Mixing
Understanding formula for hyperplanes
Clash Royale CLAN TAG#URR8PPP
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I came across this formula that classifies a vector by using a hyperplane:
Let $a,b in mathbbR^n$, then the (implicitely defined) space-dividing hyperplane $H$ is orthogonal to $b-a$. With $x in mathbbR^n$ we can see that
$$0 < (b-a)^Tx - frac12(||b||^2 - ||a||^2)$$
and we can conclude that $d(x,a)> d(x,b)$. We can then say that $x$ lies on a specific side of the hyperplane.
I dont get any insight or intuition for this inequation. So I played with it, to get some understanding:
If I move $x$ towards the hyperplane, then the term on the right approaches zero. So I get
$$(b-a)^Th = - frac12(||b||^2 - ||a||^2)$$ with $h in H$
If the $b-a$ would be normalized, the term on the right would be the distance between the origin and the hyperplane. So we get
$$d(0,H) = - frac1(||b||^2 - ||a||^2)$$
But this does not give me any interpretation or intuition, why that works.
linear-algebra geometry functional-analysis
asked Jul 21 at 19:41
dba
1749
add a comment |Â
up vote
3
down vote
favorite
I came across this formula that classifies a vector by using a hyperplane:
Let $a,b in mathbbR^n$, then the (implicitely defined) space-dividing hyperplane $H$ is orthogonal to $b-a$. With $x in mathbbR^n$ we can see that
$$0 < (b-a)^Tx - frac12(||b||^2 - ||a||^2)$$
and we can conclude that $d(x,a)> d(x,b)$. We can then say that $x$ lies on a specific side of the hyperplane.
I dont get any insight or intuition for this inequation. So I played with it, to get some understanding:
If I move $x$ towards the hyperplane, then the term on the right approaches zero. So I get
$$(b-a)^Th = - frac12(||b||^2 - ||a||^2)$$ with $h in H$
If the $b-a$ would be normalized, the term on the right would be the distance between the origin and the hyperplane. So we get
$$d(0,H) = - frac1(||b||^2 - ||a||^2)$$
But this does not give me any interpretation or intuition, why that works.
linear-algebra geometry functional-analysis
asked Jul 21 at 19:41
dba
1749
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I came across this formula that classifies a vector by using a hyperplane:
Let $a,b in mathbbR^n$, then the (implicitely defined) space-dividing hyperplane $H$ is orthogonal to $b-a$. With $x in mathbbR^n$ we can see that
$$0 < (b-a)^Tx - frac12(||b||^2 - ||a||^2)$$
and we can conclude that $d(x,a)> d(x,b)$. We can then say that $x$ lies on a specific side of the hyperplane.
I dont get any insight or intuition for this inequation. So I played with it, to get some understanding:
If I move $x$ towards the hyperplane, then the term on the right approaches zero. So I get
$$(b-a)^Th = - frac12(||b||^2 - ||a||^2)$$ with $h in H$
If the $b-a$ would be normalized, the term on the right would be the distance between the origin and the hyperplane. So we get
$$d(0,H) = - frac1(||b||^2 - ||a||^2)$$
But this does not give me any interpretation or intuition, why that works.
linear-algebra geometry functional-analysis
asked Jul 21 at 19:41
dba
1749
I came across this formula that classifies a vector by using a hyperplane:
Let $a,b in mathbbR^n$, then the (implicitely defined) space-dividing hyperplane $H$ is orthogonal to $b-a$. With $x in mathbbR^n$ we can see that
$$0 < (b-a)^Tx - frac12(||b||^2 - ||a||^2)$$
and we can conclude that $d(x,a)> d(x,b)$. We can then say that $x$ lies on a specific side of the hyperplane.
I dont get any insight or intuition for this inequation. So I played with it, to get some understanding:
If I move $x$ towards the hyperplane, then the term on the right approaches zero. So I get
$$(b-a)^Th = - frac12(||b||^2 - ||a||^2)$$ with $h in H$
If the $b-a$ would be normalized, the term on the right would be the distance between the origin and the hyperplane. So we get
$$d(0,H) = - frac1(||b||^2 - ||a||^2)$$
But this does not give me any interpretation or intuition, why that works.
