Intuition for lines which undergo linear transformations
$begingroup$
From what I understand, a 2D line $y = frac{1}{2} x$ can be specified with a vector $v = [frac{1}{2}, -1]^T$ such that the line can be written as $v^T bar{x} = 0$ (where $bar{x} = [x,y]^T$).
Now, consider a linear transformation given by the matrix $M$ (say a rotation). Denoting the line that has been transformed with this particular linear transformation as $v'$, I understand that we can write:
$$
v' = vM^{-1}
$$
Is there any intuitive explanation as to why this happens to be the case? From my (misguided) perspective, I would have thought that we might have something like $v' = Mv$ instead, which is wrong, but I don't exactly have a good intuition as to why this is wrong.
I'm hoping to get a better understanding of this, as I'd like to be able to apply these arguments to the linear transformation of more complicated geometric structures, such as conics.
linear-transformations
asked Dec 1 '18 at 18:50
Sean LeeSean Lee
1578
$endgroup$
add a comment |
$begingroup$
From what I understand, a 2D line $y = frac{1}{2} x$ can be specified with a vector $v = [frac{1}{2}, -1]^T$ such that the line can be written as $v^T bar{x} = 0$ (where $bar{x} = [x,y]^T$).
Now, consider a linear transformation given by the matrix $M$ (say a rotation). Denoting the line that has been transformed with this particular linear transformation as $v'$, I understand that we can write:
$$
v' = vM^{-1}
$$
Is there any intuitive explanation as to why this happens to be the case? From my (misguided) perspective, I would have thought that we might have something like $v' = Mv$ instead, which is wrong, but I don't exactly have a good intuition as to why this is wrong.
I'm hoping to get a better understanding of this, as I'd like to be able to apply these arguments to the linear transformation of more complicated geometric structures, such as conics.
linear-transformations
asked Dec 1 '18 at 18:50
Sean LeeSean Lee
1578
$endgroup$
add a comment |
$begingroup$
From what I understand, a 2D line $y = frac{1}{2} x$ can be specified with a vector $v = [frac{1}{2}, -1]^T$ such that the line can be written as $v^T bar{x} = 0$ (where $bar{x} = [x,y]^T$).
Now, consider a linear transformation given by the matrix $M$ (say a rotation). Denoting the line that has been transformed with this particular linear transformation as $v'$, I understand that we can write:
$$
v' = vM^{-1}
$$
Is there any intuitive explanation as to why this happens to be the case? From my (misguided) perspective, I would have thought that we might have something like $v' = Mv$ instead, which is wrong, but I don't exactly have a good intuition as to why this is wrong.
I'm hoping to get a better understanding of this, as I'd like to be able to apply these arguments to the linear transformation of more complicated geometric structures, such as conics.
linear-transformations
asked Dec 1 '18 at 18:50
Sean LeeSean Lee
1578
$endgroup$
From what I understand, a 2D line $y = frac{1}{2} x$ can be specified with a vector $v = [frac{1}{2}, -1]^T$ such that the line can be written as $v^T bar{x} = 0$ (where $bar{x} = [x,y]^T$).
Now, consider a linear transformation given by the matrix $M$ (say a rotation). Denoting the line that has been transformed with this particular linear transformation as $v'$, I understand that we can write:
$$
v' = vM^{-1}
$$
Is there any intuitive explanation as to why this happens to be the case? From my (misguided) perspective, I would have thought that we might have something like $v' = Mv$ instead, which is wrong, but I don't exactly have a good intuition as to why this is wrong.
