Upper bound of an $\|\cdot \|_\infty$ of a projector Let $A$ be a real $n\times n$ symmetric matrix of rank 1 such that $A^2=A$. Prove that $\|A \|_\infty \le \frac{\sqrt{n}+1}{2}$.
I tried to solve this using the fact that $A$ is similar to the matrix which only has 1 as the first element and all others are zeros and also comparing $\| \cdot \|_\infty$ to $\| \cdot \|_2$ and $\| \cdot \|_F$, but this didn't help because these norms are unitary invariant, and I can't conclude anything about multiplying by unitary matrices for $A$.
 A: Let $u$ be a unit vector spanning the range of $A$. Since $A$ is the orthogonal projection onto the span of $u$, we have 
$$Ax=\langle x,u\rangle u\quad\forall x\in\mathbb{R}^n$$
In particular, if $e_i$ are the standard unit vectors, then
$$Ae_i=\langle e_i,u\rangle u\quad\forall i=1,2,\dots,n$$
hence
$$\|Ae_i\|_1=|\langle e_i,u\rangle|\|u\|_1$$
where $\|\cdot\|_1$ denotes the norm in $\ell_1^n$.
 It follows that
$$\max_i\|Ae_i\|_1 = \max_i|\langle e_i,u\rangle|\|u\|_1=\|u\|_{\infty}\|u\|_1$$
where $\|\cdot\|_{\infty}$ denotes the norm in $\ell_{\infty}^n$. The problem is now to calculate the maximum of $\|u\|_{\infty}\|u\|_1$ over the unit sphere.

Claim. The maximum of $\|u\|_{\infty}\|u\|_1$ over the unit sphere $\mathbb{S}^{n-1}$ is equal to $\frac{\sqrt{n}+1}{2}$

Proof. Since both the $\ell_1^n$ and the $\ell_{\infty}^n$ norms are invariant under multiplication of the coordinates by $\pm 1$ and permutations, we may assume that the coordinates of $u$ are non-negative, and that 
$$u_1\geq u_2\geq\cdots \geq u_n\geq 0$$
so we want to maximize the function
$$f(u_1,\dots,u_n)=u_1(u_1+u_2+\cdots + u_n)$$
subject to the condition $u_1^2+\cdots +u_n^2=1$. A straightforward application of Lagrange's multipliers shows that the maximum of $f$ is indeed $\frac{\sqrt{n}+1}{2}$.
