Mixing
Linear operator cancellation
$begingroup$
I came across this by looking at Green's function motivation topic:
suppose that $L$ is a linear operator (acting on a function, distribution (as in this case) or whathever). Take the equation:
$$L left (u(x) right)=int L left (G(x,s) right)f(s)ds $$
Suppose moreover that $L$ does not act on the variable "$s$", so:
$$L(u(x)) =L int G(x,s)f(s)ds$$
Now as you can read on the text, this "suggest" the equation:
$$ u(x)=int G(x,s)f(s)ds$$
Could you give me clarification on this last step?
Is this always true,indipendently of what kind of linear operator $L$ is and indipendently on what it acts on?
E.g. , suppose that $L$ acts on vectors:
$$ Lmathbf{x}=Lmathbf{y} $$
to get the equation $mathbf{x}=mathbf{y} $ I should multiply both members by $L^{-1}$ and this looks me "prone" to the existence of the inverse.
linear-algebra
asked Jan 21 at 11:15
R.LacR.Lac
204
$endgroup$
add a comment |
$begingroup$
I came across this by looking at Green's function motivation topic:
suppose that $L$ is a linear operator (acting on a function, distribution (as in this case) or whathever). Take the equation:
$$L left (u(x) right)=int L left (G(x,s) right)f(s)ds $$
Suppose moreover that $L$ does not act on the variable "$s$", so:
$$L(u(x)) =L int G(x,s)f(s)ds$$
Now as you can read on the text, this "suggest" the equation:
$$ u(x)=int G(x,s)f(s)ds$$
Could you give me clarification on this last step?
Is this always true,indipendently of what kind of linear operator $L$ is and indipendently on what it acts on?
E.g. , suppose that $L$ acts on vectors:
$$ Lmathbf{x}=Lmathbf{y} $$
to get the equation $mathbf{x}=mathbf{y} $ I should multiply both members by $L^{-1}$ and this looks me "prone" to the existence of the inverse.
linear-algebra
asked Jan 21 at 11:15
R.LacR.Lac
204
$endgroup$
1
$begingroup$
If there exists an inverse that is true. However if you have a nontrivial nullspace you can not omit the operator. Take for example the (linear) differentiation operator.
$endgroup$
– maxmilgram
Jan 21 at 11:25
$begingroup$
Yes, E.g $ frac{partial }{partial x}left ( x+y right )=frac{partial }{partial x}left ( x right )$ is different from $x+y=x$. So according to this, we can omit the operator if the kernel has dim=0, right? This is what do you mean for "trivial" ?
$endgroup$
– R.Lac
Jan 21 at 11:55
$begingroup$
Exactly. Another thing: If the operator is fixed you could also try to restrict it's domain to achieve a trivial nullspace. In this example: if you already know that your functions $f$ and $g$ are BOTH of form $c_1sin(x) + c_2cos(x)$ you can deduce that $f=g$ even though your operator has nontrivial kernel on it's "natural" domain $C^1$.
$endgroup$
– maxmilgram
Jan 21 at 13:55
1
$begingroup$
Another way to put it: $ Lmathbf{x}=Lmathbf{y} $ implies that $ mathbf{x}=mathbf{y} +mathbf{z}$ where $mathbf{z}inker L$.
$endgroup$
– maxmilgram
Jan 21 at 16:55
add a comment |
$begingroup$
I came across this by looking at Green's function motivation topic:
suppose that $L$ is a linear operator (acting on a function, distribution (as in this case) or whathever). Take the equation:
$$L left (u(x) right)=int L left (G(x,s) right)f(s)ds $$
Suppose moreover that $L$ does not act on the variable "$s$", so:
$$L(u(x)) =L int G(x,s)f(s)ds$$
Now as you can read on the text, this "suggest" the equation:
$$ u(x)=int G(x,s)f(s)ds$$
Could you give me clarification on this last step?
Is this always true,indipendently of what kind of linear operator $L$ is and indipendently on what it acts on?
