Minimize trace of $A$ given that $A−N$ is positive semi-definite and $A$ is diagonal
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begin{array}{ll} text{minimize} & mbox{tr} (mathrm A)\ text{subject to} & mathrm A - mathrm N succeq mathrm O_nend{array} where $A$ and $N$ are pd matrices, and $A$ is diagonal.
There is a related post: Minimize trace of $A$ given that $A-N$ is positive semi-definite. . However, in that case $A$ is not diagonal thus, $tr(A)=tr(N)$ is possible, while in current case not.
For $Ain mathbb{R}^{2times2}$, I believe $min tr(A)=sum n_{ij}$, however for $Ain mathbb{R}^{3times3}$ we have inequality $min tr(A)leqsum n_{ij}$. Can you please help with analytical approach so solve it
optimization maxima-minima trace
asked Nov 19 at 3:24
Lee
907
add a comment |
up vote
1
down vote
favorite
begin{array}{ll} text{minimize} & mbox{tr} (mathrm A)\ text{subject to} & mathrm A - mathrm N succeq mathrm O_nend{array} where $A$ and $N$ are pd matrices, and $A$ is diagonal.
There is a related post: Minimize trace of $A$ given that $A-N$ is positive semi-definite. . However, in that case $A$ is not diagonal thus, $tr(A)=tr(N)$ is possible, while in current case not.
For $Ain mathbb{R}^{2times2}$, I believe $min tr(A)=sum n_{ij}$, however for $Ain mathbb{R}^{3times3}$ we have inequality $min tr(A)leqsum n_{ij}$. Can you please help with analytical approach so solve it
optimization maxima-minima trace
asked Nov 19 at 3:24
Lee
907
$A =lambda_{max}(N)I$ is feasible, so $n lambda_{max}(N)$ is an obvious bound. I do not see an easy way to get a tighter bound.
– LinAlg
Nov 19 at 14:14
If $N=begin{bmatrix}3 & 1 & -1\1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $nlambda_{max}(N)=11.1963$. Let $A=begin{bmatrix}4 & & \ & 3 & \ & &3end{bmatrix}$, then $A-Ngeq 0$ and $trace(A)=10$
– Lee
Nov 20 at 1:57
if $A=begin{bmatrix}4 & & \ & 3 & \ & &2end{bmatrix}$, then again $A-Ngeq 0$ and $trace(A)=9$. I think this is minimum, but I don't have a proof
– Lee
Nov 20 at 2:19
if $N$ is one of following structures $begin{bmatrix}+& + & +\+ & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &- \+ & - &+end{bmatrix}$, $begin{bmatrix}+& - &-\- & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& + & -\+ & + &- \- & - &+end{bmatrix}$, then I believe $min tr(A)=sum n_{ij}$.
– Lee
Nov 20 at 2:26
if $N$ is one of following structures $begin{bmatrix}+& + & -\+ & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& + &+\+ & + &- \+ & - &+end{bmatrix}$, then I believe $min tr(A)=sum |n_{ij}|-4|n_{12}| $, assuming $|n_{12}|leq |n_{13}| leq |n_{23}|$.
– Lee
Nov 20 at 9:12
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
begin{array}{ll} text{minimize} & mbox{tr} (mathrm A)\ text{subject to} & mathrm A - mathrm N succeq mathrm O_nend{array} where $A$ and $N$ are pd matrices, and $A$ is diagonal.
There is a related post: Minimize trace of $A$ given that $A-N$ is positive semi-definite. . However, in that case $A$ is not diagonal thus, $tr(A)=tr(N)$ is possible, while in current case not.
For $Ain mathbb{R}^{2times2}$, I believe $min tr(A)=sum n_{ij}$, however for $Ain mathbb{R}^{3times3}$ we have inequality $min tr(A)leqsum n_{ij}$. Can you please help with analytical approach so solve it
optimization maxima-minima trace
asked Nov 19 at 3:24
Lee
907
begin{array}{ll} text{minimize} & mbox{tr} (mathrm A)\ text{subject to} & mathrm A - mathrm N succeq mathrm O_nend{array} where $A$ and $N$ are pd matrices, and $A$ is diagonal.
