In SVM Algorithm, why vector w is orthogonal to the separating hyperplane? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Moderator Election Q&A - Questionnaire 2019 Community Moderator Election ResultsUsing SVM as a binary classifier, is the label for a data point chosen by consensus?How are Hyperplane Heatmaps created and how should they be interpreted?why is SVM cost function the norm of the hyperplane parameters, and not the mean square error?Why do we use +1 and -1 for marginal decision boundaries in SVMSVM on sparse dataWhy does an SVM model store the support vectors, and not just the separating hyperplane?SVM - why does scaling the parameters w and b result in nothing meaningful?prediction for a linear sumHow regularization parameter in SVM affects hyperplane parametersSVM radial basis generate equation for hyperplane
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In SVM Algorithm, why vector w is orthogonal to the separating hyperplane?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Moderator Election Q&A - Questionnaire
2019 Community Moderator Election ResultsUsing SVM as a binary classifier, is the label for a data point chosen by consensus?How are Hyperplane Heatmaps created and how should they be interpreted?why is SVM cost function the norm of the hyperplane parameters, and not the mean square error?Why do we use +1 and -1 for marginal decision boundaries in SVMSVM on sparse dataWhy does an SVM model store the support vectors, and not just the separating hyperplane?SVM - why does scaling the parameters w and b result in nothing meaningful?prediction for a linear sumHow regularization parameter in SVM affects hyperplane parametersSVM radial basis generate equation for hyperplane
$begingroup$
I am a beginner on Machine Learning.
In SVM, the separating hyperplane is defined as $y = w^T x + b$.
Why we say vector $w$ orthogonal to the separating hyperplane?
machine-learning svm
$endgroup$
add a comment |
$begingroup$
I am a beginner on Machine Learning.
In SVM, the separating hyperplane is defined as $y = w^T x + b$.
Why we say vector $w$ orthogonal to the separating hyperplane?
machine-learning svm
$endgroup$
2
$begingroup$
An answer to a similar question (for neural networks) is here.
$endgroup$
– bogatron
Jun 9 '15 at 16:39
$begingroup$
@bogatron - I agree with you completely. But my ones just a SVM specific answer.
$endgroup$
– untitledprogrammer
Jun 10 '15 at 19:43
2
$begingroup$
Except it isn't. Your answer is correct but there is nothing about it that is specific to SVMs (nor should there be). $w^Tx=b$ is simply a vector equation that defines a hyperplane.
$endgroup$
– bogatron
Jun 10 '15 at 22:01
add a comment |
$begingroup$
I am a beginner on Machine Learning.
In SVM, the separating hyperplane is defined as $y = w^T x + b$.
Why we say vector $w$ orthogonal to the separating hyperplane?
machine-learning svm
$endgroup$
I am a beginner on Machine Learning.
In SVM, the separating hyperplane is defined as $y = w^T x + b$.
Why we say vector $w$ orthogonal to the separating hyperplane?
machine-learning svm
machine-learning svm
edited Jun 10 '15 at 11:45
Nitesh
1,1451721
1,1451721
asked Jun 9 '15 at 14:39
Chong ZhengChong Zheng
4913
4913
2
$begingroup$
An answer to a similar question (for neural networks) is here.
$endgroup$
– bogatron
Jun 9 '15 at 16:39
$begingroup$
@bogatron - I agree with you completely. But my ones just a SVM specific answer.
$endgroup$
– untitledprogrammer
Jun 10 '15 at 19:43
2
$begingroup$
Except it isn't. Your answer is correct but there is nothing about it that is specific to SVMs (nor should there be). $w^Tx=b$ is simply a vector equation that defines a hyperplane.
$endgroup$
– bogatron
Jun 10 '15 at 22:01
add a comment |
2
$begingroup$
An answer to a similar question (for neural networks) is here.
$endgroup$
– bogatron
Jun 9 '15 at 16:39
$begingroup$
@bogatron - I agree with you completely. But my ones just a SVM specific answer.
$endgroup$
– untitledprogrammer
Jun 10 '15 at 19:43
2
$begingroup$
Except it isn't. Your answer is correct but there is nothing about it that is specific to SVMs (nor should there be). $w^Tx=b$ is simply a vector equation that defines a hyperplane.