linear-algebra geometry functional-analysis
asked Jul 21 at 19:41
dba
1749
edited Jul 21 at 19:49
asked Jul 21 at 19:41
dba
1749
asked Jul 21 at 19:41
dba
1749
asked Jul 21 at 19:41
dba
1749
1749
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
A hyperplane through the origin can be expressed via the equation $mathbf n^Tmathbf x=0$, i.e., as the set of all points with position vectors $mathbf x$ that are orthogonal to some fixed vector $mathbf n$, the normal to the hyperplane. For vectors, $mathbf n^Tmathbf x$ is just a different way to write the dot product $mathbf ncdotmathbf x$. Recall its geometric meaning: it’s a positive multiple of the length of the projection of $mathbf x$ onto $mathbf n$, which is itself some multiple of $mathbf n$. So, if the dot product is positive, this projection points in the same direction as $mathbf n$; if it’s negative, the projection points in the opposite direction. For any point $mathbf x$, then, the quantity $mathbf n^Tmathbf x$ tells you on which side of the hyperplane the point lies: if $mathbf n^Tmathbf xgt0$, the point is on the same side of the hyperplane as $mathbf n$, if it’s negative, the point is on the opposite side, and if it’s zero, the point is on the hyperplane. Moreover, the quantity $mathbf n^Tmathbf x$ is proportional to the perpendicular distance of $mathbf x$ from the hyperplane: $mathbf n$ is orthogonal to it, so this distance is the length of the projection of $mathbf x$ onto $mathbf n$.
For an arbitrary hyperplane that might not pass through the origin, we can pick some fixed point $mathbf x_0$ on the hyperplane and translate it to the origin to get the equation $mathbf n^T(mathbf x-mathbf x_0)=0$, which expands into $mathbf n^Tmathbf x-mathbf n^Tmathbf x_0=0$, for this hyperplane. As before, we can determine which side of the hyperplane relative to the direction of $mathbf n$ that any point lies on by examining the sign of $mathbf n^T(mathbf x-mathbf x_0)$ and similarly for the relative distances of arbitrary points from this hyperplane. To get a feel for what’s going on, play around with this in two dimensions, where hyperplanes are straight lines.
For your dividing hyperplane $H$, we have $mathbf n = mathbf b-mathbf a$. Since $|mathbf v|^2=mathbf v^Tmathbf v$ and the dot product is commutative, we have $(mathbf b-mathbf a)^T(mathbf b+mathbf a) = |mathbf b|^2-|mathbf a|^2$ and so can factor the equation of $H$ into $$mathbf n^Tleft(mathbf x - frac12(mathbf b+mathbf a)right) = 0.$$ From this we can read $mathbf x_0=frac12(mathbf b+mathbf a)$, the midpoint of $mathbf a$ and $mathbf b$: $H$ is the perpendicular bisector of the line segment $mathbf amathbf b$. Now, $mathbf b-mathbf a$ points from $mathbf a$ toward $mathbf b$ and so, too, toward $mathbf b$ from their midpoint. Therefore, based on the above discussion, if $(mathbf b-mathbf a)^Tmathbf x - frac12left(|mathbf b|^2-|mathbf a|^2right)gt0$, then $mathbf x$ lies on the same side of $H$ as does $mathbf b$, while if it’s negative, then it’s on the same side as $mathbf a$.
answered Jul 21 at 20:30
amd
25.9k2943
Nice. I'd put more emphasis on the relationship $$lVert brVert^2-lVert arVert^2=(b-a)^Tcdot(b+a)$$ which you certainly implied, but which doesn't stand out when skimming your post while being crucial fo the connection from midpoint to difference of squared norms.
– MvG
Jul 21 at 20:35
@MvG Good point. I took for granted that the reader knows this identity.
– amd
Jul 21 at 20:41
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
A hyperplane through the origin can be expressed via the equation $mathbf n^Tmathbf x=0$, i.e., as the set of all points with position vectors $mathbf x$ that are orthogonal to some fixed vector $mathbf n$, the normal to the hyperplane. For vectors, $mathbf n^Tmathbf x$ is just a different way to write the dot product $mathbf ncdotmathbf x$. Recall its geometric meaning: it’s a positive multiple of the length of the projection of $mathbf x$ onto $mathbf n$, which is itself some multiple of $mathbf n$. So, if the dot product is positive, this projection points in the same direction as $mathbf n$; if it’s negative, the projection points in the opposite direction. For any point $mathbf x$, then, the quantity $mathbf n^Tmathbf x$ tells you on which side of the hyperplane the point lies: if $mathbf n^Tmathbf xgt0$, the point is on the same side of the hyperplane as $mathbf n$, if it’s negative, the point is on the opposite side, and if it’s zero, the point is on the hyperplane. Moreover, the quantity $mathbf n^Tmathbf x$ is proportional to the perpendicular distance of $mathbf x$ from the hyperplane: $mathbf n$ is orthogonal to it, so this distance is the length of the projection of $mathbf x$ onto $mathbf n$.
