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 $x$:
$$fracx^T sqrtA x_2 geq fracx^T A x$$
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
edited 15 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
186
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 |
2
$begingroup$
For a positive definite diagonal matrix $A$, I want to prove that for any $x$:
$$fracx^T sqrtA x_2 geq fracx^T A x$$
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
edited 15 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
186
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 $$ frac^2A^1/2x < fracA^1/2x iff\ fracA^1/2x < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ fracB^3y < frac^2 iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– 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 $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac3 lambda I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
33 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
27 mins ago
add a comment |
2
2
2
$begingroup$
For a positive definite diagonal matrix $A$, I want to prove that for any $x$:
$$fracx^T sqrtA x_2 geq fracx^T A x$$
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
edited 15 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
186
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 $x$:
$$fracx^T sqrtA x_2 geq fracx^T A x$$
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
edited 15 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
186
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 15 mins ago
B Merlot
725
asked 2 hours ago
ReginaldReginald
186
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 15 mins ago
B Merlot
725
edited 15 mins ago
B Merlot
725
edited 15 mins ago
B Merlot
725
725
asked 2 hours ago
ReginaldReginald
186
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
186
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.
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 $x^TMx = |sqrtMx|^2$. Thus, we can rewrite your equation as $$ frac^2A^1/2x < fracA^1/2x iff\ fracA^1/2x < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ fracB^3y < frac^2 iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– 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 $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac3 lambda I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
33 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
27 mins ago
add a comment |
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 $$ frac^2A^1/2x < fracA^1/2x iff\ fracA^1/2x < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ fracB^3y < frac^2 iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– 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 $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac3 lambda I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
33 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
27 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 $$ frac^2A^1/2x < fracA^1/2x iff\ fracA^1/2x < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ fracB^3y < frac^2 iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– Omnomnomnom
1 hour ago
$begingroup$
A potentially helpful observation: Note that if $M$ is positive semidefinite, we have $x^TMx = |sqrtMx|^2$. Thus, we can rewrite your equation as $$ frac^2A^1/2x < fracA^1/2x iff\ fracA^1/2x < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ fracB^3y < frac^2 iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– 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 $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac3 lambda I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
33 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac3 lambda I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
33 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
27 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
27 mins ago
add a comment |
1 Answer
1
active
oldest
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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.
answered 26 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
$endgroup$
add a comment |
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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.
answered 26 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
$endgroup$
add a comment |
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.
answered 26 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
$endgroup$
add a comment |
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.
answered 26 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
$endgroup$
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.
answered 26 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
answered 26 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
answered 26 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
answered 26 mins ago
Alexandre EremenkoAlexandre Eremenko
51.7k6144263
51.7k6144263
add a comment |
add a comment |
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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 $$ frac^2A^1/2x < fracA^1/2x iff\ fracA^1/2x < fracA^1/2x^2 $$ with $B = A^1/4$ and $y = A^1/4y$, we can rewrite the above as $$ fracB^3y < frac^2 iff |B^3y|,,|y|^2 < |By|^3 $$
$endgroup$
– 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 $$ min |By|^6 - |B^3y|^2 quad textst quad |y|=1 $$ 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 $$ (3|By|^4 I - lambda B^4)y = 0 implies left(frac3 lambda I - B^4right)y = 0 $$
$endgroup$
– Omnomnomnom
33 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
27 mins ago