Mixing
Computing the Inverse of a two dimensional map?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
In order to find the inverse of the function $y = x^3$ where $y = f(x) = x^3$
we need $x = f^-1(y)$, which we compute it as $x = y^frac13$ so the inverse function.
But how do I calculate the inverse map of the following map? $x mapsto Ax +By + C$
and $y mapsto Dx$ ?, where $A,B,C,D$ are real numbers.
I was trying to visualize this in terms of matrices, $beginbmatrixx \ yendbmatrix mapsto beginbmatrix Ax + By + C \ Dxendbmatrix$, may that open up some new insights?
How can we guarantee the existence of the inverse for this two dimensional map?
functions inverse-function
asked Aug 14 at 5:41
BAYMAX
2,55021021
add a comment |Â
up vote
3
down vote
favorite
In order to find the inverse of the function $y = x^3$ where $y = f(x) = x^3$
we need $x = f^-1(y)$, which we compute it as $x = y^frac13$ so the inverse function.
But how do I calculate the inverse map of the following map? $x mapsto Ax +By + C$
and $y mapsto Dx$ ?, where $A,B,C,D$ are real numbers.
I was trying to visualize this in terms of matrices, $beginbmatrixx \ yendbmatrix mapsto beginbmatrix Ax + By + C \ Dxendbmatrix$, may that open up some new insights?
How can we guarantee the existence of the inverse for this two dimensional map?
functions inverse-function
asked Aug 14 at 5:41
BAYMAX
2,55021021
What are $x,y, A, B, C$? Real numbers, integers, matrices? Also, your first statement would seem to indicate $x^2 = y = f(x) = x^3$, was that intended?
– Guido A.
Aug 14 at 6:03
Thank you! I just did an edit to correct the typos.
– BAYMAX
Aug 14 at 6:09
Great, I'll give this some thought.
– Guido A.
Aug 14 at 6:10
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
In order to find the inverse of the function $y = x^3$ where $y = f(x) = x^3$
we need $x = f^-1(y)$, which we compute it as $x = y^frac13$ so the inverse function.
But how do I calculate the inverse map of the following map? $x mapsto Ax +By + C$
and $y mapsto Dx$ ?, where $A,B,C,D$ are real numbers.
I was trying to visualize this in terms of matrices, $beginbmatrixx \ yendbmatrix mapsto beginbmatrix Ax + By + C \ Dxendbmatrix$, may that open up some new insights?
How can we guarantee the existence of the inverse for this two dimensional map?
functions inverse-function
asked Aug 14 at 5:41
BAYMAX
2,55021021
In order to find the inverse of the function $y = x^3$ where $y = f(x) = x^3$
we need $x = f^-1(y)$, which we compute it as $x = y^frac13$ so the inverse function.
But how do I calculate the inverse map of the following map? $x mapsto Ax +By + C$
and $y mapsto Dx$ ?, where $A,B,C,D$ are real numbers.
I was trying to visualize this in terms of matrices, $beginbmatrixx \ yendbmatrix mapsto beginbmatrix Ax + By + C \ Dxendbmatrix$, may that open up some new insights?
How can we guarantee the existence of the inverse for this two dimensional map?
functions inverse-function
asked Aug 14 at 5:41
BAYMAX
2,55021021
edited Aug 14 at 6:13
asked Aug 14 at 5:41
BAYMAX
2,55021021
asked Aug 14 at 5:41
BAYMAX
2,55021021
asked Aug 14 at 5:41
BAYMAX
2,55021021
2,55021021
What are $x,y, A, B, C$? Real numbers, integers, matrices? Also, your first statement would seem to indicate $x^2 = y = f(x) = x^3$, was that intended?
– Guido A.
Aug 14 at 6:03
Thank you! I just did an edit to correct the typos.
– BAYMAX
Aug 14 at 6:09
Great, I'll give this some thought.
– Guido A.
Aug 14 at 6:10
add a comment |Â
What are $x,y, A, B, C$? Real numbers, integers, matrices? Also, your first statement would seem to indicate $x^2 = y = f(x) = x^3$, was that intended?
