$k^{\text{th}}$ column of $\boldsymbol{V}$. It should be noted that $\boldsymbol{V}^T$ is restricted to be a collection of $K$ normalized rows of $\boldsymbol{A}$, while there is no constraint on $\boldsymbol{U}$. Assume we are to select one sample at a time, which is the best representation of all data. Since Problem (\[eq:rewrite\]) involves a combinatorial search and is not easy to tackle, let us modify (\[eq:rewrite\]) into two consecutive problems. The first sub-problem relaxes the constraint $\boldsymbol{v}_k \in \mathbb{A}$ in (\[eq:rewrite\]) to a moderate constraint $\|\boldsymbol{v}\|=1$, and the second sub-problem reimposes the underlying constraint. These sub-problems are formulated as

$$\begin{aligned}
\label{eq:our1}
(\boldsymbol{u},\boldsymbol{v})=&\underset{\boldsymbol{u},\boldsymbol{v}}{\text{argmin}}\;\|\boldsymbol{A}-\boldsymbol{uv}^T\|_F^2 \;\small{\text{s.t.}}\; \|\boldsymbol{v}\|=1,\\
\label{eq:our2}
m^{(1)}=&\underset{m}{\text{argmax}}\;| \boldsymbol{v}^T \Tilde{\boldsymbol{a}}_m|.\end{aligned}$$ \[eq:IPM\_main\]

Here $m^{(1)}$ is the index of the first selected data point and $\boldsymbol{a}_{\boldsymbol{m}^{(1)}}$ is the selected sample. Subproblem (\[eq:our1\]) is equivalent to finding the first right singular vector of $\boldsymbol{A}$. The constraint $\|\boldsymbol{v}\|=1$ keeps $\boldsymbol{v}$ on the unit sphere to remove scale ambiguity between $\boldsymbol{u}$ and $\boldsymbol{v}$. Moreover, the unit sphere is a superset for $\mathbb{A}$ and keeps the modified problem close to the recast problem (\[eq:rewrite\]). After solving for $\boldsymbol{v}$ (which is not necessarily one of our data points), we find the data point that matches $\boldsymbol{v}$ the most (has the least 