Lower bound on $tr(A^-1)$, where $Ain mathbbR^n times n$ symmetric, positive definite.

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Let $A in mathbbR^n times n$ be a symmetric, positive definite matrix with $n geq 2$. Does the following inequality then hold?
$$
sum_i=1^n frac1lambda_i > frac1lambda_min,
$$
where the $lambda_i$'s are the eigenvalues of $A$ and $lambda_min := min_i in 1, cdots, n left( lambda_i right)$.



Thoughts so far:



A simpler inequality that clearly holds is
$$
sum_i=1^n frac1lambda_i geq fracnlambda_max,
$$
but that doesn't seem useful in this situation.



Also, I realize that the left hand size is $texttrace(A^-1)$, not sure what to do with this.



Background:



An inequality in some lecture notes that I'm reading seems to implicitly hinge on this claim. Specifically, the following is stated as a portion of a longer discourse:
$$
sigma^2 sum_i = 1^p frac1lambda_i > fracsigma^2lambda_min,
$$
where $(X^T X)$ is a positive definite matrix, $X in mathbbR^n times p$.




linear-algebra inequality eigenvalues-eigenvectors




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  • For positive matrix, $lambda_i>0$ for $i=1,2,cdots,n$.
    – Riemann
    2 days ago














up vote
1
down vote

favorite












Let $A in mathbbR^n times n$ be a symmetric, positive definite matrix with $n geq 2$. Does the following inequality then hold?
$$
sum_i=1^n frac1lambda_i > frac1lambda_min,
$$
where the $lambda_i$'s are the eigenvalues of $A$ and $lambda_min := min_i in 1, cdots, n left( lambda_i right)$.



Thoughts so far:



A simpler inequality that clearly holds is
$$
sum_i=1^n frac1lambda_i geq fracnlambda_max,
$$
but that doesn't seem useful in this situation.



Also, I realize that the left hand size is $texttrace(A^-1)$, not sure what to do with this.



Background:



An inequality in some lecture notes that I'm reading seems to implicitly hinge on this claim. Specifically, the following is stated as a portion of a longer discourse:
$$
sigma^2 sum_i = 1^p frac1lambda_i > fracsigma^2lambda_min,
$$
where $(X^T X)$ is a positive definite matrix, $X in mathbbR^n times p$.




linear-algebra inequality eigenvalues-eigenvectors




share|cite|improve this question





















  • For positive matrix, $lambda_i>0$ for $i=1,2,cdots,n$.
    – Riemann
    2 days ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $A in mathbbR^n times n$ be a symmetric, positive definite matrix with $n geq 2$. Does the following inequality then hold?
$$
sum_i=1^n frac1lambda_i > frac1lambda_min,
$$
where the $lambda_i$'s are the eigenvalues of $A$ and $lambda_min := min_i in 1, cdots, n left( lambda_i right)$.



Thoughts so far:



A simpler inequality that clearly holds is
$$
sum_i=1^n frac1lambda_i geq fracnlambda_max,
$$
but that doesn't seem useful in this situation.



Also, I realize that the left hand size is $texttrace(A^-1)$, not sure what to do with this.



Background:



An inequality in some lecture notes that I'm reading seems to implicitly hinge on this claim. Specifically, the following is stated as a portion of a longer discourse:
$$
sigma^2 sum_i = 1^p frac1lambda_i > fracsigma^2lambda_min,
$$
where $(X^T X)$ is a positive definite matrix, $X in mathbbR^n times p$.




linear-algebra inequality eigenvalues-eigenvectors




share|cite|improve this question













Let $A in mathbbR^n times n$ be a symmetric, positive definite matrix with $n geq 2$. Does the following inequality then hold?
$$
sum_i=1^n frac1lambda_i > frac1lambda_min,
$$
where the $lambda_i$'s are the eigenvalues of $A$ and $lambda_min := min_i in 1, cdots, n left( lambda_i right)$.



Thoughts so far:



A simpler inequality that clearly holds is
$$
sum_i=1^n frac1lambda_i geq fracnlambda_max,
$$
but that doesn't seem useful in this situation.



