Why can we do this in SVM optimization? I posted this question in the datascience stackexchange but no one answered so I thought I would post it here aswell:
Ok, so I've been trying to read up on how SVM:s work and started with maximal margin classifiers. At page 132 in ESL (Elements of Statistical Learning) the authors "reformulates" the optimization problem but I can't seem to understand what they are doing from 4.47 to 4.48. Does anyone know?
Here is an excerpt:

What I don't understand is why we can arbitrarly set the magnitude of beta to 1/M. What does a positively scaled multiple mean in this case? Just a multiple larger than 0?
 A: For reference, the optimization problem in hand is
\begin{gather}\tag{1}
\max_{\beta, \beta_0} M\\
\text{subject to }y_i(x_i^T\beta + \beta_0)\geq M\|\beta\|, \forall i.
\end{gather}
Assume the pair $(\widehat{\beta}, \widehat{\beta}_0)$ is a solution to this problem and let $\widehat{M}$ be the achieved objective value. Also assume that $(\beta^*, \beta_0^*)$ is a solution to $(4.48)$. We have 
\begin{equation*}
y_i(x_i^T\widehat{\beta} + \widehat{\beta}_0) \geq \widehat{M}\| \widehat{\beta} \|, \forall i
\end{equation*}
which is equivalent to
\begin{equation*}
y_i \left( x_i^T \frac{\widehat{\beta}}{\| \widehat{\beta} \|\widehat{M}} + \frac{\widehat{\beta}_0}{\| \widehat{\beta} \|\widehat{M}} \right) \geq 1,\forall i
\end{equation*}
therefore, the pair $\left( \frac{\widehat{\beta}}{\|\widehat{\beta}\|\widehat{M}}, \frac{\widehat{\beta}_0}{\| \widehat{\beta}\| \widehat{M}} \right)$ satisfies the constraints of $(4.48)$. Since $(\beta^*, \beta_0^*)$ is a solution to $(4.48)$, we get
\begin{equation*}
\| \beta^*\| \leq \left\| \frac{\widehat{\beta}}{\| \widehat{\beta} \| \widehat{M}}\right\|  = \frac{1}{\widehat{M}} \qquad\Rightarrow\qquad 1 \geq \widehat{M}\| \beta^* \|.
\end{equation*}
But also
\begin{equation*}
y_i \left( x_i^T \beta^* + \beta_0^* \right) \geq 1 \geq \widehat{M}\| \beta^*\|,\forall i
\end{equation*}
hence, $(\beta^*, \beta_0^*)$ satisfy the constraints of $(1)$.
