If I have a positive definite matrix X. How do i show that X$^2$ and X$^{-1}$ are also positive definite?












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To my understanding a positive definite matrix is a real symmetric square matrix where all eigenvalues are positive.
Therefore for a matrix A and vector v



$Av = {lambda}v$ where $lambda$ is an eigenvalue of A



so $det({lambda}{I_n} - A)=0$.



My next thought would trying to see how the formula changes when A is now A$^2$ and A$^{-1}$, but I'm getting nowhere.






linear-algebra matrices eigenvalues-eigenvectors positive-definite symmetric-matrices







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    $begingroup$


    To my understanding a positive definite matrix is a real symmetric square matrix where all eigenvalues are positive.
    Therefore for a matrix A and vector v



    $Av = {lambda}v$ where $lambda$ is an eigenvalue of A



    so $det({lambda}{I_n} - A)=0$.



    My next thought would trying to see how the formula changes when A is now A$^2$ and A$^{-1}$, but I'm getting nowhere.






    linear-algebra matrices eigenvalues-eigenvectors positive-definite symmetric-matrices







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      To my understanding a positive definite matrix is a real symmetric square matrix where all eigenvalues are positive.
      Therefore for a matrix A and vector v



      $Av = {lambda}v$ where $lambda$ is an eigenvalue of A



      so $det({lambda}{I_n} - A)=0$.



      My next thought would trying to see how the formula changes when A is now A$^2$ and A$^{-1}$, but I'm getting nowhere.






      linear-algebra matrices eigenvalues-eigenvectors positive-definite symmetric-matrices







      share|cite|improve this question









      $endgroup$




      To my understanding a positive definite matrix is a real symmetric square matrix where all eigenvalues are positive.
      Therefore for a matrix A and vector v



      $Av = {lambda}v$ where $lambda$ is an eigenvalue of A



      so $det({lambda}{I_n} - A)=0$.



      My next thought would trying to see how the formula changes when A is now A$^2$ and A$^{-1}$, but I'm getting nowhere.






      linear-algebra matrices eigenvalues-eigenvectors positive-definite symmetric-matrices




      linear-algebra matrices eigenvalues-eigenvectors positive-definite symmetric-matrices






      share|cite|improve this question













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      asked Jan 8 at 16:03









      L GL G

      248




      248






















          3 Answers
          3






          active

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          $begingroup$

          Another definition is that a symmetric matrix $M in mathbb R^{n times n}$ is positive definite if and only if $v^T M v > 0$ for all nonzero vectors $v in mathbb R^n$.



          Notice that
          begin{align}
          v^T A^2 v &= v^T A^T A v \
          &= u^T u \
          &= |u|^2
          end{align}
          where $u = Av$.
          If $v neq 0$ then $u neq 0$, so $|u|^2 > 0$. This shows that $A^2$ is positive definite.



          We can show that $A^{-1}$ is positive definite by noting that
          $$
          v^T A^{-1} v = v^T A^{-1} A A^{-1} v = u^T A u
          $$
          where $u = A^{-1}v$. If $v neq 0$ then $u neq 0$, so $u^T A u > 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks a lot this really clears it up, i forgot about that definition but it makes perfect sense
            $endgroup$
            – L G
            Jan 8 at 16:33



















          1












          $begingroup$

          For a pos.def. matrix $A$, you have (necessary and sufficient) that for any real vector $v$, that $v A v ge 0$.



          Now $ v A^{-1} A v = v (A^{-1} A A^{-1}) v =(v A^{-1}) A (A^{-1} v ) = u A u > 0 $.



          $A^2$ is even easier: $v A^2 v = v A A v = (v A) (A v) = u^2> 0$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Suppose that the eigenvalues of $A$ are $lambda_1,ldots,lambda_n$ with eigenvectors $v_1,ldots,v_n$. Note that $$A^2 v_i = lambda_i A v_i$ = lambda_i^2 v_i$$ and that $$Av_i = lambda_i v_i Rightarrow A^{-1}v_i = lambda^{-1}v_i$$ Therefore the eigenvalues of $A^2$ are $lambda_1^2,ldots,lambda_n^2$ and the eigenvalues of $A^{-1}$ are $lambda_1^{-1},ldots,lambda_n^{-1}$. This should be enough to prove that $A^2$ and $A^{-1}$ are positive definite.






            share|cite|improve this answer









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              3 Answers
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              active

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              3 Answers
              3






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              2












              $begingroup$

              Another definition is that a symmetric matrix $M in mathbb R^{n times n}$ is positive definite if and only if $v^T M v > 0$ for all nonzero vectors $v in mathbb R^n$.



