Is the following matrix positive definite?












1












$begingroup$


Is there an easy way to prove/disprove this?



$
(X)_{ij} =
begin{cases}
dfrac{k}{n} &quadtext{if i = j} \
dfrac{k(k-1)}{n(n-1)} &quadtext{Otherwise} \
end{cases}
$



where $X in R^{nXn}$ and $1 lt k lt n$






linear-algebra positive-definite







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$endgroup$

















    1












    $begingroup$


    Is there an easy way to prove/disprove this?



    $
    (X)_{ij} =
    begin{cases}
    dfrac{k}{n} &quadtext{if i = j} \
    dfrac{k(k-1)}{n(n-1)} &quadtext{Otherwise} \
    end{cases}
    $



    where $X in R^{nXn}$ and $1 lt k lt n$






    linear-algebra positive-definite







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Is there an easy way to prove/disprove this?



      $
      (X)_{ij} =
      begin{cases}
      dfrac{k}{n} &quadtext{if i = j} \
      dfrac{k(k-1)}{n(n-1)} &quadtext{Otherwise} \
      end{cases}
      $



      where $X in R^{nXn}$ and $1 lt k lt n$






      linear-algebra positive-definite







      share|cite|improve this question









      $endgroup$




      Is there an easy way to prove/disprove this?



      $
      (X)_{ij} =
      begin{cases}
      dfrac{k}{n} &quadtext{if i = j} \
      dfrac{k(k-1)}{n(n-1)} &quadtext{Otherwise} \
      end{cases}
      $



      where $X in R^{nXn}$ and $1 lt k lt n$






      linear-algebra positive-definite




      linear-algebra positive-definite






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 hours ago









      avocadoavocado

      204




      204






















          2 Answers
          2






          active

          oldest

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          3












          $begingroup$

          It's positive definite because it is in the form of $aI+bee^T$, where $e=(1,1,ldots,1)^T$ and $a,b$ are some positive scalars.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot, user1551.
            $endgroup$
            – avocado
            1 hour ago



















          1












          $begingroup$

          You can also do it by direct computation,



          Suppose $mathbf{x} := (x_1,dots,x_n) in mathbb{R}^n$, with $mathbf{x} neq mathbf{0} in mathbb{R}^n.$ Then the $i^{text{th}}$ component of $Xmathbf{x}$, which we write as $(Xmathbf{x})_i$ is given by begin{align}(Xmathbf{x})_i &= frac kn x_i + sum_{j neq i} frac{k(k-1)}{n(n-1)}x_j \
          &= frac kn left(x_i + frac{k-1}{n-1}sum_{j neq i} x_jright).
          end{align}
          So begin{align}mathbf{x}^T X mathbf{x} &= sum_{i=1}^n x_i (X mathbf{x})_i \
          &= frac kn sum_{i=1}^nleft( x_i^2 + frac{k-1}{n-1} sum_{j neq i}x_i x_j right) \
          &= frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right).
          end{align}



          In the case that $sum_{i=1}^n sum_{j neq i} x_i x_j geq 0,$ we easily get $mathbf{x}^T X mathbf{x}>0$.



          However, when $sum_{i=1}^n sum_{j neq i} x_i x_j < 0, $ we have
          begin{align}
          frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right) &> frac kn left(sum_{i=1}^n x_i^2 + sum_{i=1}^n sum_{j neq i} x_i x_j right)\
          &= frac kn left(sum_{i=1}^n x_i right)^2 \
          &> 0.
          end{align}
          So in either case, we have $mathbf{x}^T X mathbf{x} >0$. Hence, $X$ is positive definite.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks a lot, E-mu.
            $endgroup$
            – avocado
            1 hour ago













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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          It's positive definite because it is in the form of $aI+bee^T$, where $e=(1,1,ldots,1)^T$ and $a,b$ are some positive scalars.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot, user1551.
            $endgroup$
            – avocado
            1 hour ago
















          3












          $begingroup$

          It's positive definite because it is in the form of $aI+bee^T$, where $e=(1,1,ldots,1)^T$ and $a,b$ are some positive scalars.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot, user1551.
            $endgroup$
            – avocado
            1 hour ago














