Minimize trace of $A$ given that $A−N$ is positive semi-definite and $A$ is diagonal












1














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







share|cite|improve this question






















  • $A =lambda_{max}(N)I$ 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 '18 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 '18 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 '18 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 '18 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 '18 at 9:12


















1














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







share|cite|improve this question






















  • $A =lambda_{max}(N)I$ 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 '18 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 '18 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 '18 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 '18 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 '18 at 9:12
















1












1








1







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











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asked Nov 19 '18 at 3:24









Lee

1077




1077












  • $A =lambda_{max}(N)I$ 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 9:12

















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