Maximize $det X$, subject to $X_{ii}leq P_i$, where $X>0$
$begingroup$
Given $P_1,P_2,cdots,P_N$.
begin{array}{ll} text{maximize} & det X\ text{subject to} & mathrm X_{ii}leq P_i \forall i=1,2,cdots,nend{array}
$Xinmathbb{R}^{ntimes n}$, $X>0$ (i.e. positive definite).
For $n=2$, maximum is achieved when $X_{ii}=P_i$ and $X$ is diagonal matrix.
Then for $n=3$, Let $X=begin{bmatrix}X_1 & X_2\X_2^T &x_3end{bmatrix}$, where $X_1inmathbb{R}^{2times2}$.
$det(X)=det(X_1)times det(x_3-X_2^TX_1^{-1}X_2)leq det(X_1)times x_3leq P_1P_2P_3$.
Then following same logic my conjecture is $max{ det X}=P_1cdots P_n$, when $X_{ii}=P_i$.
Is it correct? If not, can you please point my mistake or give me counter example? Thanks
optimization determinant positive-definite
edited Dec 15 '18 at 3:48
Tianlalu
3,08621038
asked Dec 15 '18 at 3:17
LeeLee
330111
$endgroup$
add a comment |
$begingroup$
Given $P_1,P_2,cdots,P_N$.
begin{array}{ll} text{maximize} & det X\ text{subject to} & mathrm X_{ii}leq P_i \forall i=1,2,cdots,nend{array}
$Xinmathbb{R}^{ntimes n}$, $X>0$ (i.e. positive definite).
For $n=2$, maximum is achieved when $X_{ii}=P_i$ and $X$ is diagonal matrix.
Then for $n=3$, Let $X=begin{bmatrix}X_1 & X_2\X_2^T &x_3end{bmatrix}$, where $X_1inmathbb{R}^{2times2}$.
$det(X)=det(X_1)times det(x_3-X_2^TX_1^{-1}X_2)leq det(X_1)times x_3leq P_1P_2P_3$.
Then following same logic my conjecture is $max{ det X}=P_1cdots P_n$, when $X_{ii}=P_i$.
Is it correct? If not, can you please point my mistake or give me counter example? Thanks
optimization determinant positive-definite
edited Dec 15 '18 at 3:48
Tianlalu
3,08621038
asked Dec 15 '18 at 3:17
LeeLee
330111
$endgroup$
add a comment |
$begingroup$
Given $P_1,P_2,cdots,P_N$.
begin{array}{ll} text{maximize} & det X\ text{subject to} & mathrm X_{ii}leq P_i \forall i=1,2,cdots,nend{array}
$Xinmathbb{R}^{ntimes n}$, $X>0$ (i.e. positive definite).
For $n=2$, maximum is achieved when $X_{ii}=P_i$ and $X$ is diagonal matrix.
Then for $n=3$, Let $X=begin{bmatrix}X_1 & X_2\X_2^T &x_3end{bmatrix}$, where $X_1inmathbb{R}^{2times2}$.
$det(X)=det(X_1)times det(x_3-X_2^TX_1^{-1}X_2)leq det(X_1)times x_3leq P_1P_2P_3$.
Then following same logic my conjecture is $max{ det X}=P_1cdots P_n$, when $X_{ii}=P_i$.
Is it correct? If not, can you please point my mistake or give me counter example? Thanks
optimization determinant positive-definite
edited Dec 15 '18 at 3:48
Tianlalu
3,08621038
asked Dec 15 '18 at 3:17
LeeLee
330111
$endgroup$
Given $P_1,P_2,cdots,P_N$.
begin{array}{ll} text{maximize} & det X\ text{subject to} & mathrm X_{ii}leq P_i \forall i=1,2,cdots,nend{array}
$Xinmathbb{R}^{ntimes n}$, $X>0$ (i.e. positive definite).
For $n=2$, maximum is achieved when $X_{ii}=P_i$ and $X$ is diagonal matrix.
Then for $n=3$, Let $X=begin{bmatrix}X_1 & X_2\X_2^T &x_3end{bmatrix}$, where $X_1inmathbb{R}^{2times2}$.
$det(X)=det(X_1)times det(x_3-X_2^TX_1^{-1}X_2)leq det(X_1)times x_3leq P_1P_2P_3$.
Then following same logic my conjecture is $max{ det X}=P_1cdots P_n$, when $X_{ii}=P_i$.
