2
$\begingroup$

Let $A=(a_{ij})$ be $n \times m$ matrix. Here $a_{ij}$ are independent identical distributed random variables. 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 to get similar bound if matrix $M=PX$, 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, it can be few possibilities of the orthogonal projection, i.e. when it is orthogonal to the vector (1,0,0...) and when it is orthogonal to the vector (1,1,0,0...). What is the difference?)

0 Answers 0