$endgroup$
– bogatron
Jun 10 '15 at 22:01
2
2
$begingroup$
An answer to a similar question (for neural networks) is here.
$endgroup$
– bogatron
Jun 9 '15 at 16:39
$begingroup$
An answer to a similar question (for neural networks) is here.
$endgroup$
– bogatron
Jun 9 '15 at 16:39
$begingroup$
@bogatron - I agree with you completely. But my ones just a SVM specific answer.
$endgroup$
– untitledprogrammer
Jun 10 '15 at 19:43
$begingroup$
@bogatron - I agree with you completely. But my ones just a SVM specific answer.
$endgroup$
– untitledprogrammer
Jun 10 '15 at 19:43
2
2
$begingroup$
Except it isn't. Your answer is correct but there is nothing about it that is specific to SVMs (nor should there be). $w^Tx=b$ is simply a vector equation that defines a hyperplane.
$endgroup$
– bogatron
Jun 10 '15 at 22:01
$begingroup$
Except it isn't. Your answer is correct but there is nothing about it that is specific to SVMs (nor should there be). $w^Tx=b$ is simply a vector equation that defines a hyperplane.
$endgroup$
– bogatron
Jun 10 '15 at 22:01
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Geometrically, the vector w is directed orthogonal to the line defined by $w^T x = b$. This can be understood as follows:
First take $b = 0$. Now it is clear that all vectors, $x$, with vanishing inner product with $w$ satisfy this equation, i.e. all vectors orthogonal to w satisfy this equation.
Now translate the hyperplane away from the origin over a vector a. The equation for the plane now becomes: $(x − a)^T w = 0$, i.e. we find that for the offset $b = a^T w$, which is the projection of the vector $a$ onto the vector $w$.
Without loss of generality we may thus choose a perpendicular to the plane, in which case the length $vertvert a vertvert = vert b vert /vertvert wvertvert$ which represents the shortest, orthogonal distance between the origin and the hyperplane.
Hence the vector $w$ is said to be orthogonal to the separating hyperplane.
$endgroup$
add a comment |
$begingroup$
The reason why $w$ is normal to the hyper-plane is because we define it to be that way:
Suppose that we have a (hyper)plane in 3d space. Let $P_0$ be a point on this plane i.e. $P_0 = x_0, y_0, z_0$. Therefore the vector from the origin $(0,0,0)$ to this point is just $<x_0,y_0,z_0>$. Suppose that we have an arbitrary point $P (x,y,z)$ on the plane. The vector joining $P$ and $P_0$ is then given by:
$$ vecP - vecP_0 = <x-x_0, y-y_0, z-z_0>$$
Note that this vector lies in the plane.
Now let $hatn$ be the normal (orthogonal) vector to the plane. Therefore:
$$ hatn bullet (vecP-vecP_0) = 0$$
Therefore:
$$hatn bullet vecP- hatn bullet vecP_0 = 0$$
Note that $-hatn bullet vecP_0$ is just a number and is equal to $b$ in our case, whereas $hatn$ is just $w$ and $vecP$ is $x$. So by definition, $w$ is orthogonal to the hyperplane.
$endgroup$
add a comment |
$begingroup$
Using the algebraic definition of a vector being orthogonal to a hyperplane:
$forall x_1, x_2$ on the separating hyperplane,
$$ w^T(x_1-x_2)=(w^Tx_1 + b)-(w^Tx_2 + b)=0-0=0 smallBox.$$
$endgroup$
add a comment |
$begingroup$
Let the decision boundary be defined as $w^Tx + b = 0$. Consider the points $x_a$ and $x_b$, which lie on the decision boundary. This gives us two equations:
beginequation
w^Tx_a + b = 0 \
w^Tx_b + b = 0
endequation
Subtracting these two equations gives us $w^T.(x_a - x_b) = 0$. Note that the vector $x_a - x_b$ lies on the decision boundary, and it is directed from $x_b$ to $x_a$. Since the dot product $w^T.(x_a - x_b)$ is zero, $w^T$ must be orthogonal to $x_a - x_b$, and in turn, to the decision boundary.
