TITLE: Sum of coefficients of an orthogonal matrix
QUESTION [6 upvotes]: Let $(a_{ij})_{1 \le i,j \le n}$ be a real orthogonal matrix. Show that
  $$\left| \sum_{1 \le i,j \le n} a_{ij}\right| \le n.$$

Naively applying the Cauchy-Schwarz inequality only gives $n^{\frac{3}{2}}$ (but only relies on the columns being of norm $1$, and not orthogonality). How do we get the stronger bound $n$?

REPLY [5 votes]: Use Cauchy-Schwarz on the product $^{\operatorname t}\!UMU$ where $U = ^{\operatorname t}\! \left( \begin{matrix} 1 & 1 & \dots & 1 \end{matrix} \right)$.

REPLY [3 votes]: Let $A$ be your matrix, and let $u$ be the all ones vector (all its elements are equal to 1). Then the sum can be expressed as
$s = u^t A u$ = $u^t v$
Apply Cauchy-Schwarz to the vectors $u$ and $v$.