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On this answer the hyperplane, presumably in a perceptron classifier, is described as the dot product $\langle \vec{w_{x}},\vec{x} \rangle$, where $\vec{w_x}$ is presumably the vector of weights, and $\vec x$ an example in the training set.

My very tentative understanding is the the hyperplane was the plane defined by the examples in the training set, (or possibly the vector of weights). In other words, a vectorial or geometric (hyperplane) object, rather than a scalar.

Can I get an explanation? Thank you in advance.

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  • $\begingroup$ Looks like sloppy terminology. One form of equation for a hyperplane is $\langle\mathbf n,\mathbf x\rangle=\text{const}$, where $\mathbf n$ is a normal to the hyperplane. I’d guess the author of the answer meant that $\langle\vec{w_x},\vec x\rangle$ is constant on the hyperplane. $\endgroup$
    – amd
    Mar 8, 2017 at 19:58
  • $\begingroup$ @amd That would take some mystery out of the cryptic notation... Would you mind to write a formal answer? $\endgroup$ Mar 8, 2017 at 20:00

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The hyperplane is defined by the equation $\langle \vec{w},\vec{x} \rangle = 0$. This hyperplane partitions the training set into two sets, $\{\vec x\mid \langle\vec{w},\vec{x}_i\rangle >= 0\}$, and $\{\vec x\mid \langle\vec{w},\vec{x}_i\rangle <0\}$.

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  • $\begingroup$ I understand the concept, but not the notation. The actual $<\cdot,\cdot>$ is the problem... Is it a dot product? $\endgroup$ Mar 8, 2017 at 20:05
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    $\begingroup$ The dot product notation $\langle \vec{w},\vec{x}\rangle = 0$ is equivalent to $\sum w_ix_i = 0$, which is a linear equation in $\mathbb{R}^n$, and defines a $n-1$ dimensional subspace in $\mathbb{R}^n$, a hyperplane. $\endgroup$
    – Guangliang
    Mar 8, 2017 at 20:08
  • $\begingroup$ That was one my working "hypotheses"... So the hyperplane is orthogonal to the vector of weights? And if dotted with a training example results in zero, it is on the plane (what to do then? No update, I guess, although it won't happen in real life). Otherwise the sign dictates. Is that it? $\endgroup$ Mar 8, 2017 at 20:12
  • $\begingroup$ I see most people treating $\langle\vec{w},\vec{x}\rangle = 0$ same as it is positive. It makes no difference because $=0$ is a 'zero-measure' event in $\mathbb{R}$. $\endgroup$
    – Guangliang
    Mar 8, 2017 at 20:40
  • $\begingroup$ zero Lebesgue measure, eh? Gotcha! So my little summary is correct... $\endgroup$ Mar 8, 2017 at 20:42

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