I'm hoping to get a better understanding of this, as I'd like to be able to apply these arguments to the linear transformation of more complicated geometric structures, such as conics.
linear-transformations
linear-transformations
asked Dec 1 '18 at 18:50
Sean LeeSean Lee
1578
asked Dec 1 '18 at 18:50
Sean LeeSean Lee
1578
asked Dec 1 '18 at 18:50
Sean LeeSean Lee
1578
asked Dec 1 '18 at 18:50
Sean LeeSean Lee
1578
asked Dec 1 '18 at 18:50
Sean LeeSean Lee
1578
1578
add a comment |
add a comment |
1 Answer
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$begingroup$
Your transformation formula isn’t quite right. It’s actually $v'=M^{-T}v$—you apply the inverse transpose of $M$.
Algebraically, this can be derived using a simple substitution. If you have the point transformation $bar x' = Mbar x$, then $$v^Tbar x = v^T(M^{-1}bar x') = (v^TM^{-1})bar x' = (M^{-T}v)^Tbar x'=0.$$ This is really no different than the process that you might’ve gone through when translating and scaling graphs of functions in your pre-calculus algebra courses: In order to, say, stretch the graph of $f(x)$ by a factor of $3$, you have to replace $x$ by $x/3$—that is, you have to invert the transformation in order to substitute for the original $x$ in the function’s expression. It’s the same here: to get the equation of the line in terms of the new coordinates $bar x'$, you have to solve for the components of $bar x$ in terms of those of $bar x'$—invert $M$—in order to substitute for the former in the equation of the line.
Even though they might look similar, a coordinate vector that represents a point and one that represents a line are different—they transform differently under $M$. The terms that you may encounter for these two types of vectors are contravariant and covariant, respectively. In three dimensions, you might also see the terms polar vector and pseudovector (or axial vector). This different behavior under transformations is also connected to the concept of the dual $V^*$ of a vector space $V$, which is the space of linear functionals on $V$, but I won’t go into that here.
answered Dec 2 '18 at 1:02
amdamd
29.5k21050
$endgroup$
$begingroup$
Thank you for pointing out my error, I understand it now (:
$endgroup$
– Sean Lee
Dec 3 '18 at 1:03
$begingroup$
@SeanLee I was planning to add a bit of discussion of what’s going on geometrically, but since you’ve accepted the answer, I’ll leave it alone.
$endgroup$
– amd
Dec 3 '18 at 4:00
$begingroup$
Ah, I meant I understand the algebraic derivation, but I'd really appreciate it if you could expand on the explanation with the dual space!
$endgroup$
– Sean Lee
Dec 3 '18 at 4:06
add a comment |
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$begingroup$
Your transformation formula isn’t quite right. It’s actually $v'=M^{-T}v$—you apply the inverse transpose of $M$.
Algebraically, this can be derived using a simple substitution. If you have the point transformation $bar x' = Mbar x$, then $$v^Tbar x = v^T(M^{-1}bar x') = (v^TM^{-1})bar x' = (M^{-T}v)^Tbar x'=0.$$ This is really no different than the process that you might’ve gone through when translating and scaling graphs of functions in your pre-calculus algebra courses: In order to, say, stretch the graph of $f(x)$ by a factor of $3$, you have to replace $x$ by $x/3$—that is, you have to invert the transformation in order to substitute for the original $x$ in the function’s expression. It’s the same here: to get the equation of the line in terms of the new coordinates $bar x'$, you have to solve for the components of $bar x$ in terms of those of $bar x'$—invert $M$—in order to substitute for the former in the equation of the line.
Even though they might look similar, a coordinate vector that represents a point and one that represents a line are different—they transform differently under $M$. The terms that you may encounter for these two types of vectors are contravariant and covariant, respectively. In three dimensions, you might also see the terms polar vector and pseudovector (or axial vector). This different behavior under transformations is also connected to the concept of the dual $V^*$ of a vector space $V$, which is the space of linear functionals on $V$, but I won’t go into that here.
answered Dec 2 '18 at 1:02
amdamd
29.5k21050
$endgroup$
$begingroup$
Thank you for pointing out my error, I understand it now (:
$endgroup$
– Sean Lee
Dec 3 '18 at 1:03
$begingroup$
@SeanLee I was planning to add a bit of discussion of what’s going on geometrically, but since you’ve accepted the answer, I’ll leave it alone.