– Guido A.
Aug 14 at 6:03
Thank you! I just did an edit to correct the typos.
– BAYMAX
Aug 14 at 6:09
Great, I'll give this some thought.
– Guido A.
Aug 14 at 6:10
What are $x,y, A, B, C$? Real numbers, integers, matrices? Also, your first statement would seem to indicate $x^2 = y = f(x) = x^3$, was that intended?
– Guido A.
Aug 14 at 6:03
What are $x,y, A, B, C$? Real numbers, integers, matrices? Also, your first statement would seem to indicate $x^2 = y = f(x) = x^3$, was that intended?
– Guido A.
Aug 14 at 6:03
Thank you! I just did an edit to correct the typos.
– BAYMAX
Aug 14 at 6:09
Thank you! I just did an edit to correct the typos.
– BAYMAX
Aug 14 at 6:09
Great, I'll give this some thought.
– Guido A.
Aug 14 at 6:10
Great, I'll give this some thought.
– Guido A.
Aug 14 at 6:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
$x mapsto Ax +By + C$ and $y mapsto Dx$ can be rewritten as :
$$left(beginmatrix x' \ y' \ 1 endmatrix right)=Mleft(beginmatrix x \y\1endmatrix right)$$
Where
$$M=left(beginmatrix A & B & C \ D & 0 & 0 \ 0 & 0 &1endmatrix right).$$
If $M$ is invertible, then $left(beginmatrix x \y\1endmatrix right)=M^-1left(beginmatrix x' \ y' \ 1 endmatrix right)$.
So existence of an inverse is guaranteed by invertibility of the subjacent matrix when you have a linear system.
Which sums up here to $Dne 0 ne B$.
Note that we added a row for the affine part. By doing so, we where able to express your function as a linear system. We could have written also :
$$left(beginmatrix x' \ y' \ 0 endmatrix right)=M'left(beginmatrix x \y\1endmatrix right)$$
But the invertibility would have been more difficult to express in term of matrices.
answered Aug 14 at 6:12
nicomezi
3,6621819
Great answer. I just want to point out that, given that this is an upper triangular matrix, it is worth noting that the invertibility condition translates exactly to $A neq 0$, $D neq 0$.
– Guido A.
Aug 14 at 6:18
Yes indeed. Should I add your remark to the answer ? @GuidoA.
– nicomezi
Aug 14 at 6:21
Well, only if you want to :P It's clear from context if you know some linear algebra, but since I don't know OP's background I felt like making the remark.
– Guido A.
Aug 14 at 6:29
Perhaps using the matrix $M$ we get $y' = Dy$ ? but we require $y' = Dx$
– BAYMAX
Aug 14 at 7:04
Missread, updated.
– nicomezi
Aug 14 at 7:08
add a comment |Â
up vote
0
down vote
So you have the mapping $fcolon mathbbR^2to mathbbR^2$, $f(x,y)=(Ax+By+C,Dx)$. Claim: $f$ has an inverse iff $A,Bnot=0$. This follows from the fact that $f$ is invertible iff the linear system of equations $Ax+By=xi-C, Dx= eta$ is solvable for all $(xi,eta)in mathbbR^2$.
Missprint: The condition is $B,Dnot=0$
answered Aug 14 at 6:05
Jens Schwaiger
1,177116
If $A = B = 1$ and $C = D = 0$, we have $f(x,y) = (x+y,0)$ which is not injective.
– Guido A.
Aug 14 at 6:09
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
$x mapsto Ax +By + C$ and $y mapsto Dx$ can be rewritten as :
$$left(beginmatrix x' \ y' \ 1 endmatrix right)=Mleft(beginmatrix x \y\1endmatrix right)$$
Where
$$M=left(beginmatrix A & B & C \ D & 0 & 0 \ 0 & 0 &1endmatrix right).$$
If $M$ is invertible, then $left(beginmatrix x \y\1endmatrix right)=M^-1left(beginmatrix x' \ y' \ 1 endmatrix right)$.