There is a related post: Minimize trace of $A$ given that $A-N$ is positive semi-definite. . However, in that case $A$ is not diagonal thus, $tr(A)=tr(N)$ is possible, while in current case not.
For $Ain mathbb{R}^{2times2}$, I believe $min tr(A)=sum n_{ij}$, however for $Ain mathbb{R}^{3times3}$ we have inequality $min tr(A)leqsum n_{ij}$. Can you please help with analytical approach so solve it
optimization maxima-minima trace
optimization maxima-minima trace
asked Nov 19 at 3:24
Lee
907
asked Nov 19 at 3:24
Lee
907
asked Nov 19 at 3:24
Lee
907
asked Nov 19 at 3:24
Lee
907
asked Nov 19 at 3:24
Lee
907
907
$A =lambda_{max}(N)I$ is feasible, so $n lambda_{max}(N)$ is an obvious bound. I do not see an easy way to get a tighter bound.
– LinAlg
Nov 19 at 14:14
If $N=begin{bmatrix}3 & 1 & -1\1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $nlambda_{max}(N)=11.1963$. Let $A=begin{bmatrix}4 & & \ & 3 & \ & &3end{bmatrix}$, then $A-Ngeq 0$ and $trace(A)=10$
– Lee
Nov 20 at 1:57
if $A=begin{bmatrix}4 & & \ & 3 & \ & &2end{bmatrix}$, then again $A-Ngeq 0$ and $trace(A)=9$. I think this is minimum, but I don't have a proof
– Lee
Nov 20 at 2:19
if $N$ is one of following structures $begin{bmatrix}+& + & +\+ & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &- \+ & - &+end{bmatrix}$, $begin{bmatrix}+& - &-\- & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& + & -\+ & + &- \- & - &+end{bmatrix}$, then I believe $min tr(A)=sum n_{ij}$.
– Lee
Nov 20 at 2:26
if $N$ is one of following structures $begin{bmatrix}+& + & -\+ & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& + &+\+ & + &- \+ & - &+end{bmatrix}$, then I believe $min tr(A)=sum |n_{ij}|-4|n_{12}| $, assuming $|n_{12}|leq |n_{13}| leq |n_{23}|$.
– Lee
Nov 20 at 9:12
add a comment |
$A =lambda_{max}(N)I$ is feasible, so $n lambda_{max}(N)$ is an obvious bound. I do not see an easy way to get a tighter bound.
– LinAlg
Nov 19 at 14:14
If $N=begin{bmatrix}3 & 1 & -1\1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $nlambda_{max}(N)=11.1963$. Let $A=begin{bmatrix}4 & & \ & 3 & \ & &3end{bmatrix}$, then $A-Ngeq 0$ and $trace(A)=10$
– Lee
Nov 20 at 1:57
if $A=begin{bmatrix}4 & & \ & 3 & \ & &2end{bmatrix}$, then again $A-Ngeq 0$ and $trace(A)=9$. I think this is minimum, but I don't have a proof
– Lee
Nov 20 at 2:19
if $N$ is one of following structures $begin{bmatrix}+& + & +\+ & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &- \+ & - &+end{bmatrix}$, $begin{bmatrix}+& - &-\- & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& + & -\+ & + &- \- & - &+end{bmatrix}$, then I believe $min tr(A)=sum n_{ij}$.
– Lee
Nov 20 at 2:26
if $N$ is one of following structures $begin{bmatrix}+& + & -\+ & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& + &+\+ & + &- \+ & - &+end{bmatrix}$, then I believe $min tr(A)=sum |n_{ij}|-4|n_{12}| $, assuming $|n_{12}|leq |n_{13}| leq |n_{23}|$.
– Lee
Nov 20 at 9:12
$A =lambda_{max}(N)I$ is feasible, so $n lambda_{max}(N)$ is an obvious bound. I do not see an easy way to get a tighter bound.
– LinAlg
Nov 19 at 14:14
$A =lambda_{max}(N)I$ is feasible, so $n lambda_{max}(N)$ is an obvious bound. I do not see an easy way to get a tighter bound.
