Lipschitz constant of a matrix
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I am studying the Lipschitz continuity and trying to solve the following question:
If a function $f(x)= Ax$ is defined for $x in mathbb{R}^2$ with $A= begin{bmatrix}
a & b \
c & d
end{bmatrix}$, then find a constant L such that
begin{eqnarray*}
||Ax - Ay|| le L||x - y||, x, y in mathbb{R}^2.
end{eqnarray*}
I understand how to find the Lipschitz constant in $mathbb{R}$, but I have no idea about how to find it in $mathbb{R}^2$.
matrices multivariable-calculus continuity lipschitz-functions
asked Jan 18 at 3:54
J.RossJ.Ross
114
$endgroup$
add a comment |
$begingroup$
I am studying the Lipschitz continuity and trying to solve the following question:
If a function $f(x)= Ax$ is defined for $x in mathbb{R}^2$ with $A= begin{bmatrix}
a & b \
c & d
end{bmatrix}$, then find a constant L such that
begin{eqnarray*}
||Ax - Ay|| le L||x - y||, x, y in mathbb{R}^2.
end{eqnarray*}
I understand how to find the Lipschitz constant in $mathbb{R}$, but I have no idea about how to find it in $mathbb{R}^2$.
matrices multivariable-calculus continuity lipschitz-functions
asked Jan 18 at 3:54
J.RossJ.Ross
114
$endgroup$
$begingroup$
Presumably, you want a constant $L$ such that $| Ax_{1} -Ax_{2}| leq L | x_{1}-x_{2} |$, right? What do you know about the norm of the matrix $A$?
$endgroup$
– Brian Borchers
Jan 18 at 4:04
add a comment |
$begingroup$
I am studying the Lipschitz continuity and trying to solve the following question:
If a function $f(x)= Ax$ is defined for $x in mathbb{R}^2$ with $A= begin{bmatrix}
a & b \
c & d
end{bmatrix}$, then find a constant L such that
begin{eqnarray*}
||Ax - Ay|| le L||x - y||, x, y in mathbb{R}^2.
end{eqnarray*}
I understand how to find the Lipschitz constant in $mathbb{R}$, but I have no idea about how to find it in $mathbb{R}^2$.
matrices multivariable-calculus continuity lipschitz-functions
asked Jan 18 at 3:54
J.RossJ.Ross
114
$endgroup$
I am studying the Lipschitz continuity and trying to solve the following question:
If a function $f(x)= Ax$ is defined for $x in mathbb{R}^2$ with $A= begin{bmatrix}
a & b \
c & d
end{bmatrix}$, then find a constant L such that
begin{eqnarray*}
||Ax - Ay|| le L||x - y||, x, y in mathbb{R}^2.
end{eqnarray*}
I understand how to find the Lipschitz constant in $mathbb{R}$, but I have no idea about how to find it in $mathbb{R}^2$.
matrices multivariable-calculus continuity lipschitz-functions
matrices multivariable-calculus continuity lipschitz-functions
asked Jan 18 at 3:54
J.RossJ.Ross
114
asked Jan 18 at 3:54
J.RossJ.Ross
114
edited Jan 18 at 5:28
J.Ross
asked Jan 18 at 3:54
J.RossJ.Ross
114
asked Jan 18 at 3:54
J.RossJ.Ross
114
asked Jan 18 at 3:54
J.RossJ.Ross
114
114
$begingroup$
Presumably, you want a constant $L$ such that $| Ax_{1} -Ax_{2}| leq L | x_{1}-x_{2} |$, right? What do you know about the norm of the matrix $A$?
$endgroup$
– Brian Borchers
Jan 18 at 4:04
add a comment |
$begingroup$
Presumably, you want a constant $L$ such that $| Ax_{1} -Ax_{2}| leq L | x_{1}-x_{2} |$, right? What do you know about the norm of the matrix $A$?
$endgroup$
– Brian Borchers
Jan 18 at 4:04
$begingroup$
Presumably, you want a constant $L$ such that $| Ax_{1} -Ax_{2}| leq L | x_{1}-x_{2} |$, right? What do you know about the norm of the matrix $A$?
$endgroup$
– Brian Borchers
Jan 18 at 4:04
$begingroup$
Presumably, you want a constant $L$ such that $| Ax_{1} -Ax_{2}| leq L | x_{1}-x_{2} |$, right? What do you know about the norm of the matrix $A$?
$endgroup$
– Brian Borchers
Jan 18 at 4:04
add a comment |
1 Answer
1
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oldest
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For the Frobenius norm, given by $midmid Amidmid=sqrt{sum_{i=1}^nsum_{j=1}^mmid a_{ij}mid^2}$.
we get $midmid Ax_1-Ax_2midmidlemidmid Amidmidcdotmidmid x_1-x_2midmidle2operatorname{max}{mid amid,mid bmid,mid cmid,mid dmid}midmid x_1-x_2midmid$.
answered Jan 18 at 4:57
Chris CusterChris Custer
13.2k3827
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
For the Frobenius norm, given by $midmid Amidmid=sqrt{sum_{i=1}^nsum_{j=1}^mmid a_{ij}mid^2}$.
we get $midmid Ax_1-Ax_2midmidlemidmid Amidmidcdotmidmid x_1-x_2midmidle2operatorname{max}{mid amid,mid bmid,mid cmid,mid dmid}midmid x_1-x_2midmid$.
answered Jan 18 at 4:57
Chris CusterChris Custer
13.2k3827
$endgroup$
add a comment |
$begingroup$
For the Frobenius norm, given by $midmid Amidmid=sqrt{sum_{i=1}^nsum_{j=1}^mmid a_{ij}mid^2}$.
we get $midmid Ax_1-Ax_2midmidlemidmid Amidmidcdotmidmid x_1-x_2midmidle2operatorname{max}{mid amid,mid bmid,mid cmid,mid dmid}midmid x_1-x_2midmid$.
answered Jan 18 at 4:57
Chris CusterChris Custer
13.2k3827
$endgroup$
add a comment |
$begingroup$
For the Frobenius norm, given by $midmid Amidmid=sqrt{sum_{i=1}^nsum_{j=1}^mmid a_{ij}mid^2}$.
we get $midmid Ax_1-Ax_2midmidlemidmid Amidmidcdotmidmid x_1-x_2midmidle2operatorname{max}{mid amid,mid bmid,mid cmid,mid dmid}midmid x_1-x_2midmid$.
answered Jan 18 at 4:57
Chris CusterChris Custer
13.2k3827
$endgroup$
For the Frobenius norm, given by $midmid Amidmid=sqrt{sum_{i=1}^nsum_{j=1}^mmid a_{ij}mid^2}$.
we get $midmid Ax_1-Ax_2midmidlemidmid Amidmidcdotmidmid x_1-x_2midmidle2operatorname{max}{mid amid,mid bmid,mid cmid,mid dmid}midmid x_1-x_2midmid$.
answered Jan 18 at 4:57
Chris CusterChris Custer
13.2k3827
edited Jan 18 at 5:06
answered Jan 18 at 4:57
Chris CusterChris Custer
13.2k3827
answered Jan 18 at 4:57
Chris CusterChris Custer
13.2k3827
answered Jan 18 at 4:57
Chris CusterChris Custer
13.2k3827
13.2k3827
add a comment |
add a comment |
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$begingroup$
Presumably, you want a constant $L$ such that $| Ax_{1} -Ax_{2}| leq L | x_{1}-x_{2} |$, right? What do you know about the norm of the matrix $A$?
$endgroup$
– Brian Borchers
Jan 18 at 4:04