How do I prove that this function is a linear transformation? The question is asking me to prove that this is a linear transformation.
 
I know that for something to be a linear transformation, two conditions must hold the two properties:
L(u + v)  =  L(u) + L(v)
L(cu)  =  cL(u)
Other than that, I'm not sure how to solve this.
 A: Simply following the hint:
$$
|f(cu)-cf(u)|^2 = (f(cu)-cf(u))\cdot(f(cu)-cf(u))=\\
=f(cu)\cdot f(cu)-2cf(u)\cdot f(cu)+c^2f(u)\cdot f(u)=\\
=c^2u\cdot u-2c^2u\cdot u + c^2u\cdot u=0.
$$
Therefore, $f(cu)=cf(u)$.
Edit: I was missing the part where you prove additivity. To do that, consider the length of $f(v+u)-(f(v)+f(u))$ and do the same thing.
A: To preove linearity (both properties at once), it's enough to show that for $u,v\in {\mathbb R}^n$ and $c\in {\mathbb R}$, $f(u+cv) = f(u)+c f(v)$. Let's do this. 
Letting $a= f(u+cv)$ and $b= f(u)+cf(v)$, we have 
$$(b-a) \cdot (b-a) = b\cdot b + a \cdot a -2 a \cdot b.$$ 
Now by assumption $a\cdot a = (u+cv)\cdot (u+cv)$, and 
$$ \begin{align*} b\cdot b &= f(u) \cdot f(u) + cf(v) \cdot cf (v) + 2 f(u) \cdot cf(v) = u \cdot u + cv \cdot cv + 2 u \cdot cv\\
& = (u+cv)\cdot (u+cv).\end{align*}$$ 
Therefore $a\cdot a + b \cdot b = 2 (u+cv)\cdot (u+cv)$. 
Finally, the same argument shows that 
$a \cdot b= f(u+cv) \cdot (f(u) + cf(v))= (u+cv) \cdot u + (u+cv) \cdot cv =(u+cv) \cdot (u+cv)$,
and we're done. 