For an arbitrary hyperplane that might not pass through the origin, we can pick some fixed point $mathbf x_0$ on the hyperplane and translate it to the origin to get the equation $mathbf n^T(mathbf x-mathbf x_0)=0$, which expands into $mathbf n^Tmathbf x-mathbf n^Tmathbf x_0=0$, for this hyperplane. As before, we can determine which side of the hyperplane relative to the direction of $mathbf n$ that any point lies on by examining the sign of $mathbf n^T(mathbf x-mathbf x_0)$ and similarly for the relative distances of arbitrary points from this hyperplane. To get a feel for what’s going on, play around with this in two dimensions, where hyperplanes are straight lines.
For your dividing hyperplane $H$, we have $mathbf n = mathbf b-mathbf a$. Since $|mathbf v|^2=mathbf v^Tmathbf v$ and the dot product is commutative, we have $(mathbf b-mathbf a)^T(mathbf b+mathbf a) = |mathbf b|^2-|mathbf a|^2$ and so can factor the equation of $H$ into $$mathbf n^Tleft(mathbf x - frac12(mathbf b+mathbf a)right) = 0.$$ From this we can read $mathbf x_0=frac12(mathbf b+mathbf a)$, the midpoint of $mathbf a$ and $mathbf b$: $H$ is the perpendicular bisector of the line segment $mathbf amathbf b$. Now, $mathbf b-mathbf a$ points from $mathbf a$ toward $mathbf b$ and so, too, toward $mathbf b$ from their midpoint. Therefore, based on the above discussion, if $(mathbf b-mathbf a)^Tmathbf x - frac12left(|mathbf b|^2-|mathbf a|^2right)gt0$, then $mathbf x$ lies on the same side of $H$ as does $mathbf b$, while if it’s negative, then it’s on the same side as $mathbf a$.
answered Jul 21 at 20:30
amd
25.9k2943
Nice. I'd put more emphasis on the relationship $$lVert brVert^2-lVert arVert^2=(b-a)^Tcdot(b+a)$$ which you certainly implied, but which doesn't stand out when skimming your post while being crucial fo the connection from midpoint to difference of squared norms.
– MvG
Jul 21 at 20:35
@MvG Good point. I took for granted that the reader knows this identity.
– amd
Jul 21 at 20:41
add a comment |Â
up vote
2
down vote
accepted
A hyperplane through the origin can be expressed via the equation $mathbf n^Tmathbf x=0$, i.e., as the set of all points with position vectors $mathbf x$ that are orthogonal to some fixed vector $mathbf n$, the normal to the hyperplane. For vectors, $mathbf n^Tmathbf x$ is just a different way to write the dot product $mathbf ncdotmathbf x$. Recall its geometric meaning: it’s a positive multiple of the length of the projection of $mathbf x$ onto $mathbf n$, which is itself some multiple of $mathbf n$. So, if the dot product is positive, this projection points in the same direction as $mathbf n$; if it’s negative, the projection points in the opposite direction. For any point $mathbf x$, then, the quantity $mathbf n^Tmathbf x$ tells you on which side of the hyperplane the point lies: if $mathbf n^Tmathbf xgt0$, the point is on the same side of the hyperplane as $mathbf n$, if it’s negative, the point is on the opposite side, and if it’s zero, the point is on the hyperplane. Moreover, the quantity $mathbf n^Tmathbf x$ is proportional to the perpendicular distance of $mathbf x$ from the hyperplane: $mathbf n$ is orthogonal to it, so this distance is the length of the projection of $mathbf x$ onto $mathbf n$.
For an arbitrary hyperplane that might not pass through the origin, we can pick some fixed point $mathbf x_0$ on the hyperplane and translate it to the origin to get the equation $mathbf n^T(mathbf x-mathbf x_0)=0$, which expands into $mathbf n^Tmathbf x-mathbf n^Tmathbf x_0=0$, for this hyperplane. As before, we can determine which side of the hyperplane relative to the direction of $mathbf n$ that any point lies on by examining the sign of $mathbf n^T(mathbf x-mathbf x_0)$ and similarly for the relative distances of arbitrary points from this hyperplane. To get a feel for what’s going on, play around with this in two dimensions, where hyperplanes are straight lines.
