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 induced by inverse 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 induced by inverse matrix










2












$begingroup$


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



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



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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$endgroup$







  • 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 $$ fracA^1/4xA^1/2x < frac iff\ fracA^1/2x < fracA^1/4x $$ 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
    1 hour ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    1 hour 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
    27 mins 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
    21 mins ago
















2












$begingroup$


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



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



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 $$ fracA^1/4xA^1/2x < frac iff\ fracA^1/2x < fracA^1/4x $$ 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
    1 hour ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    1 hour 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
    27 mins 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
    21 mins ago














2












2








2





$begingroup$


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



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



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 $x$:



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



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 9 mins ago









B Merlot

725




725






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asked 2 hours ago









ReginaldReginald

186




186




New contributor




Reginald is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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 $$ fracA^1/4xA^1/2x < frac iff\ fracA^1/2x < fracA^1/4x $$ 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
    1 hour ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    1 hour 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
    27 mins 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
    21 mins 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 $$ fracA^1/4xA^1/2x < frac iff\ fracA^1/2x < fracA^1/4x $$ 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
    1 hour ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    1 hour 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
    27 mins 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
    21 mins 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 $$ fracA^1/4xA^1/2x < frac iff\ fracA^1/2x < fracA^1/4x $$ 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
1 hour 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 $$ fracA^1/4xA^1/2x < frac iff\ fracA^1/2x < fracA^1/4x $$ 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
1 hour ago




1




1




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




$begingroup$
Also, note that we fail to have strict inequality when $A = I$, for instance.
$endgroup$
– Omnomnomnom
1 hour 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
27 mins 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
27 mins 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
21 mins 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
21 mins ago











1 Answer
1






active

oldest

votes


















4












$begingroup$

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

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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









    $endgroup$

















      4












      $begingroup$

      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









      $endgroup$















        4












        4








        4





        $begingroup$

        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 20 mins ago









        Alexandre EremenkoAlexandre Eremenko

        51.7k6144263




        51.7k6144263




















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            Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e

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            Mortos em 2019Eventos de março de 2019 ←JaneiroFevereiroMarçoAbrilMaioJunhoJulhoAgostoSetembroOutubroNovembroDezembro→https://g1.globo.com/pop-arte/noticia/2019/04/01/rapper-nipsey-hussle-e-morto-a-tiros-em-los-angeles.ghtml Mortes em 2019 : ← Janeiro – Fevereiro – Março – Abril – Maio – Junho – Julho – Agosto – Setembro – Outubro – Novembro – Dezembro → Está é uma relação de pessoas notáveis que morreram durante o mês de março de 2019, listando nome, nacionalidade, ocupação e ano de nascimento. Keith Flint , músico inglês. Luke Perry , ator e dublador americano. Coutinho , futebolista brasileiro. Eurico Miranda , dirigente esportivo e político brasileiro. Charlie Whiting , diretor de provas britânico. Valery Bykovsky , cosmonauta russo. Agnès Varda , cineasta belgo-francesa. Dia Nome Profissão ou motivo de...
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