Why is $\max \{ | a_{i,j}| : 1 \le i \le m, 1 \le j \le n \} \le \sup \{||Ax|| : ||x|| = 1\}$?

Question

This is an inequality on the operator norm, but I cannot understand why it's true.

Why is $\max\{|a_{i,j}| : 1 \le i \le m, 1 \le j \le n\} \le \sup\{||Ax|| : ||x|| = 1\}$?

That is, why is the absolute value of the largest element of $A$ lesser than or equal to the square root of the largest eigenvalue of $A$?

Answer 1

For any $i_0$, choosing the vector $e_{i_0}$, which is admissible as its norm is $1$, the right hand side is $$\|Ae_{i_0}\|=\sqrt{\sum_j (a_{i_0j})^2}\geq\sqrt{a_{i_0j_0}^2}=|a_{i_0j_0}|,\ \forall\ j_0$$ where the middle equality follows from the monotonicity of the square root.