              Notice that
              begin{align}
              v^T A^2 v &= v^T A^T A v \
              &= u^T u \
              &= |u|^2
              end{align}
              where $u = Av$.
              If $v neq 0$ then $u neq 0$, so $|u|^2 > 0$. This shows that $A^2$ is positive definite.



              We can show that $A^{-1}$ is positive definite by noting that
              $$
              v^T A^{-1} v = v^T A^{-1} A A^{-1} v = u^T A u
              $$
              where $u = A^{-1}v$. If $v neq 0$ then $u neq 0$, so $u^T A u > 0$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Thanks a lot this really clears it up, i forgot about that definition but it makes perfect sense
                $endgroup$
                – L G
                Jan 8 at 16:33
















              2












              $begingroup$

              Another definition is that a symmetric matrix $M in mathbb R^{n times n}$ is positive definite if and only if $v^T M v > 0$ for all nonzero vectors $v in mathbb R^n$.



              Notice that
              begin{align}
              v^T A^2 v &= v^T A^T A v \
              &= u^T u \
              &= |u|^2
              end{align}
              where $u = Av$.
              If $v neq 0$ then $u neq 0$, so $|u|^2 > 0$. This shows that $A^2$ is positive definite.



              We can show that $A^{-1}$ is positive definite by noting that
              $$
              v^T A^{-1} v = v^T A^{-1} A A^{-1} v = u^T A u
              $$
              where $u = A^{-1}v$. If $v neq 0$ then $u neq 0$, so $u^T A u > 0$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Thanks a lot this really clears it up, i forgot about that definition but it makes perfect sense
                $endgroup$
                – L G
                Jan 8 at 16:33














              2












              2








              2





              $begingroup$

              Another definition is that a symmetric matrix $M in mathbb R^{n times n}$ is positive definite if and only if $v^T M v > 0$ for all nonzero vectors $v in mathbb R^n$.



              Notice that
              begin{align}
              v^T A^2 v &= v^T A^T A v \
              &= u^T u \
              &= |u|^2
              end{align}
              where $u = Av$.
              If $v neq 0$ then $u neq 0$, so $|u|^2 > 0$. This shows that $A^2$ is positive definite.



              We can show that $A^{-1}$ is positive definite by noting that
              $$
              v^T A^{-1} v = v^T A^{-1} A A^{-1} v = u^T A u
              $$
              where $u = A^{-1}v$. If $v neq 0$ then $u neq 0$, so $u^T A u > 0$.






              share|cite|improve this answer











              $endgroup$



              Another definition is that a symmetric matrix $M in mathbb R^{n times n}$ is positive definite if and only if $v^T M v > 0$ for all nonzero vectors $v in mathbb R^n$.



              Notice that
              begin{align}
              v^T A^2 v &= v^T A^T A v \
              &= u^T u \
              &= |u|^2
              end{align}
              where $u = Av$.
              If $v neq 0$ then $u neq 0$, so $|u|^2 > 0$. This shows that $A^2$ is positive definite.



              We can show that $A^{-1}$ is positive definite by noting that
              $$
              v^T A^{-1} v = v^T A^{-1} A A^{-1} v = u^T A u
              $$
              where $u = A^{-1}v$. If $v neq 0$ then $u neq 0$, so $u^T A u > 0$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 8 at 16:49

























              answered Jan 8 at 16:15









              littleOlittleO

              29.6k646109




              29.6k646109












              • $begingroup$
                Thanks a lot this really clears it up, i forgot about that definition but it makes perfect sense
                $endgroup$
                – L G
                Jan 8 at 16:33


















              • $begingroup$
                Thanks a lot this really clears it up, i forgot about that definition but it makes perfect sense
                $endgroup$
                – L G
                Jan 8 at 16:33
















              $begingroup$
              Thanks a lot this really clears it up, i forgot about that definition but it makes perfect sense
              $endgroup$
              – L G
              Jan 8 at 16:33




              $begingroup$
              Thanks a lot this really clears it up, i forgot about that definition but it makes perfect sense
              $endgroup$
              – L G
              Jan 8 at 16:33











              1












              $begingroup$

              For a pos.def. matrix $A$, you have (necessary and sufficient) that for any real vector $v$, that $v A v ge 0$.



