Proving inequality for positive definite matrix Planned maintenance scheduled April 23, 2019...
Proving inequality for positive definite matrix
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$begingroup$
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
edited 34 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
236
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.
$endgroup$
add a comment |
$begingroup$
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
edited 34 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
236
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
2 hours 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
52 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
46 mins ago
add a comment |
$begingroup$
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
edited 34 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
236
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
edited 34 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
236
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.
edited 34 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
236
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.
edited 34 mins ago
B Merlot
725
edited 34 mins ago
B Merlot
725
edited 34 mins ago
B Merlot
725
725
asked 2 hours ago
ReginaldReginald
236
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 2 hours ago
ReginaldReginald
236
asked 2 hours ago
ReginaldReginald
236
236
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 $x^TMx = |sqrt{M}x|^2$. 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
2 hours 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
52 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
46 mins ago
add a comment |
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
2 hours 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
52 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
46 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
2 hours 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
2 hours 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
52 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
52 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
46 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
46 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 45 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered 45 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
$endgroup$
add a comment |
$begingroup$
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.
answered 45 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
$endgroup$
add a comment |
$begingroup$
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.
answered 45 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
$endgroup$
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.
answered 45 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
answered 45 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
answered 45 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
answered 45 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
51.7k6144263
add a comment |
add a comment |
Reginald is a new contributor. Be nice, and check out our Code of Conduct.
Reginald is a new contributor. Be nice, and check out our Code of Conduct.
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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
2 hours 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
52 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
46 mins ago