# Spectral norm of a sparse Gaussian matrix

Suppose $G$ is an $m \times n$ matrix such that each entry of $G$ is a standard normal variable. We know that the spectral norm of $G$ scales as $\sqrt m + \sqrt n$. Now, given a set of indices $S$ suppose we construct a new matrix $A$ such that $A_{ij} = G_{ij}$ if $(i,j) \in S$, and 0 otherwise. Can we show that the spectral norm of $A$ is upper bounded by the spectral norm of $G$?

edit: The spectral norm is the largest singular value of the matrix: $\| G \| = \sigma_1(G)$

• Can you please add the definition of the spectral norm of a matrix to make the question more readable? – Xi'an Feb 5 '15 at 8:20
• Hi Xi'an , I just added the definition. – NSR Feb 5 '15 at 12:47

$$G = \begin{bmatrix}-1& 1& 1\\ 1 &1 &1\\ 1& 1& -1\end{bmatrix}, \qquad A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix}$$
The spectral norm of $G$ is 2, the spectral norm of $A$ is about 2.4.