Bounds on eigenvalues of symmetric $A$ where $I + A$ is positive-definite

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Let $A$ be a symmetric $n times n$ matrix such that its diagonal entries are zero, and such that $I + A$ is positive-definite.



I'd like to show that the eigenvalues of $A$ are bounded from below by $-1$ and from above by $n - 1$.



The lower bound I can get from positive-definiteness of $I + A$ and the fact that its eigenvalues are sums of eigenvalues of $I$ and $A$.



But, I don't know where to start for the upper bound. Any hints?




linear-algebra eigenvalues-eigenvectors




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  • $n$ is the dimension of the space; edited
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up vote
1
down vote

favorite
1












Let $A$ be a symmetric $n times n$ matrix such that its diagonal entries are zero, and such that $I + A$ is positive-definite.



I'd like to show that the eigenvalues of $A$ are bounded from below by $-1$ and from above by $n - 1$.



The lower bound I can get from positive-definiteness of $I + A$ and the fact that its eigenvalues are sums of eigenvalues of $I$ and $A$.



But, I don't know where to start for the upper bound. Any hints?




linear-algebra eigenvalues-eigenvectors




share|cite|improve this question





















  • $n$ is the dimension of the space; edited
    – Definitely Not A Genius
    yesterday












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





Let $A$ be a symmetric $n times n$ matrix such that its diagonal entries are zero, and such that $I + A$ is positive-definite.



I'd like to show that the eigenvalues of $A$ are bounded from below by $-1$ and from above by $n - 1$.



The lower bound I can get from positive-definiteness of $I + A$ and the fact that its eigenvalues are sums of eigenvalues of $I$ and $A$.



But, I don't know where to start for the upper bound. Any hints?




linear-algebra eigenvalues-eigenvectors




share|cite|improve this question













Let $A$ be a symmetric $n times n$ matrix such that its diagonal entries are zero, and such that $I + A$ is positive-definite.



I'd like to show that the eigenvalues of $A$ are bounded from below by $-1$ and from above by $n - 1$.



The lower bound I can get from positive-definiteness of $I + A$ and the fact that its eigenvalues are sums of eigenvalues of $I$ and $A$.



But, I don't know where to start for the upper bound. Any hints?





linear-algebra eigenvalues-eigenvectors





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday
























asked yesterday









Definitely Not A Genius

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  • $n$ is the dimension of the space; edited
    – Definitely Not A Genius
    yesterday
















  • $n$ is the dimension of the space; edited
    – Definitely Not A Genius
    yesterday















$n$ is the dimension of the space; edited
– Definitely Not A Genius
yesterday




$n$ is the dimension of the space; edited
– Definitely Not A Genius
yesterday










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Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
Then the $1+lambda_i$ are the eigenvalues of $I+A$, so are positive
as $I+A$ is positive definite.



Also $lambda_1+cdots+lambda_n=0$ as this is the trace of $A$.
So $lambda_1=-lambda_2-cdots-lambda_n<1+cdots+1=n-1$ etc.






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    1 Answer
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    up vote
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    accepted










    Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
    Then the $1+lambda_i$ are the eigenvalues of $I+A$, so are positive
    as $I+A$ is positive definite.



    Also $lambda_1+cdots+lambda_n=0$ as this is the trace of $A$.
    So $lambda_1=-lambda_2-cdots-lambda_n<1+cdots+1=n-1$ etc.






    share|cite|improve this answer

























      up vote
      4
      down vote



      accepted










      Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
      Then the $1+lambda_i$ are the eigenvalues of $I+A$, so are positive
      as $I+A$ is positive definite.



      Also $lambda_1+cdots+lambda_n=0$ as this is the trace of $A$.
      So $lambda_1=-lambda_2-cdots-lambda_n<1+cdots+1=n-1$ etc.






      share|cite|improve this answer























        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
        Then the $1+lambda_i$ are the eigenvalues of $I+A$, so are positive
        as $I+A$ is positive definite.



        Also $lambda_1+cdots+lambda_n=0$ as this is the trace of $A$.
        So $lambda_1=-lambda_2-cdots-lambda_n<1+cdots+1=n-1$ etc.






        share|cite|improve this answer













        Let $lambda_1,ldots,lambda_n$ be the eigenvalues of $A$.
        Then the $1+lambda_i$ are the eigenvalues of $I+A$, so are positive
        as $I+A$ is positive definite.



        Also $lambda_1+cdots+lambda_n=0$ as this is the trace of $A$.
        So $lambda_1=-lambda_2-cdots-lambda_n<1+cdots+1=n-1$ etc.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











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