$endgroup$
– amd
Dec 3 '18 at 4:00
$begingroup$
Ah, I meant I understand the algebraic derivation, but I'd really appreciate it if you could expand on the explanation with the dual space!
$endgroup$
– Sean Lee
Dec 3 '18 at 4:06
add a comment |
$begingroup$
Your transformation formula isn’t quite right. It’s actually $v'=M^{-T}v$—you apply the inverse transpose of $M$.
Algebraically, this can be derived using a simple substitution. If you have the point transformation $bar x' = Mbar x$, then $$v^Tbar x = v^T(M^{-1}bar x') = (v^TM^{-1})bar x' = (M^{-T}v)^Tbar x'=0.$$ This is really no different than the process that you might’ve gone through when translating and scaling graphs of functions in your pre-calculus algebra courses: In order to, say, stretch the graph of $f(x)$ by a factor of $3$, you have to replace $x$ by $x/3$—that is, you have to invert the transformation in order to substitute for the original $x$ in the function’s expression. It’s the same here: to get the equation of the line in terms of the new coordinates $bar x'$, you have to solve for the components of $bar x$ in terms of those of $bar x'$—invert $M$—in order to substitute for the former in the equation of the line.
Even though they might look similar, a coordinate vector that represents a point and one that represents a line are different—they transform differently under $M$. The terms that you may encounter for these two types of vectors are contravariant and covariant, respectively. In three dimensions, you might also see the terms polar vector and pseudovector (or axial vector). This different behavior under transformations is also connected to the concept of the dual $V^*$ of a vector space $V$, which is the space of linear functionals on $V$, but I won’t go into that here.
answered Dec 2 '18 at 1:02
amdamd
29.5k21050
$endgroup$
$begingroup$
Thank you for pointing out my error, I understand it now (:
$endgroup$
– Sean Lee
Dec 3 '18 at 1:03
$begingroup$
@SeanLee I was planning to add a bit of discussion of what’s going on geometrically, but since you’ve accepted the answer, I’ll leave it alone.
$endgroup$
– amd
Dec 3 '18 at 4:00
$begingroup$
Ah, I meant I understand the algebraic derivation, but I'd really appreciate it if you could expand on the explanation with the dual space!
$endgroup$
– Sean Lee
Dec 3 '18 at 4:06
add a comment |
$begingroup$
Your transformation formula isn’t quite right. It’s actually $v'=M^{-T}v$—you apply the inverse transpose of $M$.
Algebraically, this can be derived using a simple substitution. If you have the point transformation $bar x' = Mbar x$, then $$v^Tbar x = v^T(M^{-1}bar x') = (v^TM^{-1})bar x' = (M^{-T}v)^Tbar x'=0.$$ This is really no different than the process that you might’ve gone through when translating and scaling graphs of functions in your pre-calculus algebra courses: In order to, say, stretch the graph of $f(x)$ by a factor of $3$, you have to replace $x$ by $x/3$—that is, you have to invert the transformation in order to substitute for the original $x$ in the function’s expression. It’s the same here: to get the equation of the line in terms of the new coordinates $bar x'$, you have to solve for the components of $bar x$ in terms of those of $bar x'$—invert $M$—in order to substitute for the former in the equation of the line.
Even though they might look similar, a coordinate vector that represents a point and one that represents a line are different—they transform differently under $M$. The terms that you may encounter for these two types of vectors are contravariant and covariant, respectively. In three dimensions, you might also see the terms polar vector and pseudovector (or axial vector). This different behavior under transformations is also connected to the concept of the dual $V^*$ of a vector space $V$, which is the space of linear functionals on $V$, but I won’t go into that here.
answered Dec 2 '18 at 1:02
amdamd
29.5k21050
$endgroup$
Your transformation formula isn’t quite right. It’s actually $v'=M^{-T}v$—you apply the inverse transpose of $M$.