So existence of an inverse is guaranteed by invertibility of the subjacent matrix when you have a linear system.
Which sums up here to $Dne 0 ne B$.
Note that we added a row for the affine part. By doing so, we where able to express your function as a linear system. We could have written also :
$$left(beginmatrix x' \ y' \ 0 endmatrix right)=M'left(beginmatrix x \y\1endmatrix right)$$
But the invertibility would have been more difficult to express in term of matrices.
answered Aug 14 at 6:12
nicomezi
3,6621819
Great answer. I just want to point out that, given that this is an upper triangular matrix, it is worth noting that the invertibility condition translates exactly to $A neq 0$, $D neq 0$.
– Guido A.
Aug 14 at 6:18
Yes indeed. Should I add your remark to the answer ? @GuidoA.
– nicomezi
Aug 14 at 6:21
Well, only if you want to :P It's clear from context if you know some linear algebra, but since I don't know OP's background I felt like making the remark.
– Guido A.
Aug 14 at 6:29
Perhaps using the matrix $M$ we get $y' = Dy$ ? but we require $y' = Dx$
– BAYMAX
Aug 14 at 7:04
Missread, updated.
– nicomezi
Aug 14 at 7:08
add a comment |Â
up vote
4
down vote
accepted
$x mapsto Ax +By + C$ and $y mapsto Dx$ can be rewritten as :
$$left(beginmatrix x' \ y' \ 1 endmatrix right)=Mleft(beginmatrix x \y\1endmatrix right)$$
Where
$$M=left(beginmatrix A & B & C \ D & 0 & 0 \ 0 & 0 &1endmatrix right).$$
If $M$ is invertible, then $left(beginmatrix x \y\1endmatrix right)=M^-1left(beginmatrix x' \ y' \ 1 endmatrix right)$.
So existence of an inverse is guaranteed by invertibility of the subjacent matrix when you have a linear system.
Which sums up here to $Dne 0 ne B$.
Note that we added a row for the affine part. By doing so, we where able to express your function as a linear system. We could have written also :
$$left(beginmatrix x' \ y' \ 0 endmatrix right)=M'left(beginmatrix x \y\1endmatrix right)$$
But the invertibility would have been more difficult to express in term of matrices.
answered Aug 14 at 6:12
nicomezi
3,6621819
Great answer. I just want to point out that, given that this is an upper triangular matrix, it is worth noting that the invertibility condition translates exactly to $A neq 0$, $D neq 0$.
– Guido A.
Aug 14 at 6:18
Yes indeed. Should I add your remark to the answer ? @GuidoA.
– nicomezi
Aug 14 at 6:21
Well, only if you want to :P It's clear from context if you know some linear algebra, but since I don't know OP's background I felt like making the remark.
– Guido A.
Aug 14 at 6:29
Perhaps using the matrix $M$ we get $y' = Dy$ ? but we require $y' = Dx$
– BAYMAX
Aug 14 at 7:04
Missread, updated.
– nicomezi
Aug 14 at 7:08
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
$x mapsto Ax +By + C$ and $y mapsto Dx$ can be rewritten as :
$$left(beginmatrix x' \ y' \ 1 endmatrix right)=Mleft(beginmatrix x \y\1endmatrix right)$$
Where
$$M=left(beginmatrix A & B & C \ D & 0 & 0 \ 0 & 0 &1endmatrix right).$$
If $M$ is invertible, then $left(beginmatrix x \y\1endmatrix right)=M^-1left(beginmatrix x' \ y' \ 1 endmatrix right)$.
So existence of an inverse is guaranteed by invertibility of the subjacent matrix when you have a linear system.
Which sums up here to $Dne 0 ne B$.
