Spectral norm of $A$ matrix which equal column norm?

Question

We are given a matrix $A=(A_1, \dots, A_n)$ where $A_i \in \mathbb{R}^m$. Moreover, $||A_i||_2=1$.

This yields for example for the frobeniusnorm $||A||_F=\sqrt{\sum_{i=1}^n ||A_i||_2^2} = \sqrt{n} $ or for the infinity norm $||A||_\infty = \max_{i=1, \dots, m} \sum_{j=1}^n |(A_{j})_i|\leq \max_{i=1, \dots, m} \sum_{j=1}^n 1 \leq n.$

Does this also give us an upper bound on the spectral norm of $A$? I know that the spectral norm is the largest singular value of $A$ and an intuitive interpretation is that the spectral norm gives how much the matrix can 'scale' a vector. So I would guess there is a connection between the columns of $A$ being normalized to $1$ and the size the spectral norm can obtain, but I don't see it.

It is not diffictult to prove that the spectral norm yields a lower bound of $1$ in this case since: $||A||_2=\max_{||v||=1} ||Av||_2 \geq ||Ae_i||_2 = ||A_i||_2 = 1$.

Answer 1

Since each column of $A$ is a unit vector, for any unit $v\in\mathbb R^m$, $v^TA$ is entrywise between $-1$ and $1$. It follows that $\|v^TA\|_2\le\sqrt{n}$. Hence $\|A\|_2\le\sqrt{n}$. Equality occurs when each column of $A$ is $\pm$ column 1.

Answer 2

$\|A\|_F = \sqrt{\sum_{i=1}^n \sigma_i^2}$, where $\sigma_i$ is the $i$-th singular value of $A$.

The number of non-zero singular values equals the rank of $A$.

Thus, the spectral norm can be as large as $\sqrt n$ if the rank of $A$ is one.