Also, I realize that the left hand size is $texttrace(A^-1)$, not sure what to do with this.



Background:



An inequality in some lecture notes that I'm reading seems to implicitly hinge on this claim. Specifically, the following is stated as a portion of a longer discourse:
$$
sigma^2 sum_i = 1^p frac1lambda_i > fracsigma^2lambda_min,
$$
where $(X^T X)$ is a positive definite matrix, $X in mathbbR^n times p$.





linear-algebra inequality eigenvalues-eigenvectors





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 2 days ago









darij grinberg

9,17132958




9,17132958









asked 2 days ago









zxmkn

1686




1686











  • For positive matrix, $lambda_i>0$ for $i=1,2,cdots,n$.
    – Riemann
    2 days ago
















  • For positive matrix, $lambda_i>0$ for $i=1,2,cdots,n$.
    – Riemann
    2 days ago















For positive matrix, $lambda_i>0$ for $i=1,2,cdots,n$.
– Riemann
2 days ago




For positive matrix, $lambda_i>0$ for $i=1,2,cdots,n$.
– Riemann
2 days ago










2 Answers
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accepted










This should be obvious $$sum_i=1^n frac1lambda_i > frac1lambda_min$$



Because the eigenvalues are positive and one of the eigenvalues, say $lambda_j$ is the minimum eigenvalue. Thus $$sum_i=1^n frac1lambda_i = frac1lambda_min + sum _ine j frac1lambda_i > frac1 lambda_min$$






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    Yes.



    Since the eigenvalues are positive, all the terms in the sum are positive. Therefore, the sum is larger than any of its terms.






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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes








      up vote
      1
      down vote



      accepted










      This should be obvious $$sum_i=1^n frac1lambda_i > frac1lambda_min$$



      Because the eigenvalues are positive and one of the eigenvalues, say $lambda_j$ is the minimum eigenvalue. Thus $$sum_i=1^n frac1lambda_i = frac1lambda_min + sum _ine j frac1lambda_i > frac1 lambda_min$$






      share|cite|improve this answer



























        up vote
        1
        down vote



        accepted










        This should be obvious $$sum_i=1^n frac1lambda_i > frac1lambda_min$$



        Because the eigenvalues are positive and one of the eigenvalues, say $lambda_j$ is the minimum eigenvalue. Thus $$sum_i=1^n frac1lambda_i = frac1lambda_min + sum _ine j frac1lambda_i > frac1 lambda_min$$






        share|cite|improve this answer

























          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          This should be obvious $$sum_i=1^n frac1lambda_i > frac1lambda_min$$



          Because the eigenvalues are positive and one of the eigenvalues, say $lambda_j$ is the minimum eigenvalue. Thus $$sum_i=1^n frac1lambda_i = frac1lambda_min + sum _ine j frac1lambda_i > frac1 lambda_min$$






          share|cite|improve this answer















          This should be obvious $$sum_i=1^n frac1lambda_i > frac1lambda_min$$



          Because the eigenvalues are positive and one of the eigenvalues, say $lambda_j$ is the minimum eigenvalue. Thus $$sum_i=1^n frac1lambda_i = frac1lambda_min + sum _ine j frac1lambda_i > frac1 lambda_min$$







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago


























          answered 2 days ago









          Mohammad Riazi-Kermani

          26.9k41849




          26.9k41849




















              up vote
              4
              down vote













              Yes.



              Since the eigenvalues are positive, all the terms in the sum are positive. Therefore, the sum is larger than any of its terms.






              share|cite|improve this answer

























                up vote
                4
                down vote













                Yes.



                Since the eigenvalues are positive, all the terms in the sum are positive. Therefore, the sum is larger than any of its terms.






                share|cite|improve this answer























                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  Yes.



                  Since the eigenvalues are positive, all the terms in the sum are positive. Therefore, the sum is larger than any of its terms.






                  share|cite|improve this answer













                  Yes.



                  Since the eigenvalues are positive, all the terms in the sum are positive. Therefore, the sum is larger than any of its terms.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered 2 days ago









                  N. S.

                  97.5k5105197




                  97.5k5105197






















                       

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