Note that we added a row for the affine part. By doing so, we where able to express your function as a linear system. We could have written also :
$$left(beginmatrix x' \ y' \ 0 endmatrix right)=M'left(beginmatrix x \y\1endmatrix right)$$
But the invertibility would have been more difficult to express in term of matrices.
answered Aug 14 at 6:12
nicomezi
3,6621819
$x mapsto Ax +By + C$ and $y mapsto Dx$ can be rewritten as :
$$left(beginmatrix x' \ y' \ 1 endmatrix right)=Mleft(beginmatrix x \y\1endmatrix right)$$
Where
$$M=left(beginmatrix A & B & C \ D & 0 & 0 \ 0 & 0 &1endmatrix right).$$
If $M$ is invertible, then $left(beginmatrix x \y\1endmatrix right)=M^-1left(beginmatrix x' \ y' \ 1 endmatrix right)$.
So existence of an inverse is guaranteed by invertibility of the subjacent matrix when you have a linear system.
Which sums up here to $Dne 0 ne B$.
Note that we added a row for the affine part. By doing so, we where able to express your function as a linear system. We could have written also :
$$left(beginmatrix x' \ y' \ 0 endmatrix right)=M'left(beginmatrix x \y\1endmatrix right)$$
But the invertibility would have been more difficult to express in term of matrices.
answered Aug 14 at 6:12
nicomezi
3,6621819
edited Aug 14 at 7:08
answered Aug 14 at 6:12
nicomezi
3,6621819
answered Aug 14 at 6:12
nicomezi
3,6621819
answered Aug 14 at 6:12
nicomezi
3,6621819
3,6621819
Great answer. I just want to point out that, given that this is an upper triangular matrix, it is worth noting that the invertibility condition translates exactly to $A neq 0$, $D neq 0$.
– Guido A.
Aug 14 at 6:18
Yes indeed. Should I add your remark to the answer ? @GuidoA.
– nicomezi
Aug 14 at 6:21
Well, only if you want to :P It's clear from context if you know some linear algebra, but since I don't know OP's background I felt like making the remark.
– Guido A.
Aug 14 at 6:29
Perhaps using the matrix $M$ we get $y' = Dy$ ? but we require $y' = Dx$
– BAYMAX
Aug 14 at 7:04
Missread, updated.
– nicomezi
Aug 14 at 7:08
add a comment |Â
Great answer. I just want to point out that, given that this is an upper triangular matrix, it is worth noting that the invertibility condition translates exactly to $A neq 0$, $D neq 0$.
– Guido A.
Aug 14 at 6:18
Yes indeed. Should I add your remark to the answer ? @GuidoA.
– nicomezi
Aug 14 at 6:21
Well, only if you want to :P It's clear from context if you know some linear algebra, but since I don't know OP's background I felt like making the remark.
– Guido A.
Aug 14 at 6:29
Perhaps using the matrix $M$ we get $y' = Dy$ ? but we require $y' = Dx$
– BAYMAX
Aug 14 at 7:04
Missread, updated.
– nicomezi
Aug 14 at 7:08
Great answer. I just want to point out that, given that this is an upper triangular matrix, it is worth noting that the invertibility condition translates exactly to $A neq 0$, $D neq 0$.
– Guido A.
Aug 14 at 6:18
Great answer. I just want to point out that, given that this is an upper triangular matrix, it is worth noting that the invertibility condition translates exactly to $A neq 0$, $D neq 0$.
– Guido A.
Aug 14 at 6:18
Yes indeed. Should I add your remark to the answer ? @GuidoA.
– nicomezi
Aug 14 at 6:21
Yes indeed. Should I add your remark to the answer ? @GuidoA.
– nicomezi
Aug 14 at 6:21
Well, only if you want to :P It's clear from context if you know some linear algebra, but since I don't know OP's background I felt like making the remark.
– Guido A.
Aug 14 at 6:29
Well, only if you want to :P It's clear from context if you know some linear algebra, but since I don't know OP's background I felt like making the remark.
– Guido A.
Aug 14 at 6:29
Perhaps using the matrix $M$ we get $y' = Dy$ ? but we require $y' = Dx$
– BAYMAX
Aug 14 at 7:04
Perhaps using the matrix $M$ we get $y' = Dy$ ? but we require $y' = Dx$
– BAYMAX
Aug 14 at 7:04
Missread, updated.
