For symmetric $A$, show that $\lVert A \rVert_2 \leq \lVert A \rVert_\infty$

Question

Let $A$ be a symmetric, real matrix. Show that the following inequality holds true:
$$
\lVert A \rVert_2 \leq \lVert A \rVert_\infty
$$
where:
$$\lVert A\rVert_2 = \sqrt{\rho(A^tA)}$$
$$\lVert A\rVert_\infty = \max_i\sum_{j = 1}^n \lvert a_{ij} \rvert$$

I personally was stuck at this proof, since I am unsure what to make of the 2-Norm and its definition. My try so far is to try to simplify it:
$$\lVert A \rVert_2 = \sqrt{\rho(A^tA)} = \sqrt{\rho(A\cdot A)} = \sqrt{\lambda_{\max}(A^2)} = \lvert\lambda_{\max}(A)\rvert$$
but I am stuck there.

Answer 1

As you have said: if $A$ is symmetric, then $\|A\|_2 = \rho(A)$. On the other hand, for any (sub-multiplicative) matrix norm $\|\cdot\|$, we have $\rho(A) \leq \|A\|$.