For your dividing hyperplane $H$, we have $mathbf n = mathbf b-mathbf a$. Since $|mathbf v|^2=mathbf v^Tmathbf v$ and the dot product is commutative, we have $(mathbf b-mathbf a)^T(mathbf b+mathbf a) = |mathbf b|^2-|mathbf a|^2$ and so can factor the equation of $H$ into $$mathbf n^Tleft(mathbf x - frac12(mathbf b+mathbf a)right) = 0.$$ From this we can read $mathbf x_0=frac12(mathbf b+mathbf a)$, the midpoint of $mathbf a$ and $mathbf b$: $H$ is the perpendicular bisector of the line segment $mathbf amathbf b$. Now, $mathbf b-mathbf a$ points from $mathbf a$ toward $mathbf b$ and so, too, toward $mathbf b$ from their midpoint. Therefore, based on the above discussion, if $(mathbf b-mathbf a)^Tmathbf x - frac12left(|mathbf b|^2-|mathbf a|^2right)gt0$, then $mathbf x$ lies on the same side of $H$ as does $mathbf b$, while if it’s negative, then it’s on the same side as $mathbf a$.
answered Jul 21 at 20:30
amd
25.9k2943
Nice. I'd put more emphasis on the relationship $$lVert brVert^2-lVert arVert^2=(b-a)^Tcdot(b+a)$$ which you certainly implied, but which doesn't stand out when skimming your post while being crucial fo the connection from midpoint to difference of squared norms.
– MvG
Jul 21 at 20:35
@MvG Good point. I took for granted that the reader knows this identity.
– amd
Jul 21 at 20:41
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
A hyperplane through the origin can be expressed via the equation $mathbf n^Tmathbf x=0$, i.e., as the set of all points with position vectors $mathbf x$ that are orthogonal to some fixed vector $mathbf n$, the normal to the hyperplane. For vectors, $mathbf n^Tmathbf x$ is just a different way to write the dot product $mathbf ncdotmathbf x$. Recall its geometric meaning: it’s a positive multiple of the length of the projection of $mathbf x$ onto $mathbf n$, which is itself some multiple of $mathbf n$. So, if the dot product is positive, this projection points in the same direction as $mathbf n$; if it’s negative, the projection points in the opposite direction. For any point $mathbf x$, then, the quantity $mathbf n^Tmathbf x$ tells you on which side of the hyperplane the point lies: if $mathbf n^Tmathbf xgt0$, the point is on the same side of the hyperplane as $mathbf n$, if it’s negative, the point is on the opposite side, and if it’s zero, the point is on the hyperplane. Moreover, the quantity $mathbf n^Tmathbf x$ is proportional to the perpendicular distance of $mathbf x$ from the hyperplane: $mathbf n$ is orthogonal to it, so this distance is the length of the projection of $mathbf x$ onto $mathbf n$.
For an arbitrary hyperplane that might not pass through the origin, we can pick some fixed point $mathbf x_0$ on the hyperplane and translate it to the origin to get the equation $mathbf n^T(mathbf x-mathbf x_0)=0$, which expands into $mathbf n^Tmathbf x-mathbf n^Tmathbf x_0=0$, for this hyperplane. As before, we can determine which side of the hyperplane relative to the direction of $mathbf n$ that any point lies on by examining the sign of $mathbf n^T(mathbf x-mathbf x_0)$ and similarly for the relative distances of arbitrary points from this hyperplane. To get a feel for what’s going on, play around with this in two dimensions, where hyperplanes are straight lines.
For your dividing hyperplane $H$, we have $mathbf n = mathbf b-mathbf a$. Since $|mathbf v|^2=mathbf v^Tmathbf v$ and the dot product is commutative, we have $(mathbf b-mathbf a)^T(mathbf b+mathbf a) = |mathbf b|^2-|mathbf a|^2$ and so can factor the equation of $H$ into $$mathbf n^Tleft(mathbf x - frac12(mathbf b+mathbf a)right) = 0.$$ From this we can read $mathbf x_0=frac12(mathbf b+mathbf a)$, the midpoint of $mathbf a$ and $mathbf b$: $H$ is the perpendicular bisector of the line segment $mathbf amathbf b$. Now, $mathbf b-mathbf a$ points from $mathbf a$ toward $mathbf b$ and so, too, toward $mathbf b$ from their midpoint. Therefore, based on the above discussion, if $(mathbf b-mathbf a)^Tmathbf x - frac12left(|mathbf b|^2-|mathbf a|^2right)gt0$, then $mathbf x$ lies on the same side of $H$ as does $mathbf b$, while if it’s negative, then it’s on the same side as $mathbf a$.
