What does this notation mean? How can an isomorphism be a function?
$begingroup$
I'm extremely confused with this notation.
Could someone translate the second and third line of this example into something a little more declarative or understandable for me?
On the second line it is given that there is an isomorphism between the space of bounded linear transformations from R^2 to R^3 and the space of 3x2 matrices. This means that there exists a bijective linear transformation between the two spaces. How can the isomorphism be used in a function declaration? How can I interpret the matrix given below in terms of that definition?
Thanks so much
real-analysis functional-analysis
asked Jan 2 at 21:36
Kohler FryerKohler Fryer
1102
$endgroup$
add a comment |
$begingroup$
I'm extremely confused with this notation.
Could someone translate the second and third line of this example into something a little more declarative or understandable for me?
On the second line it is given that there is an isomorphism between the space of bounded linear transformations from R^2 to R^3 and the space of 3x2 matrices. This means that there exists a bijective linear transformation between the two spaces. How can the isomorphism be used in a function declaration? How can I interpret the matrix given below in terms of that definition?
Thanks so much
real-analysis functional-analysis
asked Jan 2 at 21:36
Kohler FryerKohler Fryer
1102
$endgroup$
2
$begingroup$
What else is an isomorphism going to be, other than a function? Check your definitions!
$endgroup$
– user3482749
Jan 2 at 21:38
2
$begingroup$
I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
$endgroup$
– Riley
Jan 3 at 0:31
add a comment |
$begingroup$
I'm extremely confused with this notation.
Could someone translate the second and third line of this example into something a little more declarative or understandable for me?
On the second line it is given that there is an isomorphism between the space of bounded linear transformations from R^2 to R^3 and the space of 3x2 matrices. This means that there exists a bijective linear transformation between the two spaces. How can the isomorphism be used in a function declaration? How can I interpret the matrix given below in terms of that definition?
Thanks so much
real-analysis functional-analysis
asked Jan 2 at 21:36
Kohler FryerKohler Fryer
1102
$endgroup$
I'm extremely confused with this notation.
Could someone translate the second and third line of this example into something a little more declarative or understandable for me?
On the second line it is given that there is an isomorphism between the space of bounded linear transformations from R^2 to R^3 and the space of 3x2 matrices. This means that there exists a bijective linear transformation between the two spaces. How can the isomorphism be used in a function declaration? How can I interpret the matrix given below in terms of that definition?
Thanks so much
real-analysis functional-analysis
real-analysis functional-analysis
asked Jan 2 at 21:36
Kohler FryerKohler Fryer
1102
asked Jan 2 at 21:36
Kohler FryerKohler Fryer
1102
asked Jan 2 at 21:36
Kohler FryerKohler Fryer
1102
asked Jan 2 at 21:36
Kohler FryerKohler Fryer
1102
asked Jan 2 at 21:36
Kohler FryerKohler Fryer
1102
1102
2
$begingroup$
What else is an isomorphism going to be, other than a function? Check your definitions!
$endgroup$
– user3482749
Jan 2 at 21:38
2
$begingroup$
I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
$endgroup$
– Riley
Jan 3 at 0:31
add a comment |
2
$begingroup$
What else is an isomorphism going to be, other than a function? Check your definitions!
$endgroup$
– user3482749
Jan 2 at 21:38
2
$begingroup$
I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
$endgroup$
– Riley
Jan 3 at 0:31
2
2
$begingroup$
What else is an isomorphism going to be, other than a function? Check your definitions!
$endgroup$
– user3482749
Jan 2 at 21:38
$begingroup$
What else is an isomorphism going to be, other than a function? Check your definitions!
$endgroup$
– user3482749
Jan 2 at 21:38
2
2
$begingroup$
I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
$endgroup$
– Riley
Jan 3 at 0:31
$begingroup$
I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
$endgroup$
– Riley
Jan 3 at 0:31
add a comment |
1 Answer
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$begingroup$
It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
$$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$
This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
$$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
{partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$
answered Jan 3 at 10:49
Christian BlatterChristian Blatter
176k8115327
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1 Answer
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$begingroup$
It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
$$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$
This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
$$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
{partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$
answered Jan 3 at 10:49
Christian BlatterChristian Blatter
176k8115327
$endgroup$
add a comment |
$begingroup$
It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
$$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$
This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
$$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
{partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$
answered Jan 3 at 10:49
Christian BlatterChristian Blatter
176k8115327
$endgroup$
add a comment |
$begingroup$
It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
$$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$
This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
$$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
{partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$
answered Jan 3 at 10:49
Christian BlatterChristian Blatter
176k8115327
$endgroup$
It is a standard fact of linear algebra that any linear map $A: >Xto Y$ between finite dimensional vector spaces with chosen bases $({bf e}_k)_{1leq kleq n}$, resp. $({bf f}_i)_{1leq ileq m}$, has a uniquely determined $(mtimes n)$-matrix $[A]$ describing this map "computation wise". For each $(i,k)in[m]times[n]$ the element $a_{ik}$ of this matrix is the $i^{rm th}$ coordinate of the vector $A{bf e}_k$:
$$A{bf e}_k=sum_{i=1}^m a_{ik},{bf f}_i .$$
This general principle is now invoked when dealing with a differentiable function $$f:>(x,y)mapsto(x^3,xy,y^2)in{mathbb R}^3 .$$ This $f$ has a derivative $$df({bf p}): >T_{bf p}mapsto T_{f({bf p})}tag{1}$$ at the point ${bf p}:=(2,3)$.
This derivative is a map $(1)$, and therefore has a matrix $[df({bf p})]in M_{3,2}({mathbb R})$. It turns out that the matrix elements are given by the partial derivatives of $f$ at ${bf p}$:
$$[df({bf p})]=left[matrix{{partial f_1overpartial x}&{partial f_1overpartial y}cr
{partial f_2overpartial x}&{partial f_2overpartial y}cr {partial f_3overpartial x}&{partial f_3overpartial y}cr}right]_{bf p} =left[matrix{3x^2&0cr
y&xcr 0&2ycr}right]_{(2,3)}=left[matrix{12&0cr 3&2cr0&6cr}right] .$$
answered Jan 3 at 10:49
Christian BlatterChristian Blatter
176k8115327
answered Jan 3 at 10:49
Christian BlatterChristian Blatter
176k8115327
answered Jan 3 at 10:49
Christian BlatterChristian Blatter
176k8115327
answered Jan 3 at 10:49
Christian BlatterChristian Blatter
176k8115327
176k8115327
add a comment |
add a comment |
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2
$begingroup$
What else is an isomorphism going to be, other than a function? Check your definitions!
$endgroup$
– user3482749
Jan 2 at 21:38
2
$begingroup$
I'm not sure without seeing the full statement, but I think it should probably read "the function $A in mathcal{B}(mathbb{R}^2,mathbb{R}^3) cong M_{3,2}(mathbb{R})$ given by ...", i.e. the function given by the matrix is not the isomorphism itself.
$endgroup$
– Riley
Jan 3 at 0:31