New contributor
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Geometrically, the vector w is directed orthogonal to the line defined by $w^T x = b$. This can be understood as follows:
First take $b = 0$. Now it is clear that all vectors, $x$, with vanishing inner product with $w$ satisfy this equation, i.e. all vectors orthogonal to w satisfy this equation.
Now translate the hyperplane away from the origin over a vector a. The equation for the plane now becomes: $(x − a)^T w = 0$, i.e. we find that for the offset $b = a^T w$, which is the projection of the vector $a$ onto the vector $w$.
Without loss of generality we may thus choose a perpendicular to the plane, in which case the length $vertvert a vertvert = vert b vert /vertvert wvertvert$ which represents the shortest, orthogonal distance between the origin and the hyperplane.
Hence the vector $w$ is said to be orthogonal to the separating hyperplane.
$endgroup$
add a comment |
$begingroup$
Geometrically, the vector w is directed orthogonal to the line defined by $w^T x = b$. This can be understood as follows:
First take $b = 0$. Now it is clear that all vectors, $x$, with vanishing inner product with $w$ satisfy this equation, i.e. all vectors orthogonal to w satisfy this equation.
Now translate the hyperplane away from the origin over a vector a. The equation for the plane now becomes: $(x − a)^T w = 0$, i.e. we find that for the offset $b = a^T w$, which is the projection of the vector $a$ onto the vector $w$.
Without loss of generality we may thus choose a perpendicular to the plane, in which case the length $vertvert a vertvert = vert b vert /vertvert wvertvert$ which represents the shortest, orthogonal distance between the origin and the hyperplane.
Hence the vector $w$ is said to be orthogonal to the separating hyperplane.
$endgroup$
add a comment |
$begingroup$
Geometrically, the vector w is directed orthogonal to the line defined by $w^T x = b$. This can be understood as follows:
First take $b = 0$. Now it is clear that all vectors, $x$, with vanishing inner product with $w$ satisfy this equation, i.e. all vectors orthogonal to w satisfy this equation.
Now translate the hyperplane away from the origin over a vector a. The equation for the plane now becomes: $(x − a)^T w = 0$, i.e. we find that for the offset $b = a^T w$, which is the projection of the vector $a$ onto the vector $w$.
Without loss of generality we may thus choose a perpendicular to the plane, in which case the length $vertvert a vertvert = vert b vert /vertvert wvertvert$ which represents the shortest, orthogonal distance between the origin and the hyperplane.
Hence the vector $w$ is said to be orthogonal to the separating hyperplane.
$endgroup$
Geometrically, the vector w is directed orthogonal to the line defined by $w^T x = b$. This can be understood as follows:
First take $b = 0$. Now it is clear that all vectors, $x$, with vanishing inner product with $w$ satisfy this equation, i.e. all vectors orthogonal to w satisfy this equation.
Now translate the hyperplane away from the origin over a vector a. The equation for the plane now becomes: $(x − a)^T w = 0$, i.e. we find that for the offset $b = a^T w$, which is the projection of the vector $a$ onto the vector $w$.
Without loss of generality we may thus choose a perpendicular to the plane, in which case the length $vertvert a vertvert = vert b vert /vertvert wvertvert$ which represents the shortest, orthogonal distance between the origin and the hyperplane.
Hence the vector $w$ is said to be orthogonal to the separating hyperplane.
edited Jun 18 '18 at 21:32
Community♦
1
1
answered Jun 9 '15 at 15:12
untitledprogrammeruntitledprogrammer
581216
581216
add a comment |
add a comment |
$begingroup$
The reason why $w$ is normal to the hyper-plane is because we define it to be that way:
Suppose that we have a (hyper)plane in 3d space. Let $P_0$ be a point on this plane i.e. $P_0 = x_0, y_0, z_0$. Therefore the vector from the origin $(0,0,0)$ to this point is just $<x_0,y_0,z_0>$. Suppose that we have an arbitrary point $P (x,y,z)$ on the plane. The vector joining $P$ and $P_0$ is then given by:
$$ vecP - vecP_0 = <x-x_0, y-y_0, z-z_0>$$
Note that this vector lies in the plane.
