Question about an inequation described by matrices
$A=(a_{ij})_{1 le i, j le n}$ is a matrix that$sum_limits{i=1}^{n} a_{ij}=1$ for every j and $sum_limits{j=1}^n a_{ij} = 1$ for every i and $a_{ij} ge 0$.And
$$begin{equation}
begin{pmatrix}
y_1 \
vdots \
y_n \
end{pmatrix}
=mathbf{A}
begin{pmatrix}
x_1 \
vdots \
x_n
end{pmatrix}
end{equation}$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$
It may somehow matter to convex function.
linear-algebra inequalities convex-analysis
edited 46 mins ago
user44191
2,5431126
asked 2 hours ago
X.T Chen
112
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$A=(a_{ij})_{1 le i, j le n}$ is a matrix that$sum_limits{i=1}^{n} a_{ij}=1$ for every j and $sum_limits{j=1}^n a_{ij} = 1$ for every i and $a_{ij} ge 0$.And
$$begin{equation}
begin{pmatrix}
y_1 \
vdots \
y_n \
end{pmatrix}
=mathbf{A}
begin{pmatrix}
x_1 \
vdots \
x_n
end{pmatrix}
end{equation}$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$
It may somehow matter to convex function.
linear-algebra inequalities convex-analysis
edited 46 mins ago
user44191
2,5431126
asked 2 hours ago
X.T Chen
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
$A=(a_{ij})_{1 le i, j le n}$ is a matrix that$sum_limits{i=1}^{n} a_{ij}=1$ for every j and $sum_limits{j=1}^n a_{ij} = 1$ for every i and $a_{ij} ge 0$.And
$$begin{equation}
begin{pmatrix}
y_1 \
vdots \
y_n \
end{pmatrix}
=mathbf{A}
begin{pmatrix}
x_1 \
vdots \
x_n
end{pmatrix}
end{equation}$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$
It may somehow matter to convex function.
linear-algebra inequalities convex-analysis
edited 46 mins ago
user44191
2,5431126
asked 2 hours ago
X.T Chen
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$A=(a_{ij})_{1 le i, j le n}$ is a matrix that$sum_limits{i=1}^{n} a_{ij}=1$ for every j and $sum_limits{j=1}^n a_{ij} = 1$ for every i and $a_{ij} ge 0$.And
$$begin{equation}
begin{pmatrix}
y_1 \
vdots \
y_n \
end{pmatrix}
=mathbf{A}
begin{pmatrix}
x_1 \
vdots \
x_n
end{pmatrix}
end{equation}$$
$y_i$ and $x_i$ are all nonnegative.Prove that : $y_1 cdots y_n ge x_1 cdots x_n$
It may somehow matter to convex function.
linear-algebra inequalities convex-analysis
linear-algebra inequalities convex-analysis
edited 46 mins ago
user44191
2,5431126
asked 2 hours ago
X.T Chen
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 46 mins ago
user44191
2,5431126
asked 2 hours ago
X.T Chen
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 46 mins ago
user44191
2,5431126
edited 46 mins ago
user44191
2,5431126
edited 46 mins ago
user44191
2,5431126
2,5431126
asked 2 hours ago
X.T Chen
112
New contributor
X.T Chen 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
X.T Chen
112
asked 2 hours ago
X.T Chen
112
112
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
X.T Chen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 1 hour ago
Iosif Pinelis
17.7k12158
add a comment |
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1 Answer
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1 Answer
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votes
By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 1 hour ago
Iosif Pinelis
17.7k12158
add a comment |
By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 1 hour ago
Iosif Pinelis
17.7k12158
add a comment |
By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 1 hour ago
Iosif Pinelis
17.7k12158
By the equivalence of conditions (ii) and (iv) in Theorem A.3 on page 14, the condition $y=Ax$ for $y=[y,dots,y_n]^T$ and $x=[x,dots,x_n]^T$ is equivalent to the condition that $sum_1^n g(y_i)gesum_1^n g(x_i)$ for all continuous concave functions $g$. Now take $g=ln$ to get the desired inequality $y_1 cdots y_n ge x_1 cdots x_n$.
answered 1 hour ago
Iosif Pinelis
17.7k12158
answered 1 hour ago
Iosif Pinelis
17.7k12158
answered 1 hour ago
Iosif Pinelis
17.7k12158
answered 1 hour ago
Iosif Pinelis
17.7k12158
17.7k12158
add a comment |
add a comment |
X.T Chen is a new contributor. Be nice, and check out our Code of Conduct.
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