Points on decision hyperplane
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1
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I'm reading a chapter on linear classifiers, and the authors define the decision hyperplane as:
$$
g(mathbf{x}) = mathbf{w}^Tmathbf{x} + w_0 = 0
$$
They say that if $mathbf{x}_1$ and $mathbf{x}_2$ are points on the decision hyperplane, then:
$$
mathbf{w}^Tmathbf{x}_1 + w_0 = 0 \
mathbf{w}^Tmathbf{x}_2 + w_0 = 0
$$
and:
$$
mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) = 0
$$
They then write the difference vector $mathbf{x}_1 - mathbf{x}_2$ lies on the decision hyperplane. Here I am lost. If it were on the decision hyperplane, wouldn't it need to satisfy the original equation? Specifically:
$$
mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0
$$
What am I missing?
linear-algebra pattern-recognition
asked Dec 2 at 20:15
Lorenz Forvang
184
add a comment |
up vote
1
down vote
favorite
I'm reading a chapter on linear classifiers, and the authors define the decision hyperplane as:
$$
g(mathbf{x}) = mathbf{w}^Tmathbf{x} + w_0 = 0
$$
They say that if $mathbf{x}_1$ and $mathbf{x}_2$ are points on the decision hyperplane, then:
$$
mathbf{w}^Tmathbf{x}_1 + w_0 = 0 \
mathbf{w}^Tmathbf{x}_2 + w_0 = 0
$$
and:
$$
mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) = 0
$$
They then write the difference vector $mathbf{x}_1 - mathbf{x}_2$ lies on the decision hyperplane. Here I am lost. If it were on the decision hyperplane, wouldn't it need to satisfy the original equation? Specifically:
$$
mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0
$$
What am I missing?
linear-algebra pattern-recognition
asked Dec 2 at 20:15
Lorenz Forvang
184
My high-school math teacher used to say "The book is wrong" when she wrote something wrong on the blackboard... Apparently, I would say this also in this situation. Indeed, if $mathbf{w}^Tmathbf{x}_1 + w_0 = 0$ and $mathbf{w}^Tmathbf{x}_2 + w_0 = 0$, then $mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0$ reduces that $w_0 = 0$... Read your book carefully, and report here your findings.
– the_candyman
Dec 2 at 20:21
The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
– amd
Dec 3 at 9:02
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm reading a chapter on linear classifiers, and the authors define the decision hyperplane as:
$$
g(mathbf{x}) = mathbf{w}^Tmathbf{x} + w_0 = 0
$$
They say that if $mathbf{x}_1$ and $mathbf{x}_2$ are points on the decision hyperplane, then:
$$
mathbf{w}^Tmathbf{x}_1 + w_0 = 0 \
mathbf{w}^Tmathbf{x}_2 + w_0 = 0
$$
and:
$$
mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) = 0
$$
They then write the difference vector $mathbf{x}_1 - mathbf{x}_2$ lies on the decision hyperplane. Here I am lost. If it were on the decision hyperplane, wouldn't it need to satisfy the original equation? Specifically:
$$
mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0
$$
What am I missing?
linear-algebra pattern-recognition
asked Dec 2 at 20:15
Lorenz Forvang
184
I'm reading a chapter on linear classifiers, and the authors define the decision hyperplane as:
$$
g(mathbf{x}) = mathbf{w}^Tmathbf{x} + w_0 = 0
$$
They say that if $mathbf{x}_1$ and $mathbf{x}_2$ are points on the decision hyperplane, then:
$$
mathbf{w}^Tmathbf{x}_1 + w_0 = 0 \
mathbf{w}^Tmathbf{x}_2 + w_0 = 0
$$
and:
$$
mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) = 0
$$
They then write the difference vector $mathbf{x}_1 - mathbf{x}_2$ lies on the decision hyperplane. Here I am lost. If it were on the decision hyperplane, wouldn't it need to satisfy the original equation? Specifically:
$$
mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0
$$
What am I missing?
linear-algebra pattern-recognition
linear-algebra pattern-recognition
asked Dec 2 at 20:15
Lorenz Forvang
184
asked Dec 2 at 20:15
Lorenz Forvang
184
edited 15 hours ago
asked Dec 2 at 20:15
Lorenz Forvang
184
asked Dec 2 at 20:15
Lorenz Forvang
184
asked Dec 2 at 20:15
Lorenz Forvang
184
184
My high-school math teacher used to say "The book is wrong" when she wrote something wrong on the blackboard... Apparently, I would say this also in this situation. Indeed, if $mathbf{w}^Tmathbf{x}_1 + w_0 = 0$ and $mathbf{w}^Tmathbf{x}_2 + w_0 = 0$, then $mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0$ reduces that $w_0 = 0$... Read your book carefully, and report here your findings.
– the_candyman
Dec 2 at 20:21
The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
– amd
Dec 3 at 9:02
add a comment |
My high-school math teacher used to say "The book is wrong" when she wrote something wrong on the blackboard... Apparently, I would say this also in this situation. Indeed, if $mathbf{w}^Tmathbf{x}_1 + w_0 = 0$ and $mathbf{w}^Tmathbf{x}_2 + w_0 = 0$, then $mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0$ reduces that $w_0 = 0$... Read your book carefully, and report here your findings.