– LinAlg
Nov 19 at 14:14
If $N=begin{bmatrix}3 & 1 & -1\1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $nlambda_{max}(N)=11.1963$. Let $A=begin{bmatrix}4 & & \ & 3 & \ & &3end{bmatrix}$, then $A-Ngeq 0$ and $trace(A)=10$
– Lee
Nov 20 at 1:57
If $N=begin{bmatrix}3 & 1 & -1\1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $nlambda_{max}(N)=11.1963$. Let $A=begin{bmatrix}4 & & \ & 3 & \ & &3end{bmatrix}$, then $A-Ngeq 0$ and $trace(A)=10$
– Lee
Nov 20 at 1:57
if $A=begin{bmatrix}4 & & \ & 3 & \ & &2end{bmatrix}$, then again $A-Ngeq 0$ and $trace(A)=9$. I think this is minimum, but I don't have a proof
– Lee
Nov 20 at 2:19
if $A=begin{bmatrix}4 & & \ & 3 & \ & &2end{bmatrix}$, then again $A-Ngeq 0$ and $trace(A)=9$. I think this is minimum, but I don't have a proof
– Lee
Nov 20 at 2:19
if $N$ is one of following structures $begin{bmatrix}+& + & +\+ & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &- \+ & - &+end{bmatrix}$, $begin{bmatrix}+& - &-\- & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& + & -\+ & + &- \- & - &+end{bmatrix}$, then I believe $min tr(A)=sum n_{ij}$.
– Lee
Nov 20 at 2:26
if $N$ is one of following structures $begin{bmatrix}+& + & +\+ & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &- \+ & - &+end{bmatrix}$, $begin{bmatrix}+& - &-\- & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& + & -\+ & + &- \- & - &+end{bmatrix}$, then I believe $min tr(A)=sum n_{ij}$.
– Lee
Nov 20 at 2:26
if $N$ is one of following structures $begin{bmatrix}+& + & -\+ & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& + &+\+ & + &- \+ & - &+end{bmatrix}$, then I believe $min tr(A)=sum |n_{ij}|-4|n_{12}| $, assuming $|n_{12}|leq |n_{13}| leq |n_{23}|$.
– Lee
Nov 20 at 9:12
if $N$ is one of following structures $begin{bmatrix}+& + & -\+ & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& + &+\+ & + &- \+ & - &+end{bmatrix}$, then I believe $min tr(A)=sum |n_{ij}|-4|n_{12}| $, assuming $|n_{12}|leq |n_{13}| leq |n_{23}|$.
– Lee
Nov 20 at 9:12
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$A =lambda_{max}(N)I$ is feasible, so $n lambda_{max}(N)$ is an obvious bound. I do not see an easy way to get a tighter bound.
– LinAlg
Nov 19 at 14:14
If $N=begin{bmatrix}3 & 1 & -1\1 & 2 &1 \-1 & 1 &2end{bmatrix}>0$, then $nlambda_{max}(N)=11.1963$. Let $A=begin{bmatrix}4 & & \ & 3 & \ & &3end{bmatrix}$, then $A-Ngeq 0$ and $trace(A)=10$
– Lee
Nov 20 at 1:57
if $A=begin{bmatrix}4 & & \ & 3 & \ & &2end{bmatrix}$, then again $A-Ngeq 0$ and $trace(A)=9$. I think this is minimum, but I don't have a proof
– Lee
Nov 20 at 2:19
if $N$ is one of following structures $begin{bmatrix}+& + & +\+ & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &- \+ & - &+end{bmatrix}$, $begin{bmatrix}+& - &-\- & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& + & -\+ & + &- \- & - &+end{bmatrix}$, then I believe $min tr(A)=sum n_{ij}$.
– Lee
Nov 20 at 2:26
if $N$ is one of following structures $begin{bmatrix}+& + & -\+ & + &+ \- & + &+end{bmatrix}$, $begin{bmatrix}+& - & +\- & + &+ \+ & + &+end{bmatrix}$, $begin{bmatrix}+& + &+\+ & + &- \+ & - &+end{bmatrix}$, then I believe $min tr(A)=sum |n_{ij}|-4|n_{12}| $, assuming $|n_{12}|leq |n_{13}| leq |n_{23}|$.
– Lee
Nov 20 at 9:12