Sharp Lower Bound for Entrywise 1-norm of a Real Semi-Orthogonal Matrix

Question

Let $A$ be a real-valued $m\times n$ ($m>n>1$) matrix such that $A^TA=I$, what is the sharp lower bound for $\|A\|_1=\|\operatorname{vec}(A)\|_1=\sum|A_{i,j}|$?

Since one can show the submultiplicativity of $\|\cdot\|_1$, we have $$\|I\|_1=\|A^TA\|_1\leq\|A^T\|_1\|A\|_1=\|A\|_1^2,$$ which implies $\|A\|_1\geq\sqrt{n}$. However, I am not sure if the equality here is actually attainable, especially for real $A$... I suspect instead $\min\|A\|_1=n$.

Answer 1

The property $A^TA=I$ means that $A$ is an isometry in the $2$-norm: $\|Ax\|_2 =\|x\|_2$ for all $x\in\mathbb{R}^n$. In particular, $\|Ae_k\|_2=1$ for the standard basis vectors $e_1,\dots, e_n$. It follows that $$ \|A\|_1 = \sum_{k=1}^n \|Ae_k\|_1 \ge \sum_{k=1}^n \|Ae_k\|_2 \ge n $$ as you expected.