Points on decision hyperplane












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – the_candyman
    Dec 2 '18 at 20:21












  • $begingroup$
    The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
    $endgroup$
    – amd
    Dec 3 '18 at 9:02
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – the_candyman
    Dec 2 '18 at 20:21












  • $begingroup$
    The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
    $endgroup$
    – amd
    Dec 3 '18 at 9:02














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 13:34







Lorenz Forvang

















asked Dec 2 '18 at 20:15









Lorenz ForvangLorenz Forvang

184




184












  • $begingroup$
    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.
    $endgroup$
    – the_candyman
    Dec 2 '18 at 20:21












  • $begingroup$
    The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
    $endgroup$
    – amd
    Dec 3 '18 at 9:02


















  • $begingroup$
    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.
    $endgroup$
    – the_candyman
    Dec 2 '18 at 20:21












  • $begingroup$
    The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
    $endgroup$
    – amd
    Dec 3 '18 at 9:02
















$begingroup$
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.
$endgroup$
– the_candyman
Dec 2 '18 at 20:21






$begingroup$
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.
$endgroup$
– the_candyman
Dec 2 '18 at 20:21














$begingroup$
The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
$endgroup$
– amd
Dec 3 '18 at 9:02




$begingroup$
The difference vector only lies on the hyperplace if $w_0=0$. Otherwise, it is parallel to the hyperplane.
$endgroup$
– amd
Dec 3 '18 at 9:02










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









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    1 Answer
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    active

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    1 Answer
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    active

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    active

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    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 20:42









        ServaesServaes

        25.6k33996




        25.6k33996






























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