Proving inequality for positive definite matrix












2












$begingroup$


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



$$frac{x^T sqrt{A} x}{|sqrt{A}x|_2} geq frac{x^T A x}{|Ax|_2}$$



So far I cannot find any counterexamples, and it intuitively makes sense since the $sqrt{cdot}$ 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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  • 3




    $begingroup$
    A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrt{M}x|^2$. Thus, we can rewrite your equation as $$ frac{|A^{1/4}x|^2}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|Ax|} iff\ frac{|Ax|}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|A^{1/4}x|^2} $$ with $B = A^{1/4}$ and $y = A^{1/4}y$, we can rewrite the above as $$ frac{|B^3y|}{|By|} < frac{|By|^2}{|y|^2} iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – 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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac{3 |By|^4}{lambda} I - B^4right)y = 0 $$
    $endgroup$
    – Omnomnomnom
    34 mins ago








  • 1




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


















2












$begingroup$


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



$$frac{x^T sqrt{A} x}{|sqrt{A}x|_2} geq frac{x^T A x}{|Ax|_2}$$



So far I cannot find any counterexamples, and it intuitively makes sense since the $sqrt{cdot}$ 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 $x^TMx = |sqrt{M}x|^2$. Thus, we can rewrite your equation as $$ frac{|A^{1/4}x|^2}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|Ax|} iff\ frac{|Ax|}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|A^{1/4}x|^2} $$ with $B = A^{1/4}$ and $y = A^{1/4}y$, we can rewrite the above as $$ frac{|B^3y|}{|By|} < frac{|By|^2}{|y|^2} iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – 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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac{3 |By|^4}{lambda} I - B^4right)y = 0 $$
    $endgroup$
    – Omnomnomnom
    34 mins ago








  • 1




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
















2












2








2





$begingroup$


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



$$frac{x^T sqrt{A} x}{|sqrt{A}x|_2} geq frac{x^T A x}{|Ax|_2}$$



So far I cannot find any counterexamples, and it intuitively makes sense since the $sqrt{cdot}$ 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 $x$:



$$frac{x^T sqrt{A} x}{|sqrt{A}x|_2} geq frac{x^T A x}{|Ax|_2}$$



So far I cannot find any counterexamples, and it intuitively makes sense since the $sqrt{cdot}$ 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









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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 16 mins ago









B Merlot

725




725






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asked 2 hours ago









ReginaldReginald

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186




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  • 3




    $begingroup$
    A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrt{M}x|^2$. Thus, we can rewrite your equation as $$ frac{|A^{1/4}x|^2}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|Ax|} iff\ frac{|Ax|}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|A^{1/4}x|^2} $$ with $B = A^{1/4}$ and $y = A^{1/4}y$, we can rewrite the above as $$ frac{|B^3y|}{|By|} < frac{|By|^2}{|y|^2} iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – 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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac{3 |By|^4}{lambda} I - B^4right)y = 0 $$
    $endgroup$
    – Omnomnomnom
    34 mins ago








  • 1




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
















  • 3




    $begingroup$
    A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrt{M}x|^2$. Thus, we can rewrite your equation as $$ frac{|A^{1/4}x|^2}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|Ax|} iff\ frac{|Ax|}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|A^{1/4}x|^2} $$ with $B = A^{1/4}$ and $y = A^{1/4}y$, we can rewrite the above as $$ frac{|B^3y|}{|By|} < frac{|By|^2}{|y|^2} iff |B^3y|,,|y|^2 < |By|^3 $$
    $endgroup$
    – 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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac{3 |By|^4}{lambda} I - B^4right)y = 0 $$
    $endgroup$
    – Omnomnomnom
    34 mins ago