– the_candyman
Dec 2 at 20:21
The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
– amd
Dec 3 at 9:02
My high-school math teacher used to say "The book is wrong" when she wrote something wrong on the blackboard... Apparently, I would say this also in this situation. Indeed, if $mathbf{w}^Tmathbf{x}_1 + w_0 = 0$ and $mathbf{w}^Tmathbf{x}_2 + w_0 = 0$, then $mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0$ reduces that $w_0 = 0$... Read your book carefully, and report here your findings.
– the_candyman
Dec 2 at 20:21
My high-school math teacher used to say "The book is wrong" when she wrote something wrong on the blackboard... Apparently, I would say this also in this situation. Indeed, if $mathbf{w}^Tmathbf{x}_1 + w_0 = 0$ and $mathbf{w}^Tmathbf{x}_2 + w_0 = 0$, then $mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0$ reduces that $w_0 = 0$... Read your book carefully, and report here your findings.
– the_candyman
Dec 2 at 20:21
The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
– amd
Dec 3 at 9:02
The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
– amd
Dec 3 at 9:02
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The statement as written is false for arbitrary $w_0$. Given
$$
mathbf{w}^Tmathbf{x}_1 + w_0 = 0 \
mathbf{w}^Tmathbf{x}_2 + w_0 = 0
$$
it follows that $mathbf{w}^Tmathbf{x}_1=mathbf{w}^Tmathbf{x}_2=-w_0$, and hence that
$$mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2)
=mathbf{w}^Tmathbf{x}_1-mathbf{w}^Tmathbf{x}_2
=(-w_0)-(-w_0)= 0.
$$
So
It is "obviously" true if and only if $w_0=0$. I would say you are not missing anything.
You'd be surprised how many errors slip through dozens of editions.
answered Dec 2 at 20:42
Servaes
21.8k33792
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The statement as written is false for arbitrary $w_0$. Given
$$
mathbf{w}^Tmathbf{x}_1 + w_0 = 0 \
mathbf{w}^Tmathbf{x}_2 + w_0 = 0
$$
it follows that $mathbf{w}^Tmathbf{x}_1=mathbf{w}^Tmathbf{x}_2=-w_0$, and hence that
$$mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2)
=mathbf{w}^Tmathbf{x}_1-mathbf{w}^Tmathbf{x}_2
=(-w_0)-(-w_0)= 0.
$$
So
It is "obviously" true if and only if $w_0=0$. I would say you are not missing anything.
You'd be surprised how many errors slip through dozens of editions.
answered Dec 2 at 20:42
Servaes
21.8k33792
add a comment |
up vote
1
down vote
accepted
The statement as written is false for arbitrary $w_0$. Given
$$
mathbf{w}^Tmathbf{x}_1 + w_0 = 0 \
mathbf{w}^Tmathbf{x}_2 + w_0 = 0
$$
it follows that $mathbf{w}^Tmathbf{x}_1=mathbf{w}^Tmathbf{x}_2=-w_0$, and hence that
$$mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2)
=mathbf{w}^Tmathbf{x}_1-mathbf{w}^Tmathbf{x}_2
=(-w_0)-(-w_0)= 0.
$$
So
It is "obviously" true if and only if $w_0=0$. I would say you are not missing anything.
You'd be surprised how many errors slip through dozens of editions.
answered Dec 2 at 20:42
Servaes
21.8k33792
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The statement as written is false for arbitrary $w_0$. Given
$$
mathbf{w}^Tmathbf{x}_1 + w_0 = 0 \
mathbf{w}^Tmathbf{x}_2 + w_0 = 0
$$
it follows that $mathbf{w}^Tmathbf{x}_1=mathbf{w}^Tmathbf{x}_2=-w_0$, and hence that
$$mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2)
=mathbf{w}^Tmathbf{x}_1-mathbf{w}^Tmathbf{x}_2
=(-w_0)-(-w_0)= 0.
$$
So
It is "obviously" true if and only if $w_0=0$. I would say you are not missing anything.
You'd be surprised how many errors slip through dozens of editions.
answered Dec 2 at 20:42
Servaes
21.8k33792
The statement as written is false for arbitrary $w_0$. Given
$$
mathbf{w}^Tmathbf{x}_1 + w_0 = 0 \
mathbf{w}^Tmathbf{x}_2 + w_0 = 0
$$
it follows that $mathbf{w}^Tmathbf{x}_1=mathbf{w}^Tmathbf{x}_2=-w_0$, and hence that
$$mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2)
=mathbf{w}^Tmathbf{x}_1-mathbf{w}^Tmathbf{x}_2
=(-w_0)-(-w_0)= 0.
$$
So
It is "obviously" true if and only if $w_0=0$. I would say you are not missing anything.
You'd be surprised how many errors slip through dozens of editions.
answered Dec 2 at 20:42
Servaes
21.8k33792
answered Dec 2 at 20:42
Servaes
21.8k33792
answered Dec 2 at 20:42
Servaes
21.8k33792
answered Dec 2 at 20:42
Servaes
21.8k33792
21.8k33792
add a comment |
add a comment |
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My high-school math teacher used to say "The book is wrong" when she wrote something wrong on the blackboard... Apparently, I would say this also in this situation. Indeed, if $mathbf{w}^Tmathbf{x}_1 + w_0 = 0$ and $mathbf{w}^Tmathbf{x}_2 + w_0 = 0$, then $mathbf{w}^T(mathbf{x}_1 - mathbf{x}_2) + w_0 = 0$ reduces that $w_0 = 0$... Read your book carefully, and report here your findings.
– the_candyman
Dec 2 at 20:21
The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
– amd
Dec 3 at 9:02