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












6












$begingroup$


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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  • 6




    $begingroup$
    No, it must also be positive definite.
    $endgroup$
    – hard2fathom
    Jan 6 at 15:01










  • $begingroup$
    @hard2fathom, thank you for your answer! What is this?
    $endgroup$
    – Math123
    Jan 6 at 15:06










  • $begingroup$
    Related: stats.stackexchange.com/questions/72790/…
    $endgroup$
    – Julius
    Jan 6 at 17:47
















6












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    No, it must also be positive definite.
    $endgroup$
    – hard2fathom
    Jan 6 at 15:01










  • $begingroup$
    @hard2fathom, thank you for your answer! What is this?
    $endgroup$
    – Math123
    Jan 6 at 15:06










  • $begingroup$
    Related: stats.stackexchange.com/questions/72790/…
    $endgroup$
    – Julius
    Jan 6 at 17:47














6












6








6


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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




correlation mathematical-statistics multivariate-analysis covariance covariance-matrix






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 15:30









kjetil b halvorsen

29.5k980216




29.5k980216










asked Jan 6 at 14:57









Math123Math123

311




311








  • 6




    $begingroup$
    No, it must also be positive definite.
    $endgroup$
    – hard2fathom
    Jan 6 at 15:01










  • $begingroup$
    @hard2fathom, thank you for your answer! What is this?
    $endgroup$
    – Math123
    Jan 6 at 15:06










  • $begingroup$
    Related: stats.stackexchange.com/questions/72790/…
    $endgroup$
    – Julius
    Jan 6 at 17:47














  • 6




    $begingroup$
    No, it must also be positive definite.
    $endgroup$
    – hard2fathom
    Jan 6 at 15:01










  • $begingroup$
    @hard2fathom, thank you for your answer! What is this?
    $endgroup$
    – Math123
    Jan 6 at 15:06










  • $begingroup$
    Related: stats.stackexchange.com/questions/72790/…
    $endgroup$
    – Julius
    Jan 6 at 17:47








6




6




$begingroup$
No, it must also be positive definite.
$endgroup$
– hard2fathom
Jan 6 at 15:01




$begingroup$
No, it must also be positive definite.
$endgroup$
– hard2fathom
Jan 6 at 15:01












$begingroup$
@hard2fathom, thank you for your answer! What is this?
$endgroup$
– Math123
Jan 6 at 15:06




$begingroup$
@hard2fathom, thank you for your answer! What is this?
$endgroup$
– Math123
Jan 6 at 15:06












$begingroup$
Related: stats.stackexchange.com/questions/72790/…
$endgroup$
– Julius
Jan 6 at 17:47




$begingroup$
Related: stats.stackexchange.com/questions/72790/…
$endgroup$
– Julius
Jan 6 at 17:47










1 Answer
1






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oldest

votes


















6












$begingroup$

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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    active

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    6












    $begingroup$

    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









    $endgroup$


















      6












      $begingroup$

      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









      $endgroup$
















        6












        6








        6





        $begingroup$

        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









        $endgroup$



        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 Jan 6 at 16:05









        kjetil b halvorsenkjetil b halvorsen

        29.5k980216




        29.5k980216






























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