– nicomezi
Aug 14 at 7:08
Missread, updated.
– nicomezi
Aug 14 at 7:08
add a comment |Â
up vote
0
down vote
So you have the mapping $fcolon mathbbR^2to mathbbR^2$, $f(x,y)=(Ax+By+C,Dx)$. Claim: $f$ has an inverse iff $A,Bnot=0$. This follows from the fact that $f$ is invertible iff the linear system of equations $Ax+By=xi-C, Dx= eta$ is solvable for all $(xi,eta)in mathbbR^2$.
Missprint: The condition is $B,Dnot=0$
answered Aug 14 at 6:05
Jens Schwaiger
1,177116
If $A = B = 1$ and $C = D = 0$, we have $f(x,y) = (x+y,0)$ which is not injective.
– Guido A.
Aug 14 at 6:09
add a comment |Â
up vote
0
down vote
So you have the mapping $fcolon mathbbR^2to mathbbR^2$, $f(x,y)=(Ax+By+C,Dx)$. Claim: $f$ has an inverse iff $A,Bnot=0$. This follows from the fact that $f$ is invertible iff the linear system of equations $Ax+By=xi-C, Dx= eta$ is solvable for all $(xi,eta)in mathbbR^2$.
Missprint: The condition is $B,Dnot=0$
answered Aug 14 at 6:05
Jens Schwaiger
1,177116
If $A = B = 1$ and $C = D = 0$, we have $f(x,y) = (x+y,0)$ which is not injective.
– Guido A.
Aug 14 at 6:09
add a comment |Â
up vote
0
down vote
up vote
0
down vote
So you have the mapping $fcolon mathbbR^2to mathbbR^2$, $f(x,y)=(Ax+By+C,Dx)$. Claim: $f$ has an inverse iff $A,Bnot=0$. This follows from the fact that $f$ is invertible iff the linear system of equations $Ax+By=xi-C, Dx= eta$ is solvable for all $(xi,eta)in mathbbR^2$.
Missprint: The condition is $B,Dnot=0$
answered Aug 14 at 6:05
Jens Schwaiger
1,177116
So you have the mapping $fcolon mathbbR^2to mathbbR^2$, $f(x,y)=(Ax+By+C,Dx)$. Claim: $f$ has an inverse iff $A,Bnot=0$. This follows from the fact that $f$ is invertible iff the linear system of equations $Ax+By=xi-C, Dx= eta$ is solvable for all $(xi,eta)in mathbbR^2$.
Missprint: The condition is $B,Dnot=0$
answered Aug 14 at 6:05
Jens Schwaiger
1,177116
edited Aug 14 at 6:21
answered Aug 14 at 6:05
Jens Schwaiger
1,177116
answered Aug 14 at 6:05
Jens Schwaiger
1,177116
answered Aug 14 at 6:05
Jens Schwaiger
1,177116
1,177116
If $A = B = 1$ and $C = D = 0$, we have $f(x,y) = (x+y,0)$ which is not injective.
– Guido A.
Aug 14 at 6:09
add a comment |Â
If $A = B = 1$ and $C = D = 0$, we have $f(x,y) = (x+y,0)$ which is not injective.
– Guido A.
Aug 14 at 6:09
If $A = B = 1$ and $C = D = 0$, we have $f(x,y) = (x+y,0)$ which is not injective.
– Guido A.
Aug 14 at 6:09
If $A = B = 1$ and $C = D = 0$, we have $f(x,y) = (x+y,0)$ which is not injective.
– Guido A.
Aug 14 at 6:09
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2882103%2fcomputing-the-inverse-of-a-two-dimensional-map%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
What are $x,y, A, B, C$? Real numbers, integers, matrices? Also, your first statement would seem to indicate $x^2 = y = f(x) = x^3$, was that intended?
– Guido A.
Aug 14 at 6:03
Thank you! I just did an edit to correct the typos.
– BAYMAX
Aug 14 at 6:09
Great, I'll give this some thought.
– Guido A.
Aug 14 at 6:10