answered Jul 21 at 20:30
amd
25.9k2943
A hyperplane through the origin can be expressed via the equation $mathbf n^Tmathbf x=0$, i.e., as the set of all points with position vectors $mathbf x$ that are orthogonal to some fixed vector $mathbf n$, the normal to the hyperplane. For vectors, $mathbf n^Tmathbf x$ is just a different way to write the dot product $mathbf ncdotmathbf x$. Recall its geometric meaning: it’s a positive multiple of the length of the projection of $mathbf x$ onto $mathbf n$, which is itself some multiple of $mathbf n$. So, if the dot product is positive, this projection points in the same direction as $mathbf n$; if it’s negative, the projection points in the opposite direction. For any point $mathbf x$, then, the quantity $mathbf n^Tmathbf x$ tells you on which side of the hyperplane the point lies: if $mathbf n^Tmathbf xgt0$, the point is on the same side of the hyperplane as $mathbf n$, if it’s negative, the point is on the opposite side, and if it’s zero, the point is on the hyperplane. Moreover, the quantity $mathbf n^Tmathbf x$ is proportional to the perpendicular distance of $mathbf x$ from the hyperplane: $mathbf n$ is orthogonal to it, so this distance is the length of the projection of $mathbf x$ onto $mathbf n$.
For an arbitrary hyperplane that might not pass through the origin, we can pick some fixed point $mathbf x_0$ on the hyperplane and translate it to the origin to get the equation $mathbf n^T(mathbf x-mathbf x_0)=0$, which expands into $mathbf n^Tmathbf x-mathbf n^Tmathbf x_0=0$, for this hyperplane. As before, we can determine which side of the hyperplane relative to the direction of $mathbf n$ that any point lies on by examining the sign of $mathbf n^T(mathbf x-mathbf x_0)$ and similarly for the relative distances of arbitrary points from this hyperplane. To get a feel for what’s going on, play around with this in two dimensions, where hyperplanes are straight lines.
For your dividing hyperplane $H$, we have $mathbf n = mathbf b-mathbf a$. Since $|mathbf v|^2=mathbf v^Tmathbf v$ and the dot product is commutative, we have $(mathbf b-mathbf a)^T(mathbf b+mathbf a) = |mathbf b|^2-|mathbf a|^2$ and so can factor the equation of $H$ into $$mathbf n^Tleft(mathbf x - frac12(mathbf b+mathbf a)right) = 0.$$ From this we can read $mathbf x_0=frac12(mathbf b+mathbf a)$, the midpoint of $mathbf a$ and $mathbf b$: $H$ is the perpendicular bisector of the line segment $mathbf amathbf b$. Now, $mathbf b-mathbf a$ points from $mathbf a$ toward $mathbf b$ and so, too, toward $mathbf b$ from their midpoint. Therefore, based on the above discussion, if $(mathbf b-mathbf a)^Tmathbf x - frac12left(|mathbf b|^2-|mathbf a|^2right)gt0$, then $mathbf x$ lies on the same side of $H$ as does $mathbf b$, while if it’s negative, then it’s on the same side as $mathbf a$.
answered Jul 21 at 20:30
amd
25.9k2943
edited Jul 21 at 20:47
answered Jul 21 at 20:30
amd
25.9k2943
answered Jul 21 at 20:30
amd
25.9k2943
answered Jul 21 at 20:30
amd
25.9k2943
25.9k2943
Nice. I'd put more emphasis on the relationship $$lVert brVert^2-lVert arVert^2=(b-a)^Tcdot(b+a)$$ which you certainly implied, but which doesn't stand out when skimming your post while being crucial fo the connection from midpoint to difference of squared norms.
– MvG
Jul 21 at 20:35
@MvG Good point. I took for granted that the reader knows this identity.
– amd
Jul 21 at 20:41
add a comment |Â
Nice. I'd put more emphasis on the relationship $$lVert brVert^2-lVert arVert^2=(b-a)^Tcdot(b+a)$$ which you certainly implied, but which doesn't stand out when skimming your post while being crucial fo the connection from midpoint to difference of squared norms.
– MvG
Jul 21 at 20:35
@MvG Good point. I took for granted that the reader knows this identity.
– amd
Jul 21 at 20:41
Nice. I'd put more emphasis on the relationship $$lVert brVert^2-lVert arVert^2=(b-a)^Tcdot(b+a)$$ which you certainly implied, but which doesn't stand out when skimming your post while being crucial fo the connection from midpoint to difference of squared norms.
– MvG
Jul 21 at 20:35
Nice. I'd put more emphasis on the relationship $$lVert brVert^2-lVert arVert^2=(b-a)^Tcdot(b+a)$$ which you certainly implied, but which doesn't stand out when skimming your post while being crucial fo the connection from midpoint to difference of squared norms.
– MvG
Jul 21 at 20:35
@MvG Good point. I took for granted that the reader knows this identity.
– amd
Jul 21 at 20:41
@MvG Good point. I took for granted that the reader knows this identity.
– amd
Jul 21 at 20:41
add a comment |Â
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