A Matrix Inequality for positive definite matrices The Next CEO of Stack OverflowArithmetic-geometric mean of positive matricesEigenvalue distribution of positive-definite analytic functionDecomposition of positive definite matrices.A spectral inequality for positive-definite matrices determinant inequality for symmetric positive definite matricesDeterminant inequality involving Hermitian, positive definite matricesEigenvalues of a partitioned self-adjoint matrixA log inequality for positive definite trace-one matricesExpectation of square root of positive definite matrixProve that matrix is positive definite

A Matrix Inequality for positive definite matrices



The Next CEO of Stack OverflowArithmetic-geometric mean of positive matricesEigenvalue distribution of positive-definite analytic functionDecomposition of positive definite matrices.A spectral inequality for positive-definite matrices determinant inequality for symmetric positive definite matricesDeterminant inequality involving Hermitian, positive definite matricesEigenvalues of a partitioned self-adjoint matrixA log inequality for positive definite trace-one matricesExpectation of square root of positive definite matrixProve that matrix is positive definite










4












$begingroup$


Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^frac12$ ?



PS. I think the answer is No. But I could not find any counterexample!






linear-algebra matrices operator-theory







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
    $endgroup$
    – lcv
    3 hours ago










  • $begingroup$
    @lcv. can you please elaborate your notations?
    $endgroup$
    – Samya Kumar Ray
    2 hours ago















4












$begingroup$


Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^frac12$ ?



PS. I think the answer is No. But I could not find any counterexample!






linear-algebra matrices operator-theory







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
    $endgroup$
    – lcv
    3 hours ago










  • $begingroup$
    @lcv. can you please elaborate your notations?
    $endgroup$
    – Samya Kumar Ray
    2 hours ago













4












4








4





$begingroup$


Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^frac12$ ?



PS. I think the answer is No. But I could not find any counterexample!






linear-algebra matrices operator-theory







share|cite|improve this question











$endgroup$




Let $X$ and $Y$ be positive semi-definite self-adjoint complex matrices of same finite order. The, is it true that $|X-Y|leq X+Y$ where for any matrix $A$, $|A|$ is defined to be $|A|:=(A^*A)^frac12$ ?



PS. I think the answer is No. But I could not find any counterexample!






linear-algebra matrices operator-theory




linear-algebra matrices operator-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 mins ago









Denis Serre

29.6k795197




29.6k795197










asked 4 hours ago









Samya Kumar RaySamya Kumar Ray

364




364







  • 1




    $begingroup$
    Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
    $endgroup$
    – lcv
    3 hours ago










  • $begingroup$
    @lcv. can you please elaborate your notations?
    $endgroup$
    – Samya Kumar Ray
    2 hours ago












  • 1




    $begingroup$
    Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
    $endgroup$
    – lcv
    3 hours ago










  • $begingroup$
    @lcv. can you please elaborate your notations?
    $endgroup$
    – Samya Kumar Ray
    2 hours ago







1




1




$begingroup$
Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
$endgroup$
– lcv
3 hours ago




$begingroup$
Consider $X,Y$ to be of trace 1 and their difference to be block diagonal $sigma^x,sigma^x$. Now $|X-Y|=$ identity matrix. Take the trace of both sides, the LHS gives 4 the RHS is 2.
$endgroup$
– lcv
3 hours ago












$begingroup$
@lcv. can you please elaborate your notations?
$endgroup$
– Samya Kumar Ray
2 hours ago




$begingroup$
@lcv. can you please elaborate your notations?
$endgroup$
– Samya Kumar Ray
2 hours ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

The answer is No. Here is a counter-example:
$$X=beginpmatrix 9 & 3 \ 3 & 1 endpmatrix,qquad Y=beginpmatrix 1 & 3 \ 3 & 9 endpmatrix.$$






share|cite|improve this answer











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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    The answer is No. Here is a counter-example:
    $$X=beginpmatrix 9 & 3 \ 3 & 1 endpmatrix,qquad Y=beginpmatrix 1 & 3 \ 3 & 9 endpmatrix.$$






    share|cite|improve this answer











    $endgroup$

















      6












      $begingroup$

      The answer is No. Here is a counter-example:
      $$X=beginpmatrix 9 & 3 \ 3 & 1 endpmatrix,qquad Y=beginpmatrix 1 & 3 \ 3 & 9 endpmatrix.$$






      share|cite|improve this answer











      $endgroup$















        6












        6








        6





        $begingroup$

        The answer is No. Here is a counter-example:
        $$X=beginpmatrix 9 & 3 \ 3 & 1 endpmatrix,qquad Y=beginpmatrix 1 & 3 \ 3 & 9 endpmatrix.$$






        share|cite|improve this answer











        $endgroup$



        The answer is No. Here is a counter-example:
        $$X=beginpmatrix 9 & 3 \ 3 & 1 endpmatrix,qquad Y=beginpmatrix 1 & 3 \ 3 & 9 endpmatrix.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 17 mins ago

























        answered 1 hour ago









        Denis SerreDenis Serre

        29.6k795197




        29.6k795197



























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