Now let $hatn$ be the normal (orthogonal) vector to the plane. Therefore:
$$ hatn bullet (vecP-vecP_0) = 0$$
Therefore:
$$hatn bullet vecP- hatn bullet vecP_0 = 0$$
Note that $-hatn bullet vecP_0$ is just a number and is equal to $b$ in our case, whereas $hatn$ is just $w$ and $vecP$ is $x$. So by definition, $w$ is orthogonal to the hyperplane.
$endgroup$
add a comment |
$begingroup$
The reason why $w$ is normal to the hyper-plane is because we define it to be that way:
Suppose that we have a (hyper)plane in 3d space. Let $P_0$ be a point on this plane i.e. $P_0 = x_0, y_0, z_0$. Therefore the vector from the origin $(0,0,0)$ to this point is just $<x_0,y_0,z_0>$. Suppose that we have an arbitrary point $P (x,y,z)$ on the plane. The vector joining $P$ and $P_0$ is then given by:
$$ vecP - vecP_0 = <x-x_0, y-y_0, z-z_0>$$
Note that this vector lies in the plane.
Now let $hatn$ be the normal (orthogonal) vector to the plane. Therefore:
$$ hatn bullet (vecP-vecP_0) = 0$$
Therefore:
$$hatn bullet vecP- hatn bullet vecP_0 = 0$$
Note that $-hatn bullet vecP_0$ is just a number and is equal to $b$ in our case, whereas $hatn$ is just $w$ and $vecP$ is $x$. So by definition, $w$ is orthogonal to the hyperplane.
$endgroup$
add a comment |
$begingroup$
The reason why $w$ is normal to the hyper-plane is because we define it to be that way:
Suppose that we have a (hyper)plane in 3d space. Let $P_0$ be a point on this plane i.e. $P_0 = x_0, y_0, z_0$. Therefore the vector from the origin $(0,0,0)$ to this point is just $<x_0,y_0,z_0>$. Suppose that we have an arbitrary point $P (x,y,z)$ on the plane. The vector joining $P$ and $P_0$ is then given by:
$$ vecP - vecP_0 = <x-x_0, y-y_0, z-z_0>$$
Note that this vector lies in the plane.
Now let $hatn$ be the normal (orthogonal) vector to the plane. Therefore:
$$ hatn bullet (vecP-vecP_0) = 0$$
Therefore:
$$hatn bullet vecP- hatn bullet vecP_0 = 0$$
Note that $-hatn bullet vecP_0$ is just a number and is equal to $b$ in our case, whereas $hatn$ is just $w$ and $vecP$ is $x$. So by definition, $w$ is orthogonal to the hyperplane.
$endgroup$
The reason why $w$ is normal to the hyper-plane is because we define it to be that way:
Suppose that we have a (hyper)plane in 3d space. Let $P_0$ be a point on this plane i.e. $P_0 = x_0, y_0, z_0$. Therefore the vector from the origin $(0,0,0)$ to this point is just $<x_0,y_0,z_0>$. Suppose that we have an arbitrary point $P (x,y,z)$ on the plane. The vector joining $P$ and $P_0$ is then given by:
$$ vecP - vecP_0 = <x-x_0, y-y_0, z-z_0>$$
Note that this vector lies in the plane.
Now let $hatn$ be the normal (orthogonal) vector to the plane. Therefore:
$$ hatn bullet (vecP-vecP_0) = 0$$
Therefore:
$$hatn bullet vecP- hatn bullet vecP_0 = 0$$
Note that $-hatn bullet vecP_0$ is just a number and is equal to $b$ in our case, whereas $hatn$ is just $w$ and $vecP$ is $x$. So by definition, $w$ is orthogonal to the hyperplane.
