Proving inequality for positive definite matrix
$begingroup$
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 1 hour ago
B Merlot
725
asked 3 hours ago
ReginaldReginald
286
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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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$endgroup$
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 1 hour ago
B Merlot
725
asked 3 hours ago
ReginaldReginald
286
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
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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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
1 hour 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
1 hour 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 1 hour ago
B Merlot
725
asked 3 hours ago
ReginaldReginald
286
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 1 hour ago
B Merlot
725
asked 3 hours ago
ReginaldReginald
286
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 1 hour ago
B Merlot
725
asked 3 hours ago
ReginaldReginald
286
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 1 hour ago
B Merlot
725
edited 1 hour ago
B Merlot
725
edited 1 hour ago
B Merlot
725
725
asked 3 hours ago
ReginaldReginald
286
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
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.
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.
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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
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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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
1 hour 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
1 hour 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
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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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
1 hour 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
1 hour 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
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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
add a comment |
1 Answer
1
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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 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
$endgroup$
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 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
$endgroup$
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 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
$endgroup$
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 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
$endgroup$
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 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
51.8k6144264
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
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 $$ min |By|^6 - |B^3y|^2 quad text{st} quad |y|=1 $$ 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
1 hour 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
1 hour ago