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 $G$?

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: $\| G \| = \sigma_1(G)$

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 $G$?

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 $G$?