E.g. , suppose that $L$ acts on vectors:
$$ Lmathbf{x}=Lmathbf{y} $$
to get the equation $mathbf{x}=mathbf{y} $ I should multiply both members by $L^{-1}$ and this looks me "prone" to the existence of the inverse.
linear-algebra
asked Jan 21 at 11:15
R.LacR.Lac
204
$endgroup$
I came across this by looking at Green's function motivation topic:
suppose that $L$ is a linear operator (acting on a function, distribution (as in this case) or whathever). Take the equation:
$$L left (u(x) right)=int L left (G(x,s) right)f(s)ds $$
Suppose moreover that $L$ does not act on the variable "$s$", so:
$$L(u(x)) =L int G(x,s)f(s)ds$$
Now as you can read on the text, this "suggest" the equation:
$$ u(x)=int G(x,s)f(s)ds$$
Could you give me clarification on this last step?
Is this always true,indipendently of what kind of linear operator $L$ is and indipendently on what it acts on?
E.g. , suppose that $L$ acts on vectors:
$$ Lmathbf{x}=Lmathbf{y} $$
to get the equation $mathbf{x}=mathbf{y} $ I should multiply both members by $L^{-1}$ and this looks me "prone" to the existence of the inverse.
linear-algebra
linear-algebra
asked Jan 21 at 11:15
R.LacR.Lac
204
asked Jan 21 at 11:15
R.LacR.Lac
204
asked Jan 21 at 11:15
R.LacR.Lac
204
asked Jan 21 at 11:15
R.LacR.Lac
204
asked Jan 21 at 11:15
R.LacR.Lac
204
204
1
$begingroup$
If there exists an inverse that is true. However if you have a nontrivial nullspace you can not omit the operator. Take for example the (linear) differentiation operator.
$endgroup$
– maxmilgram
Jan 21 at 11:25
$begingroup$
Yes, E.g $ frac{partial }{partial x}left ( x+y right )=frac{partial }{partial x}left ( x right )$ is different from $x+y=x$. So according to this, we can omit the operator if the kernel has dim=0, right? This is what do you mean for "trivial" ?
$endgroup$
– R.Lac
Jan 21 at 11:55
$begingroup$
Exactly. Another thing: If the operator is fixed you could also try to restrict it's domain to achieve a trivial nullspace. In this example: if you already know that your functions $f$ and $g$ are BOTH of form $c_1sin(x) + c_2cos(x)$ you can deduce that $f=g$ even though your operator has nontrivial kernel on it's "natural" domain $C^1$.
$endgroup$
– maxmilgram
Jan 21 at 13:55
1
$begingroup$
Another way to put it: $ Lmathbf{x}=Lmathbf{y} $ implies that $ mathbf{x}=mathbf{y} +mathbf{z}$ where $mathbf{z}inker L$.
$endgroup$
– maxmilgram
Jan 21 at 16:55
add a comment |
1
$begingroup$
If there exists an inverse that is true. However if you have a nontrivial nullspace you can not omit the operator. Take for example the (linear) differentiation operator.
$endgroup$
– maxmilgram
Jan 21 at 11:25
$begingroup$
Yes, E.g $ frac{partial }{partial x}left ( x+y right )=frac{partial }{partial x}left ( x right )$ is different from $x+y=x$. So according to this, we can omit the operator if the kernel has dim=0, right? This is what do you mean for "trivial" ?
$endgroup$
– R.Lac
Jan 21 at 11:55
$begingroup$
Exactly. Another thing: If the operator is fixed you could also try to restrict it's domain to achieve a trivial nullspace. In this example: if you already know that your functions $f$ and $g$ are BOTH of form $c_1sin(x) + c_2cos(x)$ you can deduce that $f=g$ even though your operator has nontrivial kernel on it's "natural" domain $C^1$.
$endgroup$
– maxmilgram
Jan 21 at 13:55
1
$begingroup$
Another way to put it: $ Lmathbf{x}=Lmathbf{y} $ implies that $ mathbf{x}=mathbf{y} +mathbf{z}$ where $mathbf{z}inker L$.
