I'm taking a geometry class and had a question regarding a homework problem I have. It's related to the magnitudes of vectors and proving their properties. Previously we learned about inner products and their properties (e.g. positive definiteness).
The problem is:
If we define the magnitude of a vector to be $\Vert \mathbf{v} \Vert = \langle \mathbf{v}, \mathbf{v} \rangle^{1/2}$, the magnitude has the following properties:
- $\Vert \mathbf{v} \Vert \ge 0$ for all $\mathbf{v}$, and the necessary and sufficient condition for $\Vert \mathbf{v} \Vert = 0$ is $\mathbf{v} = \mathbf{0}$.
- For all $\alpha \in \Bbb{R}$, $\Vert \alpha \mathbf{v} \Vert = \vert \alpha \vert \Vert \mathbf{v} \Vert$
There are other properties but for now I only have to prove these two.
My approach:
We know that the inner product of vector $\mathbf{v}$ (i.e. $\langle \mathbf{v}, \mathbf{v} \rangle$) is positive definite. Therefore, we can also assume that $\Vert \mathbf{v} \Vert$ is also positive definite since $\Vert \mathbf{v} \Vert = \langle \mathbf{v}, \mathbf{v} \rangle^{1/2}$.
$\Vert \alpha \mathbf{v} \Vert = \langle \alpha \mathbf{v}, \alpha \mathbf{v} \rangle^{1/2} = \left( \alpha^2 \langle \mathbf{v}, \mathbf{v} \rangle \right)^{1/2} = \vert \alpha \vert \langle \mathbf{v}, \mathbf{v} \rangle^{1/2} = \vert \alpha \vert \Vert \mathbf{v} \Vert$
Are my approaches correct? I'm especially a bit dubious about the way I proved property #1 because it seems quite hand-wavy.
Any tips or feedback are appreciated!