Proving inequality for positive definite matrix Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Eigenvalues of A+B where A is symmetric positive definite and B is diagonalA spectral inequality for positive-definite matrices Showing positive stability of a matrix constructed from a positive matrixCondition number after some “non standard” transformProve that matrix is positive definiteInequality between nuclear norm and operator norm for positive definite matricesStability of a matrix productInverse of a matrix and the inverse of its diagonalsMaximum rotation made by a symmetric positive definite matrix?Angle inequality for inverse of PD diagonal matrix

Proving inequality for positive definite matrix



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Eigenvalues of A+B where A is symmetric positive definite and B is diagonalA spectral inequality for positive-definite matrices Showing positive stability of a matrix constructed from a positive matrixCondition number after some “non standard” transformProve that matrix is positive definiteInequality between nuclear norm and operator norm for positive definite matricesStability of a matrix productInverse of a matrix and the inverse of its diagonalsMaximum rotation made by a symmetric positive definite matrix?Angle inequality for inverse of PD diagonal matrix










5












$begingroup$


For a positive definite diagonal matrix $A$, I want to prove that for any $x$:



$$fracx^T sqrtA xsqrtAx geq fracx^T A x_2$$



So far I cannot find any counterexamples, and it intuitively makes sense since the $sqrtcdot$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this.



EDIT: changed $>$ to $geq$






linear-algebra







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Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$







  • 3




    $begingroup$
    A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrtMx|^2$. Thus, we can rewrite your equation as $$ frac^2 < fracA^1/2x iff\ frac < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < fracBy iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – Omnomnomnom
    4 hours ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    3 hours ago






  • 1




    $begingroup$
    More thoughts that are insufficient for an answer: Since both sides scale with $|y|$, it suffices to consider the inequality in the case that $|y| = 1$. That is: $$ |B^3y| leq |By|^3 $$ To that end: consider $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ Let $f(y) = |By|^6 - |B^3y|^2$, and let $g(y) = |y|^2$. We compute the Lagrangian $$ 2B^2(3|By|^4 I - lambda B^4)y $$ Now, $B$ positive definite. So, setting the Lagrangian to zero yields $$ (3|By|^4 I - lambda B^4)y = 0 implies left(fracBylambda I - B^4right)y = 0 $$
    $endgroup$
    – Omnomnomnom
    3 hours ago







  • 1




    $begingroup$
    Applying the Hölder-von Neumann inequality yields $$ |By|^2 = operatornametr[B^2yy^T] leq operatornametr[B^3]^1/1.5operatornametr[(yy^T)^3]^1/3 = operatornametr[B^3]^2/3|y|^2/3 $$ which is close to what we're looking for, but not quite there
    $endgroup$
    – Omnomnomnom
    2 hours ago
















5












$begingroup$


For a positive definite diagonal matrix $A$, I want to prove that for any $x$:



$$fracx^T sqrtA xsqrtAx geq fracx^T A x_2$$



So far I cannot find any counterexamples, and it intuitively makes sense since the $sqrtcdot$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this.



EDIT: changed $>$ to $geq$






linear-algebra







share|cite|improve this question









New contributor




Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 3




    $begingroup$
    A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrtMx|^2$. Thus, we can rewrite your equation as $$ frac^2 < fracA^1/2x iff\ frac < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < fracBy iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – Omnomnomnom
    4 hours ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    3 hours ago






  • 1




    $begingroup$
    More thoughts that are insufficient for an answer: Since both sides scale with $|y|$, it suffices to consider the inequality in the case that $|y| = 1$. That is: $$ |B^3y| leq |By|^3 $$ To that end: consider $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ Let $f(y) = |By|^6 - |B^3y|^2$, and let $g(y) = |y|^2$. We compute the Lagrangian $$ 2B^2(3|By|^4 I - lambda B^4)y $$ Now, $B$ positive definite. So, setting the Lagrangian to zero yields $$ (3|By|^4 I - lambda B^4)y = 0 implies left(fracBylambda I - B^4right)y = 0 $$
    $endgroup$
    – Omnomnomnom
    3 hours ago







  • 1




    $begingroup$
    Applying the Hölder-von Neumann inequality yields $$ |By|^2 = operatornametr[B^2yy^T] leq operatornametr[B^3]^1/1.5operatornametr[(yy^T)^3]^1/3 = operatornametr[B^3]^2/3|y|^2/3 $$ which is close to what we're looking for, but not quite there
    $endgroup$
    – Omnomnomnom
    2 hours ago














5












5








5


1



$begingroup$


For a positive definite diagonal matrix $A$, I want to prove that for any $x$:



$$fracx^T sqrtA xsqrtAx geq fracx^T A x_2$$



So far I cannot find any counterexamples, and it intuitively makes sense since the $sqrtcdot$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this.



