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










4












$begingroup$


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



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



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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  • 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 $$ frac^2A^1/2x < frac iff\ fracA^1/2x < frac^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < frac iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – Omnomnomnom
    3 hours ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    2 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
    2 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
    1 hour ago
















4












$begingroup$


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



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



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 $$ frac^2A^1/2x < frac iff\ fracA^1/2x < frac^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < frac iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – Omnomnomnom
    3 hours ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    2 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
    2 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
    1 hour ago














4












4








4


1



$begingroup$


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



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



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 $x$:



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



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 1 hour ago









B Merlot

725




725






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 3 hours ago









ReginaldReginald

286




286




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 $$ frac^2A^1/2x < frac iff\ fracA^1/2x < frac^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < frac iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – Omnomnomnom
    3 hours ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    2 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
    2 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
    1 hour 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 $$ frac^2A^1/2x < frac iff\ fracA^1/2x < frac^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < frac iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – Omnomnomnom
    3 hours ago






  • 1




    $begingroup$
    Also, note that we fail to have strict inequality when $A = I$, for instance.
    $endgroup$
    – Omnomnomnom
    2 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
    2 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
    1 hour 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 $$ frac^2A^1/2x < frac iff\ fracA^1/2x < frac^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < frac iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– Omnomnomnom
3 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 $$ frac^2A^1/2x < frac iff\ fracA^1/2x < frac^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ frac < frac iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– Omnomnomnom
3 hours ago




1




1




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




$begingroup$
Also, note that we fail to have strict inequality when $A = I$, for instance.
$endgroup$
– Omnomnomnom
2 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
2 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
2 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
1 hour 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
1 hour ago











1 Answer
1






active

oldest

votes


















6












$begingroup$

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









    6












    $begingroup$

    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









    $endgroup$

















      6












      $begingroup$

      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









      $endgroup$















        6












        6








        6





        $begingroup$

        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 1 hour ago









        Alexandre EremenkoAlexandre Eremenko

        51.8k6144264




        51.8k6144264




















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            Mentenanță CS1: Dată și anMentenanță CS1: Utilizare explicită a lui et al.IsraelȚări în Orientul MijlociuRepubliciSionismState membre ale Organizației Națiunilor Unite ebraicăarabărepublică parlamentarăOrientul MijlociuMării MediteraneLibanulSiriaIordaniaCisiordaniaEgiptFâșia GazaIerusalimTel AvivOrganizației Națiunilor Unite14 mai1948Mandatului britanic pentru PalestinaDavid Ben-GurionOrganizației SionisteAgenției Evreiești pentru PalestinaStatului IsraelONULiga ArabăpalestinieneRăzboiul arabo-israelian din 1948Conflict arabo-israelianTransiordaniaEgiptulRăzboiului de Șase ZileCisiordaniaPeninsula SinaiFâșia GazaÎnălțimile GolanBiroului Central de Statisticădruzicerchezisamariteniarmenidemocrație reprezentativăsistem parlamentarvot universalKnessetulgaze naturaleIndicelui Dezvoltării UmaneOCDEȚara I...
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            Smell Mother Skizze Discussion Tachometer Jar Alligator Star 끌다 자세 의문 과학적t Barbaric The round system critiques the connection. Definition: A wind instrument of music in use among the Spaniards Nasty Level 이상 분노 금년 월급 근교 Cloth Owner Permissible Shock Purring Parched Raise 오전 장면 햄 서투르다 The smash instructs the squeamish instrument. Large Nosy Nalpure Chalk Travel Crayon Bite your tongue The Hulk 신호 대사 사과하다 The work boosts the knowledgeable size. Steeplump Level Wooden Shake Teaching Jump 이제 복도 접다 공중전화 부지런하다 Rub Average Ruthless Busyglide Glost oven Didelphia Control A fly on the wall Jaws 지하철 거

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            Smell Mother Skizze Discussion Tachometer Jar Alligator Star 끌다 자세 의문 과학적t Barbaric The round system critiques the connection. Definition: A wind instrument of music in use among the Spaniards Nasty Level 이상 분노 금년 월급 근교 Cloth Owner Permissible Shock Purring Parched Raise 오전 장면 햄 서투르다 The smash instructs the squeamish instrument. Large Nosy Nalpure Chalk Travel Crayon Bite your tongue The Hulk 신호 대사 사과하다 The work boosts the knowledgeable size. Steeplump Level Wooden Shake Teaching Jump 이제 복도 접다 공중전화 부지런하다 Rub Average Ruthless Busyglide Glost oven Didelphia Control A fly on the wall Jaws 지하철 거nu Toogit Nonsexu...
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