answered Sep 4 '18 at 14:09
Shehryar MalikShehryar Malik
112
112
add a comment |
add a comment |
$begingroup$
Using the algebraic definition of a vector being orthogonal to a hyperplane:
$forall x_1, x_2$ on the separating hyperplane,
$$ w^T(x_1-x_2)=(w^Tx_1 + b)-(w^Tx_2 + b)=0-0=0 smallBox.$$
$endgroup$
add a comment |
$begingroup$
Using the algebraic definition of a vector being orthogonal to a hyperplane:
$forall x_1, x_2$ on the separating hyperplane,
$$ w^T(x_1-x_2)=(w^Tx_1 + b)-(w^Tx_2 + b)=0-0=0 smallBox.$$
$endgroup$
add a comment |
$begingroup$
Using the algebraic definition of a vector being orthogonal to a hyperplane:
$forall x_1, x_2$ on the separating hyperplane,
$$ w^T(x_1-x_2)=(w^Tx_1 + b)-(w^Tx_2 + b)=0-0=0 smallBox.$$
$endgroup$
Using the algebraic definition of a vector being orthogonal to a hyperplane:
$forall x_1, x_2$ on the separating hyperplane,
$$ w^T(x_1-x_2)=(w^Tx_1 + b)-(w^Tx_2 + b)=0-0=0 smallBox.$$
answered Feb 17 '18 at 0:11
IndominusIndominus
1105
1105
add a comment |
add a comment |
$begingroup$
Let the decision boundary be defined as $w^Tx + b = 0$. Consider the points $x_a$ and $x_b$, which lie on the decision boundary. This gives us two equations:
beginequation
w^Tx_a + b = 0 \
w^Tx_b + b = 0
endequation
Subtracting these two equations gives us $w^T.(x_a - x_b) = 0$. Note that the vector $x_a - x_b$ lies on the decision boundary, and it is directed from $x_b$ to $x_a$. Since the dot product $w^T.(x_a - x_b)$ is zero, $w^T$ must be orthogonal to $x_a - x_b$, and in turn, to the decision boundary.
New contributor
$endgroup$
add a comment |
$begingroup$
Let the decision boundary be defined as $w^Tx + b = 0$. Consider the points $x_a$ and $x_b$, which lie on the decision boundary. This gives us two equations:
beginequation
w^Tx_a + b = 0 \
w^Tx_b + b = 0
endequation
Subtracting these two equations gives us $w^T.(x_a - x_b) = 0$. Note that the vector $x_a - x_b$ lies on the decision boundary, and it is directed from $x_b$ to $x_a$. Since the dot product $w^T.(x_a - x_b)$ is zero, $w^T$ must be orthogonal to $x_a - x_b$, and in turn, to the decision boundary.
New contributor
$endgroup$
add a comment |
$begingroup$
Let the decision boundary be defined as $w^Tx + b = 0$. Consider the points $x_a$ and $x_b$, which lie on the decision boundary. This gives us two equations:
beginequation
w^Tx_a + b = 0 \
w^Tx_b + b = 0
endequation
Subtracting these two equations gives us $w^T.(x_a - x_b) = 0$. Note that the vector $x_a - x_b$ lies on the decision boundary, and it is directed from $x_b$ to $x_a$. Since the dot product $w^T.(x_a - x_b)$ is zero, $w^T$ must be orthogonal to $x_a - x_b$, and in turn, to the decision boundary.
New contributor
$endgroup$
Let the decision boundary be defined as $w^Tx + b = 0$. Consider the points $x_a$ and $x_b$, which lie on the decision boundary. This gives us two equations:
beginequation
w^Tx_a + b = 0 \
w^Tx_b + b = 0
endequation
Subtracting these two equations gives us $w^T.(x_a - x_b) = 0$. Note that the vector $x_a - x_b$ lies on the decision boundary, and it is directed from $x_b$ to $x_a$. Since the dot product $w^T.(x_a - x_b)$ is zero, $w^T$ must be orthogonal to $x_a - x_b$, and in turn, to the decision boundary.
New contributor
New contributor
answered 48 mins ago
adityagaydhaniadityagaydhani
12
12
New contributor
New contributor
add a comment |
add a comment |
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2
$begingroup$
An answer to a similar question (for neural networks) is here.
$endgroup$
– bogatron
Jun 9 '15 at 16:39
$begingroup$
@bogatron - I agree with you completely. But my ones just a SVM specific answer.
$endgroup$
– untitledprogrammer
Jun 10 '15 at 19:43
2
$begingroup$
Except it isn't. Your answer is correct but there is nothing about it that is specific to SVMs (nor should there be). $w^Tx=b$ is simply a vector equation that defines a hyperplane.
$endgroup$
– bogatron
Jun 10 '15 at 22:01