Algebraically, this can be derived using a simple substitution. If you have the point transformation $bar x' = Mbar x$, then $$v^Tbar x = v^T(M^{-1}bar x') = (v^TM^{-1})bar x' = (M^{-T}v)^Tbar x'=0.$$ This is really no different than the process that you might’ve gone through when translating and scaling graphs of functions in your pre-calculus algebra courses: In order to, say, stretch the graph of $f(x)$ by a factor of $3$, you have to replace $x$ by $x/3$—that is, you have to invert the transformation in order to substitute for the original $x$ in the function’s expression. It’s the same here: to get the equation of the line in terms of the new coordinates $bar x'$, you have to solve for the components of $bar x$ in terms of those of $bar x'$—invert $M$—in order to substitute for the former in the equation of the line.
Even though they might look similar, a coordinate vector that represents a point and one that represents a line are different—they transform differently under $M$. The terms that you may encounter for these two types of vectors are contravariant and covariant, respectively. In three dimensions, you might also see the terms polar vector and pseudovector (or axial vector). This different behavior under transformations is also connected to the concept of the dual $V^*$ of a vector space $V$, which is the space of linear functionals on $V$, but I won’t go into that here.
answered Dec 2 '18 at 1:02
amdamd
29.5k21050
answered Dec 2 '18 at 1:02
amdamd
29.5k21050
answered Dec 2 '18 at 1:02
amdamd
29.5k21050
answered Dec 2 '18 at 1:02
amdamd
29.5k21050
29.5k21050
$begingroup$
Thank you for pointing out my error, I understand it now (:
$endgroup$
– Sean Lee
Dec 3 '18 at 1:03
$begingroup$
@SeanLee I was planning to add a bit of discussion of what’s going on geometrically, but since you’ve accepted the answer, I’ll leave it alone.
$endgroup$
– amd
Dec 3 '18 at 4:00
$begingroup$
Ah, I meant I understand the algebraic derivation, but I'd really appreciate it if you could expand on the explanation with the dual space!
$endgroup$
– Sean Lee
Dec 3 '18 at 4:06
add a comment |
$begingroup$
Thank you for pointing out my error, I understand it now (:
$endgroup$
– Sean Lee
Dec 3 '18 at 1:03
$begingroup$
@SeanLee I was planning to add a bit of discussion of what’s going on geometrically, but since you’ve accepted the answer, I’ll leave it alone.
$endgroup$
– amd
Dec 3 '18 at 4:00
$begingroup$
Ah, I meant I understand the algebraic derivation, but I'd really appreciate it if you could expand on the explanation with the dual space!
$endgroup$
– Sean Lee
Dec 3 '18 at 4:06
$begingroup$
Thank you for pointing out my error, I understand it now (:
$endgroup$
– Sean Lee
Dec 3 '18 at 1:03
$begingroup$
Thank you for pointing out my error, I understand it now (:
$endgroup$
– Sean Lee
Dec 3 '18 at 1:03
$begingroup$
@SeanLee I was planning to add a bit of discussion of what’s going on geometrically, but since you’ve accepted the answer, I’ll leave it alone.
$endgroup$
– amd
Dec 3 '18 at 4:00
$begingroup$
@SeanLee I was planning to add a bit of discussion of what’s going on geometrically, but since you’ve accepted the answer, I’ll leave it alone.
$endgroup$
– amd
Dec 3 '18 at 4:00
$begingroup$
Ah, I meant I understand the algebraic derivation, but I'd really appreciate it if you could expand on the explanation with the dual space!
$endgroup$
– Sean Lee
Dec 3 '18 at 4:06
$begingroup$
Ah, I meant I understand the algebraic derivation, but I'd really appreciate it if you could expand on the explanation with the dual space!
$endgroup$
– Sean Lee
Dec 3 '18 at 4:06
add a comment |
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