Show operator norm attains value under Euclidean norm in $\mathbb{R^n}$

Question

I've been trying to show that for $A \in M_{m\times n}(\mathbb{R})$, there is some $x \in \mathbb{R^n}$ with $||x|| = 1$ such that $||Ax|| = ||A||$, the operator norm of $A$.

The definition of operator norm used is: $||A|| = sup\{||Ax|| : ||x|| = 1\}$

I've tried proving this by contradiction by constructing a sequence of non-decreasing norms that converges, but I was not able to generate a contradiction.

Any help would be very appreciated!

Answer 1

One way of getting the required $x$ is to note that $$\|Ax\|^2=\langle Ax,Ax\rangle = x^TA^TAx$$ So one needs to maximize this expression subject to the condition $\|x\|=1$. Since $A^TA$ is symmetric, it follows that any such $x$ is an eigenvector of the largest eigenvalue of $A^TA$.