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






New contributor




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Check out our Code of Conduct.









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




















            Reginald is a new contributor. Be nice, and check out our Code of Conduct.









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            When We Were Young (canção de Adele) Índice Antecedentes e lançamento | Faixas e formatos | Performances e covers | Desempenho nas tabelas musicais | Créditos | Histórico de lançamento | Referências Menu de navegação«Best albums of 2015»«Adele - When We Were Young (Radio Date: 22-01-2016)»«When We Were Young - Single by Adele»«Adele: Inside Her Private Life and Triumphant Return»«adele interview: world exclusive first interview in three years»«Tobias Jesso Jr: since Adele tweeted his song, he's even bigger than his dad»«Adele interviews Tobias Jesso Jr: 'I think that a couple of the ideas we had could be rap songs'»«Adele on Her Return: 'I Was So Frightened That No One Cared'»«Tobias Jesso Jr. on working with Adele: 'I was as nervous as shit'»«How Ariel Rechtshaid Pushed Adele To Her Limit»«Adele Previews 'When We Were Young' on '60 Minutes' Teaser, Tops Trending 140»«Adele Performs New '25' Ballad, "When We Were Young" Live: Watch»«Which '25' Song Should Be Adele's Next Single?»«Adele's 'When We Were Young' Confirmed As Second Single From '25'»«Adele's new single artwork for 'When We Were Young' is perfectly adorable»«Adele at the BBC review: honest, funny and spectacular – Celebrity News News»«'Saturday Night Live': Adele Sings 'Hello' and 'When We Were Young'»«'Adele: Live in New York City' NBC Special – Set List Revealed!»«Adele Closes Out the 2016 Brit Awards With 'When We Were Young'»«What Is The Adele Live Tour Set List? There's No Way She'd Leave Out These 8 Songs»«See Demi Lovato's Soaring Cover of Adele's 'When We Were Young'»«'The Voice': 5 Best Moments From Week 1 Blind Auditions»«Adele – When We Were Young (Media Control Charts)»«Adele – When We Were Young (Entertainment Monitoring Africa)»«Adele – When We Were Young (ARIA Charts)»«Adele – When We Were Young (Ö3 Austria Top 40)»«Adele – When We Were Young (Ultratop 50)»«Adele – When We Were Young (Ultratop 40)»«Adele – When We Were Young (Canadian Hot 100)»«Adele – When We Were Young (Tracklisten)»«Adele – When We Were Young (The Official Charts Company)»«Adele – Hello (IFPI Slovenská Republika)»«Adele – When We Were Young (Productores de Música de España)»«Adele – When We Were Young (Billboard Hot 100)»«Adele – When We Were Young (Pop Songs)»«Adele – When We Were Young (Adult Pop Songs)»«Adele – When We Were Young (Hot Adult Contemporary Charts)»«Adele – When We Were Young (Hot Dance Club Songs)»«Adele – When We Were Young (Rock Airplay)»«Adele – When We Were Young (IFPI Finlândia)»«Adele – When We Were Young (Syndicat National de l'Éditon Phonographique)»«Adele – When We Were Young (Magyar Hanglemezkiadók Szövetsége)»«Adele – When We Were Young (Irish Recorded Music Association)»«Adele – When We Were Young (Mexico Airplay)»«Adele – When We Were Young (VG-lista)»«Adele – When We Were Young (NZ Top 40 Singles)»«Adele – When We Were Young (MegaCharts)»«Adele – When We Were Young (Związek Producentów Audio Video)»«Adele – When We Were Young (Portugal Digital Songs)»«Adele – When We Were Young (UK Indie Singles Chart)»«Adele – When We Were Young (UK Singles Chart)»«Adele – When We Were Young (Sverigetopplistan)»«Adele – When We Were Young (Schweizer Hitparade)»«Adele – When We Were Young (Portugal Digital Songs)»«Music Canada – Gold/Platinum – When We Were Young»«NZ Top 40 Singles Chart»«Certified Awards»e

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            Singles de AdeleCanções compostas por AdeleSingles de 2016Singles de balada cançãoinglesaAdeleálbum de estúdioTobias Jesso, Jr.LondresXL Recordingsdigitalmentebloqueio de escritorLos AngelesTobias Jesso Jr.Philip GlassAriel RechtshaidBillboardColumbia RecordsAdele at the BBCBBC OneAdele Live in New York CityNBCGlastonbury FestivalAdele Live 2016Demi LovatoRolling StoneLuke Evans "When We Were Young" Single de Adele do álbum 25 Lançamento 22 de janeiro de 2016  ( 2016-01-22 ) Formato(s) Download digital Gravação 2015; Dean Street Studios (Londres, Inglaterra) Gênero(s) Soul [ 1 ] Duração 4 : 50 Gravadora(s) XL Composição Adele Adkins, Tobias Jesso, Jr. Produção Ariel Rechtshaid Cronologia de singles de Adele "Hello" (2015) "Send My Love (To Your New Lover)" (2016) ...
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