Let $A \in \mathbb{R}^{n \times n}$ be a nonsymmetric diagonally dominant matrix with $a_{ij} < 0$ $\forall i \ne j$ and $a_{ii}>0$.

Let the singular value decomposition of $A$ be $A=U \Sigma V^T$ where the columns of $U$ and $V$ are noted $u_i$ and $v_i$ respectively.

I can find the following lower bound for the trace of $U V^T$ \begin{align} tr(UV^T) \ge \frac{1}{2} \end{align}

The question is: **Is the bound sharp?**

Some simple experiments suggest a bound depending on the rank of $A$.

Proof of $u_i^Tv_i > 0$ \begin{align} U \Sigma V^T =& A \\ U \Sigma =& A V \\ u_i \sigma_i =& A v_i \\ v_i^T u_i \sigma_i =& v_i^T A v_i \\ u_i^T v_i =& \frac{v_i^T A v_i}{\sigma_i} > 0 \end{align}

Bound proof: \begin{align} \sum_i \sigma_i =& tr\left(\Sigma \right) = tr\left(U^T A V \right) = tr\left(V U^T A \right) = \sum_i \left(V U^T a_i \right)_i \le \sum_i \left\|V U^T a_i \right\|_2 \\ =& \sum_i \left\|a_i \right\|_2 \le \sum_i \sum_j \left|a_{ij} \right| \le 2 \sum_i a_{ii} = 2~tr\left(A\right) = 2~tr\left(\sum_i \sigma_i u_i v_i^T \right) \\ =& 2\sum_i \sigma_i u_i^ T v_i \le 2 \sqrt{\sum_i \sigma_i^2} \sqrt{\sum_i \left(u_i^ T v_i\right)^2} \le 2 \sum_i \sigma_i \sum_i u_i^T v_i = 2 \sum_i \sigma_i tr\left(U^T V\right) \end{align}