Eigenvalues of element-wise ordered matrices

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Let $A = (a_ij)$ and $B = (b_ij)$ be two positive-definite matrices such that
$$a_ij geq b_ij text for all $i,j$ and for all $i$, a_ii > b_ii,.$$



That is, $A geq B$ elementwise and the diagonals of $A$ are greater than the diagonals of $B$. In that case, can we say that $lambda_A,i > lambda_B,i$ for $i = 1, dots, p$?



Intuitively, I think this should be true, but I am not able to show it mainly because I don't know if $A-B$ is positive definite.




matrices inequality eigenvalues-eigenvectors positive-definite




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    Check [1, 2 ; 2, 1]...
    – user251257
    Jul 16 at 19:01














up vote
1
down vote

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Let $A = (a_ij)$ and $B = (b_ij)$ be two positive-definite matrices such that
$$a_ij geq b_ij text for all $i,j$ and for all $i$, a_ii > b_ii,.$$



That is, $A geq B$ elementwise and the diagonals of $A$ are greater than the diagonals of $B$. In that case, can we say that $lambda_A,i > lambda_B,i$ for $i = 1, dots, p$?



Intuitively, I think this should be true, but I am not able to show it mainly because I don't know if $A-B$ is positive definite.




matrices inequality eigenvalues-eigenvectors positive-definite




share|cite|improve this question















  • 1




    Check [1, 2 ; 2, 1]...
    – user251257
    Jul 16 at 19:01












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $A = (a_ij)$ and $B = (b_ij)$ be two positive-definite matrices such that
$$a_ij geq b_ij text for all $i,j$ and for all $i$, a_ii > b_ii,.$$



That is, $A geq B$ elementwise and the diagonals of $A$ are greater than the diagonals of $B$. In that case, can we say that $lambda_A,i > lambda_B,i$ for $i = 1, dots, p$?



Intuitively, I think this should be true, but I am not able to show it mainly because I don't know if $A-B$ is positive definite.




matrices inequality eigenvalues-eigenvectors positive-definite




share|cite|improve this question











Let $A = (a_ij)$ and $B = (b_ij)$ be two positive-definite matrices such that
$$a_ij geq b_ij text for all $i,j$ and for all $i$, a_ii > b_ii,.$$



That is, $A geq B$ elementwise and the diagonals of $A$ are greater than the diagonals of $B$. In that case, can we say that $lambda_A,i > lambda_B,i$ for $i = 1, dots, p$?



Intuitively, I think this should be true, but I am not able to show it mainly because I don't know if $A-B$ is positive definite.





matrices inequality eigenvalues-eigenvectors positive-definite





share|cite|improve this question










share|cite|improve this question




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asked Jul 16 at 18:55









Greenparker

203112




203112







  • 1




    Check [1, 2 ; 2, 1]...
    – user251257
    Jul 16 at 19:01












  • 1




    Check [1, 2 ; 2, 1]...
    – user251257
    Jul 16 at 19:01







1




1




Check [1, 2 ; 2, 1]...
– user251257
Jul 16 at 19:01




Check [1, 2 ; 2, 1]...
– user251257
Jul 16 at 19:01










1 Answer
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Counter-example:
$$
A=beginpmatrix5 & 0\
0 & 5
endpmatrixqquad B=beginpmatrixphantom+4 & -3\
-3 & phantom+4
endpmatrix
$$
Addendum: In case you are interested, it is worthwhile to point out that this disproves the claim in a slightly stronger setting since $A$ and $B$ are Stieltjes matrices (a subset of the SPD matrices).






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    1 Answer
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    1 Answer
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    active

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    active

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    up vote
    3
    down vote



    accepted










    Counter-example:
    $$
    A=beginpmatrix5 & 0\
    0 & 5
    endpmatrixqquad B=beginpmatrixphantom+4 & -3\
    -3 & phantom+4
    endpmatrix
    $$
    Addendum: In case you are interested, it is worthwhile to point out that this disproves the claim in a slightly stronger setting since $A$ and $B$ are Stieltjes matrices (a subset of the SPD matrices).






    share|cite|improve this answer



























      up vote
      3
      down vote



      accepted










      Counter-example:
      $$
      A=beginpmatrix5 & 0\
      0 & 5
      endpmatrixqquad B=beginpmatrixphantom+4 & -3\
      -3 & phantom+4
      endpmatrix
      $$
      Addendum: In case you are interested, it is worthwhile to point out that this disproves the claim in a slightly stronger setting since $A$ and $B$ are Stieltjes matrices (a subset of the SPD matrices).






      share|cite|improve this answer

























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        Counter-example:
        $$
        A=beginpmatrix5 & 0\
        0 & 5
        endpmatrixqquad B=beginpmatrixphantom+4 & -3\
        -3 & phantom+4
        endpmatrix
        $$
        Addendum: In case you are interested, it is worthwhile to point out that this disproves the claim in a slightly stronger setting since $A$ and $B$ are Stieltjes matrices (a subset of the SPD matrices).






        share|cite|improve this answer















        Counter-example:
        $$
        A=beginpmatrix5 & 0\
        0 & 5
        endpmatrixqquad B=beginpmatrixphantom+4 & -3\
        -3 & phantom+4
        endpmatrix
        $$
        Addendum: In case you are interested, it is worthwhile to point out that this disproves the claim in a slightly stronger setting since $A$ and $B$ are Stieltjes matrices (a subset of the SPD matrices).







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 16 at 19:17


























        answered Jul 16 at 19:05









        parsiad

        16k32253




        16k32253






















             

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