Why is $|\lambda u|=|\lambda||u|$ My text book in linear algebra - out of the blue - claims that:
$|\lambda u|=|\lambda||u|$
Where u is a vector and $\lambda$ is a constant.
I would understand if || were used to denote absolute numbers, but in this book, || is used to denote length (so the dot product of the vector v with itself is written as $|v|^2$)
I'm positively dumbfounded. How do I prove this and why would av scalar like $\lambda$ even have a length in the first place?
EDIT:
This is used in a later section to prove that the projection of v on u equals:
$$|\frac{u \cdotp v}{|u|^2}u|=\frac{|u \cdotp v|}{|u|^2}|u|=\frac{|u \cdotp v|}{|u|}$$ so it definitly seems they're referring to lengths, and not absolute numbers. Otherwise I can't really see that working.
 A: Perhaps this makes it clear: $|\lambda u|_V=|\lambda|_{\mathbb R}\,|u|_V$.
It reads: the length of $\lambda u$ in the vector space $V$ is the product to the absolute value of $\lambda$ in $\mathbb R$ by the length of $u$ in $V$.
Using subscripts as above sometimes helps to make sense of what is going on.
For instance, if $T:V\to W$ is a linear transformation and $v_1,v_2 \in V$, then
$$
T(v_1+ v_2) = T(v_1) + T(v_2) 
$$
actually means
$$
T(v_1+_V v_2) = T(v_1) +_W T(v_2) 
$$
A: You may know about the dot product that $(\lambda u)\cdot (\mu v)=\lambda\mu (u\cdot v)$.
Hence
$$ |\lambda u|=\sqrt{\lambda u\cdot \lambda u)}=\sqrt{\lambda ^2 (u\cdot u)}=|\lambda|\sqrt{u\cdot u}=|\lambda|\,|u|.$$
A: It is part of the definition of an inner product $\langle \;,\; \rangle$ on a complex vector space $V$ that

*

*$\langle \lambda v, w\rangle = \lambda\langle v, w\rangle,$ and

*$\langle v, \lambda w\rangle = \bar{\lambda}\langle v, w\rangle,$
for all $\lambda \in \Bbb C$ and $v, w \in V$. (As usual, $\bar{\lambda}$ is the complex conjugate of $\lambda$.)
After this, a norm is defined as
$$\|v\| = \sqrt{\langle v, v\rangle}.$$
This then gives us
\begin{align}
\|\lambda v\| &= \sqrt{\langle \lambda v, \lambda v\rangle}\\
&= \sqrt{\lambda\langle v, \lambda v\rangle}\\
&= \sqrt{\lambda\bar{\lambda}\langle v, v\rangle}\\
&= \sqrt{|\lambda|^2\langle v, v\rangle}\\
&= |\lambda|\sqrt{\langle v, v\rangle}\\
&= |\lambda|\|v\|.
\end{align}

In case you're working over $\Bbb R$, you just have $\lambda$ everywhere instead of $\bar{\lambda}$.
