Operator norm $ ( \ell_2 \to \ell_1)$ Let $X$ be a ﬁnite dimensional normed vector space and $Y$ an arbitrary normed vector space. $ T:X→Y$.
I want to  calculate $\|T\|$ for where $X = K^n$, equipped with the Euclidean norm $\|\cdot\|_2$, $Y := \ell_1(\mathbb{N})$ and $Tx := (x_1,\ldots,x_n,0,0,\ldots) \in \ell_1(\mathbb{N})$, for all $x = (x_1,\ldots,x_n) \in K^n$.
I do not know how to continue
$$ ||T∥_2 = \sup \limits_{x \neq 0} \frac{∥Tx∥_1}{∥x∥_2} =  \sup \limits_{x \neq 0} \frac{∥(x_1,…,x_n,0,0,…)∥_1}{∥(x_1,…,x_n)∥_2} = \sup \limits_{x \neq 0} \frac{|x_1|+…+|x_n|}{(|x_1|^2+…+|x_n|^2)^{\frac{1}{2}}}= ? $$

Using Cauchy-Schwarz inequality, we have 
$$
\sum_{i=1}^n\left\lvert x_i\right\rvert=\sum_{i=1}^n\left\lvert x_i\right\rvert\cdot 1\leqslant \left(\sum_{i=1}^nx_i^2\right)^{1/2}\left(\sum_{i=1}^n1\right)^{1/2}=\sqrt n\left(\sum_{i=1}^nx_i^2\right)^{1/2}
$$
hence $\left\lVert T\right\rVert\leqslant \sqrt n$.
For the opposite inequality, look at the case where $x_i=1$ for all $i\in\{1,\dots,n\}$.