$endgroup$
– maxmilgram
Jan 21 at 16:55
1
1
$begingroup$
If there exists an inverse that is true. However if you have a nontrivial nullspace you can not omit the operator. Take for example the (linear) differentiation operator.
$endgroup$
– maxmilgram
Jan 21 at 11:25
$begingroup$
If there exists an inverse that is true. However if you have a nontrivial nullspace you can not omit the operator. Take for example the (linear) differentiation operator.
$endgroup$
– maxmilgram
Jan 21 at 11:25
$begingroup$
Yes, E.g $ frac{partial }{partial x}left ( x+y right )=frac{partial }{partial x}left ( x right )$ is different from $x+y=x$. So according to this, we can omit the operator if the kernel has dim=0, right? This is what do you mean for "trivial" ?
$endgroup$
– R.Lac
Jan 21 at 11:55
$begingroup$
Yes, E.g $ frac{partial }{partial x}left ( x+y right )=frac{partial }{partial x}left ( x right )$ is different from $x+y=x$. So according to this, we can omit the operator if the kernel has dim=0, right? This is what do you mean for "trivial" ?
$endgroup$
– R.Lac
Jan 21 at 11:55
$begingroup$
Exactly. Another thing: If the operator is fixed you could also try to restrict it's domain to achieve a trivial nullspace. In this example: if you already know that your functions $f$ and $g$ are BOTH of form $c_1sin(x) + c_2cos(x)$ you can deduce that $f=g$ even though your operator has nontrivial kernel on it's "natural" domain $C^1$.
$endgroup$
– maxmilgram
Jan 21 at 13:55
$begingroup$
Exactly. Another thing: If the operator is fixed you could also try to restrict it's domain to achieve a trivial nullspace. In this example: if you already know that your functions $f$ and $g$ are BOTH of form $c_1sin(x) + c_2cos(x)$ you can deduce that $f=g$ even though your operator has nontrivial kernel on it's "natural" domain $C^1$.
$endgroup$
– maxmilgram
Jan 21 at 13:55
1
1
$begingroup$
Another way to put it: $ Lmathbf{x}=Lmathbf{y} $ implies that $ mathbf{x}=mathbf{y} +mathbf{z}$ where $mathbf{z}inker L$.
$endgroup$
– maxmilgram
Jan 21 at 16:55
$begingroup$
Another way to put it: $ Lmathbf{x}=Lmathbf{y} $ implies that $ mathbf{x}=mathbf{y} +mathbf{z}$ where $mathbf{z}inker L$.
$endgroup$
– maxmilgram
Jan 21 at 16:55
add a comment |
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$begingroup$
If there exists an inverse that is true. However if you have a nontrivial nullspace you can not omit the operator. Take for example the (linear) differentiation operator.
$endgroup$
– maxmilgram
Jan 21 at 11:25
$begingroup$
Yes, E.g $ frac{partial }{partial x}left ( x+y right )=frac{partial }{partial x}left ( x right )$ is different from $x+y=x$. So according to this, we can omit the operator if the kernel has dim=0, right? This is what do you mean for "trivial" ?
$endgroup$
– R.Lac
Jan 21 at 11:55
$begingroup$
Exactly. Another thing: If the operator is fixed you could also try to restrict it's domain to achieve a trivial nullspace. In this example: if you already know that your functions $f$ and $g$ are BOTH of form $c_1sin(x) + c_2cos(x)$ you can deduce that $f=g$ even though your operator has nontrivial kernel on it's "natural" domain $C^1$.
$endgroup$
– maxmilgram
Jan 21 at 13:55
1
$begingroup$
Another way to put it: $ Lmathbf{x}=Lmathbf{y} $ implies that $ mathbf{x}=mathbf{y} +mathbf{z}$ where $mathbf{z}inker L$.
$endgroup$
– maxmilgram
Jan 21 at 16:55