Is the following matrix positive definite?












3












$begingroup$


Is there an easy way to prove / disprove this?



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



where $X in mathbb{R}^{n times n}$ and $1 < k < n$






linear-algebra positive-definite symmetric-matrices







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$endgroup$

















    3












    $begingroup$


    Is there an easy way to prove / disprove this?



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



    where $X in mathbb{R}^{n times n}$ and $1 < k < n$






    linear-algebra positive-definite symmetric-matrices







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Is there an easy way to prove / disprove this?



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



      where $X in mathbb{R}^{n times n}$ and $1 < k < n$






      linear-algebra positive-definite symmetric-matrices







      share|cite|improve this question











      $endgroup$




      Is there an easy way to prove / disprove this?



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



      where $X in mathbb{R}^{n times n}$ and $1 < k < n$






      linear-algebra positive-definite symmetric-matrices




      linear-algebra positive-definite symmetric-matrices






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago







      avocado

















      asked 2 days ago









      avocadoavocado

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          $begingroup$

          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.






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            5












            $begingroup$

            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











            $endgroup$













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              2 Answers
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              2 Answers
              2






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              active

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              9












              $begingroup$

              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









              $endgroup$


















                9












                $begingroup$

                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









                $endgroup$
















                  9












                  9








                  9





                  $begingroup$

                  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 2 days ago









                  user1551user1551

                  72.1k566127




                  72.1k566127























                      5












                      $begingroup$

                      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











                      $endgroup$


















                        5












                        $begingroup$

                        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











                        $endgroup$
















                          5












                          5








                          5





                          $begingroup$

                          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 2 days ago

























                          answered 2 days ago









                          E-muE-mu

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