              Now $ v A^{-1} A v = v (A^{-1} A A^{-1}) v =(v A^{-1}) A (A^{-1} v ) = u A u > 0 $.



              $A^2$ is even easier: $v A^2 v = v A A v = (v A) (A v) = u^2> 0$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                For a pos.def. matrix $A$, you have (necessary and sufficient) that for any real vector $v$, that $v A v ge 0$.



                Now $ v A^{-1} A v = v (A^{-1} A A^{-1}) v =(v A^{-1}) A (A^{-1} v ) = u A u > 0 $.



                $A^2$ is even easier: $v A^2 v = v A A v = (v A) (A v) = u^2> 0$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  For a pos.def. matrix $A$, you have (necessary and sufficient) that for any real vector $v$, that $v A v ge 0$.



                  Now $ v A^{-1} A v = v (A^{-1} A A^{-1}) v =(v A^{-1}) A (A^{-1} v ) = u A u > 0 $.



                  $A^2$ is even easier: $v A^2 v = v A A v = (v A) (A v) = u^2> 0$






                  share|cite|improve this answer









                  $endgroup$



                  For a pos.def. matrix $A$, you have (necessary and sufficient) that for any real vector $v$, that $v A v ge 0$.



                  Now $ v A^{-1} A v = v (A^{-1} A A^{-1}) v =(v A^{-1}) A (A^{-1} v ) = u A u > 0 $.



                  $A^2$ is even easier: $v A^2 v = v A A v = (v A) (A v) = u^2> 0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 16:18









                  AndreasAndreas

                  7,8661037




                  7,8661037























                      0












                      $begingroup$

                      Suppose that the eigenvalues of $A$ are $lambda_1,ldots,lambda_n$ with eigenvectors $v_1,ldots,v_n$. Note that $$A^2 v_i = lambda_i A v_i$ = lambda_i^2 v_i$$ and that $$Av_i = lambda_i v_i Rightarrow A^{-1}v_i = lambda^{-1}v_i$$ Therefore the eigenvalues of $A^2$ are $lambda_1^2,ldots,lambda_n^2$ and the eigenvalues of $A^{-1}$ are $lambda_1^{-1},ldots,lambda_n^{-1}$. This should be enough to prove that $A^2$ and $A^{-1}$ are positive definite.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Suppose that the eigenvalues of $A$ are $lambda_1,ldots,lambda_n$ with eigenvectors $v_1,ldots,v_n$. Note that $$A^2 v_i = lambda_i A v_i$ = lambda_i^2 v_i$$ and that $$Av_i = lambda_i v_i Rightarrow A^{-1}v_i = lambda^{-1}v_i$$ Therefore the eigenvalues of $A^2$ are $lambda_1^2,ldots,lambda_n^2$ and the eigenvalues of $A^{-1}$ are $lambda_1^{-1},ldots,lambda_n^{-1}$. This should be enough to prove that $A^2$ and $A^{-1}$ are positive definite.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Suppose that the eigenvalues of $A$ are $lambda_1,ldots,lambda_n$ with eigenvectors $v_1,ldots,v_n$. Note that $$A^2 v_i = lambda_i A v_i$ = lambda_i^2 v_i$$ and that $$Av_i = lambda_i v_i Rightarrow A^{-1}v_i = lambda^{-1}v_i$$ Therefore the eigenvalues of $A^2$ are $lambda_1^2,ldots,lambda_n^2$ and the eigenvalues of $A^{-1}$ are $lambda_1^{-1},ldots,lambda_n^{-1}$. This should be enough to prove that $A^2$ and $A^{-1}$ are positive definite.






                          share|cite|improve this answer









                          $endgroup$



                          Suppose that the eigenvalues of $A$ are $lambda_1,ldots,lambda_n$ with eigenvectors $v_1,ldots,v_n$. Note that $$A^2 v_i = lambda_i A v_i$ = lambda_i^2 v_i$$ and that $$Av_i = lambda_i v_i Rightarrow A^{-1}v_i = lambda^{-1}v_i$$ Therefore the eigenvalues of $A^2$ are $lambda_1^2,ldots,lambda_n^2$ and the eigenvalues of $A^{-1}$ are $lambda_1^{-1},ldots,lambda_n^{-1}$. This should be enough to prove that $A^2$ and $A^{-1}$ are positive definite.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 8 at 16:16









                          ODFODF

                          1,461510




                          1,461510






























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