          3












          3








          3





          $begingroup$

          It's positive definite because it is in the form of $aI+bee^T$, where $e=(1,1,ldots,1)^T$ and $a,b$ are some positive scalars.






          share|cite|improve this answer









          $endgroup$



          It's positive definite because it is in the form of $aI+bee^T$, where $e=(1,1,ldots,1)^T$ and $a,b$ are some positive scalars.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 1 hour ago









          user1551user1551

          72k566126




          72k566126












          • $begingroup$
            Thanks a lot, user1551.
            $endgroup$
            – avocado
            1 hour ago


















          • $begingroup$
            Thanks a lot, user1551.
            $endgroup$
            – avocado
            1 hour ago
















          $begingroup$
          Thanks a lot, user1551.
          $endgroup$
          – avocado
          1 hour ago




          $begingroup$
          Thanks a lot, user1551.
          $endgroup$
          – avocado
          1 hour ago











          1












          $begingroup$

          You can also do it by direct computation,



          Suppose $mathbf{x} := (x_1,dots,x_n) in mathbb{R}^n$, with $mathbf{x} neq mathbf{0} in mathbb{R}^n.$ Then the $i^{text{th}}$ component of $Xmathbf{x}$, which we write as $(Xmathbf{x})_i$ is given by begin{align}(Xmathbf{x})_i &= frac kn x_i + sum_{j neq i} frac{k(k-1)}{n(n-1)}x_j \
          &= frac kn left(x_i + frac{k-1}{n-1}sum_{j neq i} x_jright).
          end{align}
          So begin{align}mathbf{x}^T X mathbf{x} &= sum_{i=1}^n x_i (X mathbf{x})_i \
          &= frac kn sum_{i=1}^nleft( x_i^2 + frac{k-1}{n-1} sum_{j neq i}x_i x_j right) \
          &= frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right).
          end{align}



          In the case that $sum_{i=1}^n sum_{j neq i} x_i x_j geq 0,$ we easily get $mathbf{x}^T X mathbf{x}>0$.



          However, when $sum_{i=1}^n sum_{j neq i} x_i x_j < 0, $ we have
          begin{align}
          frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right) &> frac kn left(sum_{i=1}^n x_i^2 + sum_{i=1}^n sum_{j neq i} x_i x_j right)\
          &= frac kn left(sum_{i=1}^n x_i right)^2 \
          &> 0.
          end{align}
          So in either case, we have $mathbf{x}^T X mathbf{x} >0$. Hence, $X$ is positive definite.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks a lot, E-mu.
            $endgroup$
            – avocado
            1 hour ago


















          1












          $begingroup$

          You can also do it by direct computation,



          Suppose $mathbf{x} := (x_1,dots,x_n) in mathbb{R}^n$, with $mathbf{x} neq mathbf{0} in mathbb{R}^n.$ Then the $i^{text{th}}$ component of $Xmathbf{x}$, which we write as $(Xmathbf{x})_i$ is given by begin{align}(Xmathbf{x})_i &= frac kn x_i + sum_{j neq i} frac{k(k-1)}{n(n-1)}x_j \
          &= frac kn left(x_i + frac{k-1}{n-1}sum_{j neq i} x_jright).
          end{align}
          So begin{align}mathbf{x}^T X mathbf{x} &= sum_{i=1}^n x_i (X mathbf{x})_i \
          &= frac kn sum_{i=1}^nleft( x_i^2 + frac{k-1}{n-1} sum_{j neq i}x_i x_j right) \
          &= frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right).
          end{align}



          In the case that $sum_{i=1}^n sum_{j neq i} x_i x_j geq 0,$ we easily get $mathbf{x}^T X mathbf{x}>0$.