  • 1




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










3




3




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




$begingroup$
A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrt{M}x|^2$. Thus, we can rewrite your equation as $$ frac{|A^{1/4}x|^2}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|Ax|} iff\ frac{|Ax|}{|A^{1/2}x|} < frac{|A^{1/2}x|^2}{|A^{1/4}x|^2} $$ with $B = A^{1/4}$ and $y = A^{1/4}y$, we can rewrite the above as $$ frac{|B^3y|}{|By|} < frac{|By|^2}{|y|^2} iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– 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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac{3 |By|^4}{lambda} I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
34 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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac{3 |By|^4}{lambda} I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
34 mins ago






1




1




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






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












1 Answer
1






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oldest

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4












$begingroup$

Your inequality says



$$frac{sumsqrt{lambda_j}x_j^2}{left(sumlambda_j x_j^2right)^{1/2}}geq
frac{sumlambda_jx_j^2}{left(sumlambda_j^2x_j^2right)^{1/2}},$$
or after a simple transformation
$$sumlambda_j x_j^2leqleft(sumsqrt{lambda_j}x_j^2right)^{2/3}
left(sumlambda_j^2x_j^2right)^{1/3}$$
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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    active

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    4












    $begingroup$

    Your inequality says



    $$frac{sumsqrt{lambda_j}x_j^2}{left(sumlambda_j x_j^2right)^{1/2}}geq
    frac{sumlambda_jx_j^2}{left(sumlambda_j^2x_j^2right)^{1/2}},$$
    or after a simple transformation
    $$sumlambda_j x_j^2leqleft(sumsqrt{lambda_j}x_j^2right)^{2/3}
    left(sumlambda_j^2x_j^2right)^{1/3}$$
    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



      $$frac{sumsqrt{lambda_j}x_j^2}{left(sumlambda_j x_j^2right)^{1/2}}geq
      frac{sumlambda_jx_j^2}{left(sumlambda_j^2x_j^2right)^{1/2}},$$
      or after a simple transformation
      $$sumlambda_j x_j^2leqleft(sumsqrt{lambda_j}x_j^2right)^{2/3}
      left(sumlambda_j^2x_j^2right)^{1/3}$$
      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



        $$frac{sumsqrt{lambda_j}x_j^2}{left(sumlambda_j x_j^2right)^{1/2}}geq
        frac{sumlambda_jx_j^2}{left(sumlambda_j^2x_j^2right)^{1/2}},$$
        or after a simple transformation
        $$sumlambda_j x_j^2leqleft(sumsqrt{lambda_j}x_j^2right)^{2/3}
        left(sumlambda_j^2x_j^2right)^{1/3}$$
        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



        $$frac{sumsqrt{lambda_j}x_j^2}{left(sumlambda_j x_j^2right)^{1/2}}geq
        frac{sumlambda_jx_j^2}{left(sumlambda_j^2x_j^2right)^{1/2}},$$
        or after a simple transformation
        $$sumlambda_j x_j^2leqleft(sumsqrt{lambda_j}x_j^2right)^{2/3}
        left(sumlambda_j^2x_j^2right)^{1/3}$$
        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 27 mins ago









        Alexandre EremenkoAlexandre Eremenko

        51.7k6144263




        51.7k6144263






















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            Península de Txukotka

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            Península de Txukotka Tipus Península Ubicació 66° N, 172° O  /  66°N , 172°O  / 66; -172 Mar mar dels Txuktxis La península de Txukotka a l'extrem oriental de Sibèria La península de Txukotka o península dels Txuktxis (en rus Чуко́тский полуо́стров , Txukotski poluóstrov ) és una península que es troba a l'extrem oriental d'Euràsia. Contingut 1 Descripció 2 Població 3 Vegeu també 4 Bibliografia Descripció L'estret de Bering, entre la península de Txukotka i la península de Seward a Alaska Foto de satèl·lit de la zona amb la península de Txukotka a l'esquerra El seu punt més extrem és el cap Dezhnev a la vora del nucli d'Uelen, un dels poquíssims llocs habitats de la zona. Al nord limita amb el mar dels Txuktxis i amb el mar de Bering al sud i amb l'estret de Bering a l'est. A la costa sud es troben la badia de Sant Lawrence i la badia de Kresta. La penínsu...
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