Given the definition of vectorial norm:
it as a function such that:
$$\Vert \cdot \Vert : \mathbb{R}^n \rightarrow [0, +\infty)$$ for every vector $\mathbf{x} \in \mathbb{R}^n$ are valid the following properties:
a) $\Vert \mathbf{x} \Vert \ge 0, \quad \Vert \mathbf{x} \Vert = 0 \Leftrightarrow \mathbf{x} = 0$;
b) $ \Vert \alpha \mathbf{x} \Vert = \vert \alpha \vert \cdot \Vert \mathbf{x} \Vert, \quad \forall \alpha \in \mathbb{R}$;
c) $ \Vert \mathbf{x + y} \Vert = \Vert \mathbf{x} \Vert + \Vert \mathbf{y} \Vert, \quad \forall \mathbf{y} \in \mathbb{R}^n$
and given the definition of euclidean norm:
$$\Vert \mathbf{x} \Vert_2 = \sqrt{\mathbf{x}^T\mathbf{x}} = \sqrt{\sum_{i=1}^n x_i^2} $$
I have to prove that the propery b) is valid also for the euclidean norm : I write what my textbook says, but there are some passages I don't understand, so:
$\forall \lambda \in \mathbb{R}$ holds the following: $$\begin{array}{lcl} \Vert \lambda \mathbf{x} \Vert_2 & = & \sqrt{\sum_{i=1}^n \lambda^2 x_i^2} \\ & = & \sqrt{\lambda^2\sum_{i=1}^n x_i^2} \\ & = & \vert\lambda\vert \sqrt{\sum_{i=1}^n x_i^2} \\ & = & \vert \lambda \vert \cdot \Vert \mathbf{x} \Vert_2 \end{array}$$
but, I think there are many passages have been skipped, I have done myself:
$\lambda \mathbf{x}$ is a scalar multiplication so:
$$\begin{array}{lcl} \Vert \lambda \mathbf{x} \Vert_2 & = & \Vert \lambda \cdot (x_1, x_2, \ldots, x_n )^T\Vert_2 \\ &=& \Vert (\lambda x_1, \lambda x_2, \ldots, \lambda x_n)^T \Vert_2 \end{array} $$
but from here I can't continue, :
1) why it obtains that square root?;
2) why we have $\vert{\lambda} \vert$ with that abs value after the transport outside the square root?
Please, can you help me? Many thanks for your patience!
Hints:
For your question 1), recall what the definition of the Euclidean norm of a vector $(a_1, \ldots, a_n)^T$ is!
For 2), recall that $\sqrt{\lambda^2} = \color{red}{|}\lambda\color{red}{|}$ for any $\lambda\in \mathbb{R}$. We need the absolute values because $\sqrt{\cdot}$ refers to the non-negative square root. For example, $\sqrt{(-2)^2}= \sqrt{4} = 2 = \lvert -2 \rvert$.