Let $A=(a_{ij})$ be an $n \times m$ matrix. Here the $a_{ij}$ are independent identically distributed random variables. The covariance matrix $M=\frac 1n AA^T$. It is known that the maximum eigenvalue of a symmetric matrix is bounded by its largest diagonal element. Thus, the Euclidean norm $$ \|M\|=\sup_{\|x\|=1}\langle Mx,x\rangle \geq \sup_i M_{ii}. $$
How can I get a similar bound if matrix the $M=PX$, where $X$ is some random vector in $\mathbb R^n$ and $P$ is an orthogonal projection in $\mathbb R^n$?
(Also, if I am right, there can be a few possibilities for the orthogonal projection, i.e. when it is orthogonal to the vector $(1,0,0,\ldots)$ and when it is orthogonal to the vector $(1,1,0,0,\ldots)$. What is the difference?)