          However, when $sum_{i=1}^n sum_{j neq i} x_i x_j < 0, $ we have
          begin{align}
          frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right) &> frac kn left(sum_{i=1}^n x_i^2 + sum_{i=1}^n sum_{j neq i} x_i x_j right)\
          &= frac kn left(sum_{i=1}^n x_i right)^2 \
          &> 0.
          end{align}
          So in either case, we have $mathbf{x}^T X mathbf{x} >0$. Hence, $X$ is positive definite.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks a lot, E-mu.
            $endgroup$
            – avocado
            1 hour ago
















          1












          1








          1





          $begingroup$

          You can also do it by direct computation,



          Suppose $mathbf{x} := (x_1,dots,x_n) in mathbb{R}^n$, with $mathbf{x} neq mathbf{0} in mathbb{R}^n.$ Then the $i^{text{th}}$ component of $Xmathbf{x}$, which we write as $(Xmathbf{x})_i$ is given by begin{align}(Xmathbf{x})_i &= frac kn x_i + sum_{j neq i} frac{k(k-1)}{n(n-1)}x_j \
          &= frac kn left(x_i + frac{k-1}{n-1}sum_{j neq i} x_jright).
          end{align}
          So begin{align}mathbf{x}^T X mathbf{x} &= sum_{i=1}^n x_i (X mathbf{x})_i \
          &= frac kn sum_{i=1}^nleft( x_i^2 + frac{k-1}{n-1} sum_{j neq i}x_i x_j right) \
          &= frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right).
          end{align}



          In the case that $sum_{i=1}^n sum_{j neq i} x_i x_j geq 0,$ we easily get $mathbf{x}^T X mathbf{x}>0$.



          However, when $sum_{i=1}^n sum_{j neq i} x_i x_j < 0, $ we have
          begin{align}
          frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right) &> frac kn left(sum_{i=1}^n x_i^2 + sum_{i=1}^n sum_{j neq i} x_i x_j right)\
          &= frac kn left(sum_{i=1}^n x_i right)^2 \
          &> 0.
          end{align}
          So in either case, we have $mathbf{x}^T X mathbf{x} >0$. Hence, $X$ is positive definite.






          share|cite|improve this answer











          $endgroup$



          You can also do it by direct computation,



          Suppose $mathbf{x} := (x_1,dots,x_n) in mathbb{R}^n$, with $mathbf{x} neq mathbf{0} in mathbb{R}^n.$ Then the $i^{text{th}}$ component of $Xmathbf{x}$, which we write as $(Xmathbf{x})_i$ is given by begin{align}(Xmathbf{x})_i &= frac kn x_i + sum_{j neq i} frac{k(k-1)}{n(n-1)}x_j \
          &= frac kn left(x_i + frac{k-1}{n-1}sum_{j neq i} x_jright).
          end{align}
          So begin{align}mathbf{x}^T X mathbf{x} &= sum_{i=1}^n x_i (X mathbf{x})_i \
          &= frac kn sum_{i=1}^nleft( x_i^2 + frac{k-1}{n-1} sum_{j neq i}x_i x_j right) \
          &= frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right).
          end{align}



          In the case that $sum_{i=1}^n sum_{j neq i} x_i x_j geq 0,$ we easily get $mathbf{x}^T X mathbf{x}>0$.



          However, when $sum_{i=1}^n sum_{j neq i} x_i x_j < 0, $ we have
          begin{align}
          frac kn left(sum_{i=1}^n x_i^2 + frac{k-1}{n-1}sum_{i=1}^n sum_{j neq i} x_i x_j right) &> frac kn left(sum_{i=1}^n x_i^2 + sum_{i=1}^n sum_{j neq i} x_i x_j right)\
          &= frac kn left(sum_{i=1}^n x_i right)^2 \
          &> 0.
          end{align}
          So in either case, we have $mathbf{x}^T X mathbf{x} >0$. Hence, $X$ is positive definite.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          E-muE-mu

          515314




          515314












          • $begingroup$
            Thanks a lot, E-mu.
            $endgroup$
            – avocado
            1 hour ago




















          • $begingroup$
            Thanks a lot, E-mu.
            $endgroup$
            – avocado
            1 hour ago


















          $begingroup$
          Thanks a lot, E-mu.
          $endgroup$
          – avocado
          1 hour ago






          $begingroup$
          Thanks a lot, E-mu.
          $endgroup$
          – avocado
          1 hour ago




















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