -0.9& 1 & -0.9 \
        -0.9 & -0.9 & 1 end{pmatrix}
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        kjetil b halvorsenkjetil b halvorsen

        28.9k980208




        28.9k980208






















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