Prove: $\|\lambda v\| = |\lambda| \cdot \|v\| $, for a vector space $V$ with an inner product and $\lambda \in F$,
How do we prove this?
I understand the geometric meaning is that if you multiply a vector by a scalar then you make it length greater by that scalar times, and even if you multiply by a minus-signed scalar then you actually lengthen it by the same $|\lambda|$ but to the other direction. But how do we algebraically prove this?
It comes nearly immediately from the definitions.
Let $\langle \cdot, \cdot \rangle$ be the inner product, which is linear in the first slot and conjugate linear in the second slot. Then \begin{align*} \|\lambda v\| &= \sqrt{ \langle \lambda v , \lambda v\rangle}\\ &= \sqrt{\lambda \cdot \langle v, \lambda v\rangle }\\ &= \sqrt{\lambda \cdot \overline{\lambda} \cdot \langle v,v\rangle}\\ &= \sqrt{ |\lambda|^2 \cdot \langle v,v\rangle}\\ &= |\lambda| \cdot \sqrt{\langle v,v\rangle}\\ &= |\lambda|\cdot \|v\| \end{align*}