We have a homogeneous linear system in $x \in \mathbb R^n$
$$A x = 0_m$$
where $A \in \mathbb R^{m \times n}$. We also have constraints on the last $n_2$ entries of $x$. Hence, we write
$$\begin{bmatrix} A_1 & A_2\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix} = 0_m$$
where $A_1 \in \mathbb R^{m \times n_1}$ and $A_2 \in \mathbb R^{m \times n_2}$.
We would like to minimize $\|Ax\|_2$ subject to the equality constraint $\|x_2\|_2 = 1$. Hence, we have the following quadratically constrained quadratic program (QCQP)
$$\begin{array}{ll} \text{minimize} & \|A_1 x_1 + A_2 x_2\|_2^2\\ \text{subject to} & \|x_2\|_2^2 = 1\end{array}$$
Let the Lagrangian be
$$\mathcal{L} (x_1, x_2, \lambda) := \begin{bmatrix} x_1\\ x_2\end{bmatrix}^T \begin{bmatrix} A_1^T A_1 & A_1^T A_2\\ A_2^T A_1 & A_2^T A_2\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix} - \lambda (x_2^T x_2 - 1)$$
Taking the partial derivatives and finding where they vanish, we obtain three equations
$$A_1^T A_1 x_1 + A_1^T A_2 x_2 = 0_{n_1} \qquad \qquad A_2^T A_1 x_1 + A_2^T A_2 x_2 = \lambda x_2 \qquad \qquad x_2^T x_2 = 1$$
Assuming that $A_1$ has full column rank, then $A_1^T A_1$ is invertible. From the 1st equation, we have
$$x_1 = -(A_1^T A_1)^{-1} A_1^T A_2 x_2$$
and
$$A_1 x_1 = - \underbrace{A_1 (A_1^T A_1)^{-1} A_1^T}_{=:P_1} A_2 x_2 = -P_1 A_2 x_2$$
where $P_1$ is the projection matrix that projects onto the column space of $A_1$. Note that $I_m - P_1$ is the projection matrix that projects onto the left null space of $A_1$, which is orthogonal to the column space of $A_1$. As $I_m - P_1$ is a projection matrix, $(I_m - P_1)^2 = I_m - P_1$, which we will use. Hence,
$$\begin{array}{rl} \|A_1 x_1 + A_2 x_2\|_2^2 &= \|(I_m - P_1) A_2 x_2\|_2^2\\\\ &\geq \sigma_{\min}^2 ((I_m - P_1) A_2) \|x_2\|_2^2\\\\ &= \sigma_{\min}^2 ((I_m - P_1) A_2)\\\\ &= \lambda_{\min} (A_2^T (I_m - P_1)^2 A_2)\\\\ &= \lambda_{\min} (A_2^T (I_m - P_1) A_2)\end{array}$$
where we used the equality constraint $\|x_2\|_2^2 = 1$. Thus, we conclude that
the minimum is attained at the intersection of the eigenspace of the minimum eigenvalue of $A_2^T (I_m - P_1) A_2$ with the unit Euclidean sphere in $\mathbb R^{n_2}$.
$x_2$ is a normalized eigenvector of $A_2^T (I_m - P_1) A_2$. It is also a (normalized) right singular vector of $(I_m - P_1) A_2$. Once we have $x_2$, we obtain $x_1$ via $x_1 = -(A_1^T A_1)^{-1} A_1^T A_2 x_2$.
We can arrive at the same conclusion using a different approach. Using $A_1 x_1 = - P_1 A_2 x_2$, the 2nd equation produced by the vanishing gradient of the Lagrangian then becomes
$$A_2^T (I_m - P_1) A_2 x_2 = \lambda x_2$$
Thus, $(\lambda, x_2)$ is an eigenpair of $A_2^T (I_m - P_1) A_2$. Which eigenpair? The eigenpair corresponding to the minimum eigenvalue of $A_2^T (I_m - P_1) A_2$, and in which the eigenvector has unit $2$-norm.
In MATLAB, use function inv to invert matrices and function eig to find eigenvalues and eigenvectors. If $A_1$ does not have full column rank, then use function pinv to compute the pseudoinverse of $A_1^T A_1$.
Lastly, there are several errors in Huang & Barth [PDF]. For example, in equation (7), the Lagrangian function is obviously incorrectly typed, as the $2$-norms are missing.
