my attempt
$$\lim_{p\to\infty}\|A\|_{p} = \lim_{p\to\infty} (|A_1|^p + \cdots + |A_n|^p)^{1/p}$$
$$= \lim_{p\to\infty} \left( \sum_{i=1}^{n} |x_i|^p \right)^{1/p}$$
I know the infinity norm is max row but I'm unsure how to get to that.
Turn above to
$$\|A\|_{\infty} = \max_j \sum_{i=1}^{\infty}|a_{ij}|$$
Not sure which properties to use to solve this I'm very confused. Any help is appreciated
It seems like you are confusing about operator norm and vector norm. To clarify, the p-norm of a matrix/operator is defined to be $\Vert A \Vert_{p,\text{op}}= \sup_{\Vert x \Vert_{p,\text{vec}}=1} \Vert A x \Vert_{p,\text{vec}} $.
And the limiting result holds for vector norm $\Vert \cdot \Vert_p = \Vert \cdot \Vert_{p,\text{vec}} $.
Proof:
$\Vert x \Vert_p = (\sum_i \vert x_i \vert^p)^{\frac{1}{p}} \le (\sum_i \max_i \vert x_i\vert^p )^{\frac{1}{p}} = n^{\frac{1}{p}} \max_i \vert x_i\vert \to \max_i \vert x_i \vert = \Vert x \Vert_\infty$ since $n$ is a fixed number, so $\varlimsup_{p \to \infty} \Vert x \Vert_p \le \Vert x \Vert_\infty$.
On the other hand, $\Vert x \Vert_p = (\sum_i \vert x_i \vert^p)^{\frac{1}{p}} \ge (\max_i \vert x_i \vert^p)^{\frac{1}{p}} = \max_i \vert x_i \vert = \Vert x \Vert_\infty$ So $\varliminf_{p \to \infty}\Vert x \Vert_p \ge \Vert x \Vert_\infty$.