EDIT: changed $>$ to $geq$






linear-algebra







share|cite|improve this question









New contributor




Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




For a positive definite diagonal matrix $A$, I want to prove that for any $x$:



$$fracx^T sqrtA xsqrtAx geq fracx^T A x_2$$



So far I cannot find any counterexamples, and it intuitively makes sense since the $sqrtcdot$ operator should bring the eigenvalues of $A$ closer to $1$, but I can't prove this.



EDIT: changed $>$ to $geq$






linear-algebra




linear-algebra






share|cite|improve this question









New contributor




Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









B Merlot

775




775






New contributor




Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 4 hours ago









ReginaldReginald

336




336




New contributor




Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 3




    $begingroup$
    A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrtMx|^2$. Thus, we can rewrite your equation as $$ frac^2 < fracA^1/2x iff\ frac < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < fracBy iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – Omnomnomnom
    4 hours ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    3 hours ago






  • 1




    $begingroup$
    More thoughts that are insufficient for an answer: Since both sides scale with $|y|$, it suffices to consider the inequality in the case that $|y| = 1$. That is: $$ |B^3y| leq |By|^3 $$ To that end: consider $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ Let $f(y) = |By|^6 - |B^3y|^2$, and let $g(y) = |y|^2$. We compute the Lagrangian $$ 2B^2(3|By|^4 I - lambda B^4)y $$ Now, $B$ positive definite. So, setting the Lagrangian to zero yields $$ (3|By|^4 I - lambda B^4)y = 0 implies left(fracBylambda I - B^4right)y = 0 $$
    $endgroup$
    – Omnomnomnom
    3 hours ago







  • 1




    $begingroup$
    Applying the Hölder-von Neumann inequality yields $$ |By|^2 = operatornametr[B^2yy^T] leq operatornametr[B^3]^1/1.5operatornametr[(yy^T)^3]^1/3 = operatornametr[B^3]^2/3|y|^2/3 $$ which is close to what we're looking for, but not quite there
    $endgroup$
    – Omnomnomnom
    2 hours ago













  • 3




    $begingroup$
    A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrtMx|^2$. Thus, we can rewrite your equation as $$ frac^2 < fracA^1/2x iff\ frac < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < fracBy iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – Omnomnomnom
    4 hours ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    3 hours ago






  • 1




    $begingroup$
    More thoughts that are insufficient for an answer: Since both sides scale with $|y|$, it suffices to consider the inequality in the case that $|y| = 1$. That is: $$ |B^3y| leq |By|^3 $$ To that end: consider $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ Let $f(y) = |By|^6 - |B^3y|^2$, and let $g(y) = |y|^2$. We compute the Lagrangian $$ 2B^2(3|By|^4 I - lambda B^4)y $$ Now, $B$ positive definite. So, setting the Lagrangian to zero yields $$ (3|By|^4 I - lambda B^4)y = 0 implies left(fracBylambda I - B^4right)y = 0 $$
    $endgroup$
    – Omnomnomnom
    3 hours ago







  • 1




    $begingroup$
    Applying the Hölder-von Neumann inequality yields $$ |By|^2 = operatornametr[B^2yy^T] leq operatornametr[B^3]^1/1.5operatornametr[(yy^T)^3]^1/3 = operatornametr[B^3]^2/3|y|^2/3 $$ which is close to what we're looking for, but not quite there
    $endgroup$
    – Omnomnomnom
    2 hours ago








3




3




$begingroup$
A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrtMx|^2$. Thus, we can rewrite your equation as $$ frac^2 < fracA^1/2x iff\ frac < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < fracBy iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– Omnomnomnom
4 hours ago




$begingroup$
A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrtMx|^2$. Thus, we can rewrite your equation as $$ frac^2 < fracA^1/2x iff\ frac < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < fracBy iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– Omnomnomnom
4 hours ago




1




1




$begingroup$
Also, note that we fail to have strict inequality when $A = I$, for instance.
$endgroup$
– Omnomnomnom
3 hours ago




$begingroup$
Also, note that we fail to have strict inequality when $A = I$, for instance.
$endgroup$
– Omnomnomnom
3 hours ago




1




1




$begingroup$
More thoughts that are insufficient for an answer: Since both sides scale with $|y|$, it suffices to consider the inequality in the case that $|y| = 1$. That is: $$ |B^3y| leq |By|^3 $$ To that end: consider $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ Let $f(y) = |By|^6 - |B^3y|^2$, and let $g(y) = |y|^2$. We compute the Lagrangian $$ 2B^2(3|By|^4 I - lambda B^4)y $$ Now, $B$ positive definite. So, setting the Lagrangian to zero yields $$ (3|By|^4 I - lambda B^4)y = 0 implies left(fracBylambda I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
3 hours ago





