TITLE: What is the proof that SVM can be used to solve the least squares problem with norm equality constraint?
QUESTION [5 upvotes]: I've seen it claimed that the solution to the minimization problem:
$$\begin{align*}
\arg \min_{b} \quad & {\left\| A b \right\|}_{2}^{2} \\
\text{subject to} \quad & {\left\| b \right\|}_{2} = 1
\end{align*}$$
is given by first finding the singular value decomposition of A, $$\textbf{A} = \bf{U \Sigma V}$$
And then taking the column of $\bf{V}$ corresponding to the smallest singular value.
Can someone present a proof that this is so?

REPLY [2 votes]: Note that if $A = U\Sigma V$, then $A^* A = V^*\Sigma^*U^*U\Sigma V = V^*\Sigma^*\Sigma V$. Therefore, the eigenvalues of $A^*A$ are equal to square of the absolute values of the singular values of $A$. Also the minmization problem can be seen as
$$ \underset{b: \;||b||=1}{argmin} \; ||Ab||= \underset{b: \;||b||=1}{argmin}\;||Ab||^2 = \underset{b: \;||b||=1}{argmin}\; b^*(A^*A) b $$
This is equivalent to finding the eigenvector corresponding to the minimum eigenvalue of $A^TA$, which is precisely the column in $V$ corresponding to the minimum eigenvalue in the diagonal matrix $\Sigma^*\Sigma$ (which is in turn the absolute square of the minimum singular value of $A$).