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
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.
$endgroup$
add a comment |
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
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 $x^TMx = |sqrtMx|^2$. 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
add a comment |
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
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 $x$:
$$fracx^T sqrtA x geq fracx^T A x_2$$
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
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.
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
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 $$ frac^2A^1/2x < frac iff\ fracA^1/2x < frac^2 $$ 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
add a comment |
1 Answer
1
active
oldest
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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.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
$endgroup$
add a comment |
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1 Answer
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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.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
$endgroup$
add a comment |
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.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
$endgroup$
add a comment |
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.
answered 1 hour ago
Alexandre EremenkoAlexandre Eremenko
51.8k6144264
$endgroup$
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 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 |
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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