$begingroup$
More thoughts that are insufficient for an answer: Since both sides scale with $|y|$, it suffices to consider the inequality in the case that $|y| = 1$. That is: $$ |B^3y| leq |By|^3 $$ To that end: consider $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ Let $f(y) = |By|^6 - |B^3y|^2$, and let $g(y) = |y|^2$. We compute the Lagrangian $$ 2B^2(3|By|^4 I - lambda B^4)y $$ Now, $B$ positive definite. So, setting the Lagrangian to zero yields $$ (3|By|^4 I - lambda B^4)y = 0 implies left(fracBylambda I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
3 hours ago





1




1




$begingroup$
Applying the Hölder-von Neumann inequality yields $$ |By|^2 = operatornametr[B^2yy^T] leq operatornametr[B^3]^1/1.5operatornametr[(yy^T)^3]^1/3 = operatornametr[B^3]^2/3|y|^2/3 $$ which is close to what we're looking for, but not quite there
$endgroup$
– Omnomnomnom
2 hours ago





$begingroup$
Applying the Hölder-von Neumann inequality yields $$ |By|^2 = operatornametr[B^2yy^T] leq operatornametr[B^3]^1/1.5operatornametr[(yy^T)^3]^1/3 = operatornametr[B^3]^2/3|y|^2/3 $$ which is close to what we're looking for, but not quite there
$endgroup$
– Omnomnomnom
2 hours ago











1 Answer
1






active

oldest

votes


















8












$begingroup$

Your inequality says



$$fracsumsqrtlambda_jx_j^2left(sumlambda_j x_j^2right)^1/2geq
fracsumlambda_jx_j^2left(sumlambda_j^2x_j^2right)^1/2,$$
or after a simple transformation
$$sumlambda_j x_j^2leqleft(sumsqrtlambda_jx_j^2right)^2/3
left(sumlambda_j^2x_j^2right)^1/3$$
And this is Holder's inequality with
$p=3/2$ and $q=3$. The strict inequality does not always hold.






share|cite|improve this answer









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    1 Answer
    1






    active

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    active

    oldest

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    active

    oldest

    votes









    8












    $begingroup$

    Your inequality says



    $$fracsumsqrtlambda_jx_j^2left(sumlambda_j x_j^2right)^1/2geq
    fracsumlambda_jx_j^2left(sumlambda_j^2x_j^2right)^1/2,$$
    or after a simple transformation
    $$sumlambda_j x_j^2leqleft(sumsqrtlambda_jx_j^2right)^2/3
    left(sumlambda_j^2x_j^2right)^1/3$$
    And this is Holder's inequality with
    $p=3/2$ and $q=3$. The strict inequality does not always hold.






    share|cite|improve this answer









    $endgroup$

















      8












      $begingroup$

      Your inequality says



      $$fracsumsqrtlambda_jx_j^2left(sumlambda_j x_j^2right)^1/2geq
      fracsumlambda_jx_j^2left(sumlambda_j^2x_j^2right)^1/2,$$
      or after a simple transformation
      $$sumlambda_j x_j^2leqleft(sumsqrtlambda_jx_j^2right)^2/3
      left(sumlambda_j^2x_j^2right)^1/3$$
      And this is Holder's inequality with
      $p=3/2$ and $q=3$. The strict inequality does not always hold.






      share|cite|improve this answer









      $endgroup$















        8












        8








        8





        $begingroup$

        Your inequality says



        $$fracsumsqrtlambda_jx_j^2left(sumlambda_j x_j^2right)^1/2geq
        fracsumlambda_jx_j^2left(sumlambda_j^2x_j^2right)^1/2,$$
        or after a simple transformation
        $$sumlambda_j x_j^2leqleft(sumsqrtlambda_jx_j^2right)^2/3
        left(sumlambda_j^2x_j^2right)^1/3$$
        And this is Holder's inequality with
        $p=3/2$ and $q=3$. The strict inequality does not always hold.






        share|cite|improve this answer









        $endgroup$



        Your inequality says



        $$fracsumsqrtlambda_jx_j^2left(sumlambda_j x_j^2right)^1/2geq
        fracsumlambda_jx_j^2left(sumlambda_j^2x_j^2right)^1/2,$$
        or after a simple transformation
        $$sumlambda_j x_j^2leqleft(sumsqrtlambda_jx_j^2right)^2/3
        left(sumlambda_j^2x_j^2right)^1/3$$
        And this is Holder's inequality with
        $p=3/2$ and $q=3$. The strict inequality does not always hold.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Alexandre EremenkoAlexandre Eremenko

        51.8k6144264




        51.8k6144264




















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