# Row norms of a tall matrix with orthonormal columns

Let $X$ be a $n \times k$ matrix with $n > k$. If the columns of $X$ are orthonormal, then I want to show that the row norms are bounded by 1. My current solutions involves completing $X$ into an orthogonal matrix and then using the fact that $X^T X = X X^T = I$. I would like a more direct argument such as assuming that a row has norm larger than one leads to an immediate contradiction.

This is just the Pythagorean theorem. Denote the column vectors by $$f_1, \dots, f_k$$ Let $$e_j$$ be the usual unit vector with $$1$$ in the $$j$$-th position. Then $$e_j = \sum_{l = 1}^k \langle e_j, f_l\rangle \ f_l + u_j,$$ where $$u_j$$ is a vector perpendicular to the span of $$f_1, \dots, f_k$$. Hence $$1 = ||e_j||^2 = \sum_{l = 1}^k |\langle e_j, f_l\rangle |^2 + ||u_j||^2.$$ Hence $$\sum_{l = 1}^k |\langle e_j, f_l\rangle |^2 \le 1$$. Note that the $$(j, l)$$ entry of the matrix is $$\langle f_l, e_j \rangle$$.
• why do you add $n$ for $e_j$? is $n$ a vector?, if yes could you write its expression? or the $n$ is the number of rows of $X$? I didn't understand the definition of your $n$ – Marso Feb 7 at 15:24
The squared row norms of a matrix $$X$$ with orthonormal columns are also known as the leverage scores of $$X$$. See section 2.4 of Sketching as a Tool for Numerical Linear Algebra for another proof that each of these squared row norms is less than or equal to 1.