Is it correct? If not, can you please point my mistake or give me counter example? Thanks
optimization determinant positive-definite
optimization determinant positive-definite
edited Dec 15 '18 at 3:48
Tianlalu
3,08621038
asked Dec 15 '18 at 3:17
LeeLee
330111
edited Dec 15 '18 at 3:48
Tianlalu
3,08621038
asked Dec 15 '18 at 3:17
LeeLee
330111
edited Dec 15 '18 at 3:48
Tianlalu
3,08621038
edited Dec 15 '18 at 3:48
Tianlalu
3,08621038
edited Dec 15 '18 at 3:48
Tianlalu
3,08621038
3,08621038
asked Dec 15 '18 at 3:17
LeeLee
330111
asked Dec 15 '18 at 3:17
LeeLee
330111
asked Dec 15 '18 at 3:17
LeeLee
330111
330111
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
$$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
$$(X^{-1})_{ii} - lambda_i = 0$$
$$(X^{-1})_{ij} = 0 quad (i neq j)$$
$$lambda_i (X_{ii} - P_i)=0$$
$$X_{ii} leq P_i$$
$$lambda geq 0.$$
The point you found satisfies these conditions and is therefore optimal.
answered Dec 15 '18 at 16:34
LinAlgLinAlg
9,5491521
$endgroup$
$begingroup$
what if rank$ X< n$? I think in this case these points are not optimal
$endgroup$
– Lee
Dec 18 '18 at 9:24
1
$begingroup$
@Lee a positive definite matrix has full rank.
$endgroup$
– LinAlg
Dec 18 '18 at 15:13
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
$$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
$$(X^{-1})_{ii} - lambda_i = 0$$
$$(X^{-1})_{ij} = 0 quad (i neq j)$$
$$lambda_i (X_{ii} - P_i)=0$$
$$X_{ii} leq P_i$$
$$lambda geq 0.$$
The point you found satisfies these conditions and is therefore optimal.
answered Dec 15 '18 at 16:34
LinAlgLinAlg
9,5491521
$endgroup$
$begingroup$
what if rank$ X< n$? I think in this case these points are not optimal
$endgroup$
– Lee
Dec 18 '18 at 9:24
1
$begingroup$
@Lee a positive definite matrix has full rank.
$endgroup$
– LinAlg
Dec 18 '18 at 15:13
add a comment |
$begingroup$
Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
$$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
$$(X^{-1})_{ii} - lambda_i = 0$$
$$(X^{-1})_{ij} = 0 quad (i neq j)$$
$$lambda_i (X_{ii} - P_i)=0$$
$$X_{ii} leq P_i$$
$$lambda geq 0.$$
The point you found satisfies these conditions and is therefore optimal.
answered Dec 15 '18 at 16:34
LinAlgLinAlg
9,5491521
$endgroup$
$begingroup$
what if rank$ X< n$? I think in this case these points are not optimal
$endgroup$
– Lee
Dec 18 '18 at 9:24
1
$begingroup$
@Lee a positive definite matrix has full rank.
$endgroup$
– LinAlg
Dec 18 '18 at 15:13
add a comment |
$begingroup$
Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
$$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
$$(X^{-1})_{ii} - lambda_i = 0$$
$$(X^{-1})_{ij} = 0 quad (i neq j)$$
$$lambda_i (X_{ii} - P_i)=0$$
$$X_{ii} leq P_i$$
$$lambda geq 0.$$
The point you found satisfies these conditions and is therefore optimal.
answered Dec 15 '18 at 16:34
LinAlgLinAlg
9,5491521
$endgroup$
Your problem is equivalent to maximizing $log det X$ such that $X_{ii} leq P_i$ and $X$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is
$$L(X,lambda) = logdet X - sum_i lambda_i (X_{ii} - P_i)$$
The derivative of $logdet X$ is $(X^{-1})^T$, so the KKT conditions are:
$$(X^{-1})_{ii} - lambda_i = 0$$
$$(X^{-1})_{ij} = 0 quad (i neq j)$$
$$lambda_i (X_{ii} - P_i)=0$$
$$X_{ii} leq P_i$$
$$lambda geq 0.$$
The point you found satisfies these conditions and is therefore optimal.
answered Dec 15 '18 at 16:34
LinAlgLinAlg
9,5491521
answered Dec 15 '18 at 16:34
LinAlgLinAlg
9,5491521
answered Dec 15 '18 at 16:34
LinAlgLinAlg
9,5491521
answered Dec 15 '18 at 16:34
LinAlgLinAlg
9,5491521
9,5491521
$begingroup$
what if rank$ X< n$? I think in this case these points are not optimal
$endgroup$
– Lee
Dec 18 '18 at 9:24
1
$begingroup$
@Lee a positive definite matrix has full rank.
$endgroup$
– LinAlg
Dec 18 '18 at 15:13
add a comment |
$begingroup$
what if rank$ X< n$? I think in this case these points are not optimal
$endgroup$
– Lee
Dec 18 '18 at 9:24
1
$begingroup$
@Lee a positive definite matrix has full rank.
$endgroup$
– LinAlg
Dec 18 '18 at 15:13
$begingroup$
what if rank$ X< n$? I think in this case these points are not optimal
$endgroup$
– Lee
Dec 18 '18 at 9:24
$begingroup$
what if rank$ X< n$? I think in this case these points are not optimal
$endgroup$
– Lee
Dec 18 '18 at 9:24
1
1
$begingroup$
@Lee a positive definite matrix has full rank.
$endgroup$
– LinAlg
Dec 18 '18 at 15:13
$begingroup$
@Lee a positive definite matrix has full rank.
$endgroup$
– LinAlg
Dec 18 '18 at 15:13
add a comment |
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