In each of the following cases determine if it is a linear map or not? In each of the following justify why it is a linear map or not :
Let $f:V\rightarrow W$ be a linear map. 
(a)
$f\left( u,v\right) =(u-v,vu), W=V=\mathbb{R}^{2}$
(b)
$f\left(u,v,w\right)=(u+w,3w), V=\mathbb{R}^{3}, W=\mathbb{R}^{2}$
(c)
$f\left( u,v\right)=(u-v,u+v,1), V=\mathbb{R}^{2}, W=\mathbb{R}^{3}$
I understand that in order for them to be linear maps they need to respect scalar multiplication and vector addition. However how do we sow that they do not hold ?
 A: If $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ is a linear map, then for any $\vec{v},\vec{w}$ in $\mathbb{R}^n$, $f(\vec{v}+\vec{w})=f(\vec{v})+f(\vec{w})$, and $f(\lambda \vec{v})=\lambda f(\vec{v})$.


*

*part(a)


Let $\vec{u}=(r,s), \vec{v}=(a,b)$.
$f(\vec{u}+\vec{v})=f(r+a,s+b)=(r+a-s-b,(r+a)(s+b))$
$f(\vec{u})+f(\vec{v})=f(r,s)+f(a,b)=(r-a+t-b,rs+ab)$
$f(\vec{u}+\vec{v})\neq f(\vec{u})+f(\vec{v})$
Therefore this is not a linear map.


*

*part(b)


Let $\vec{u}=(r,s,t), \vec{v}=(a,b,c),\vec{w}=(o,p,q)$.
$f(\vec{u}+\vec{v}+\vec{w})=f(r+a+o,s+b+p,t+c+q)=(r+a+o+t+c+q,3(t+c+q))$
$f(\vec{u})+f(\vec{v})+f(\vec{w})=f(r,s,t)+f(a,b,c)+f(o,p,q)=(r+t,3t)+(a+c,3c)+(o+q,3q)=(r+a+o+t+c+q,3t+3c+3q)$
$f(\vec{u}+\vec{v}+\vec{w})=f(\vec{u})+f(\vec{v})+f(\vec{w})$
$f(\lambda\vec{u})=f(\lambda r,\lambda s,\lambda t)=(\lambda r+\lambda t,3\lambda t)$
$\lambda f(\vec{u})=\lambda f(r,s,t)=\lambda (r+t,3t)=(\lambda (r+t),3\lambda t)$
$f(\lambda \vec{u})=\lambda f(\vec{u})$
Therefore this is a linear map.


*

*part(c)


Let $\vec{u}=(r,s),\vec{v}=(a,b)$
$f(\vec{u}+\vec{v})=f(r+a,s+b)=(r+a-s-b,r+a+s+b,1)$
$f(\vec{u})+f(\vec{v})=f(r,s)+f(a,b)=(r-s,r+s,1)+(a-b,a+b,1)=(r-s+a-b,r+s+a+b,2)$
$f(\vec{u}+\vec{v})\neq f(\vec{u})+f(\vec{v})$.
Therefore this is not a linear map.
A: To show that a map is not linear, you can show that the map does not satisfy a property that all linear maps must satisfy (for instance, that f(0) = 0), or show by example that the map does not preserve vector addition or scalar multiplication (it suffices to show that one is not preserved). 
For instance, in the case of your third function, it does not satisfy f(0) = 0, so it is not linear. 
That strategy does not work for the first function, since f(0) = 0, however it is easy to show by example that it does not preserve scalar multiplication. If it were linear then it must satisfy $f(2,2) = 2f(1,1)$, but it easy to check that this is not the case. 
The second function is linear. 
A: To try and prove that functions are nonlinear, you try and prove they are linear and see that something goes wrong. 
i.e. that $f(c\vec{x})\ne cf(\vec{x})$, as it should. Similarly for additivity. 
As an exercise, try and prove that $f(x)=x^2$ is linear, and see what happens. Then try and connect it to the graph of the function, what happens to the output when you translate the argument by a constant? What about when you multiply the argument by some scalar? 
A: In each case we check the criteria
$a(u,v)=(au,av) to (au-av,a^2uv)=a(u-v,auv)$ not equal $ a(u-v,uv)$ So this is not a linear transform.
$a(u,v,w)=(au,av,aw) to (au+aw,3aw)=a(u+w,3w)$ so the first condition holds. $(x,y,z)+(u,v,w)=(x+u,y+v,w+z) to (x+z,3z)+(u+w,3w)=(x+u+w+z,3w+3z)$ so the second condition holds.
$(au,av)to (au-av,au+av,1)$ not equal $a(u-v,u+v,1)$ so this is not a linear mapping. There's a shortcut for this one namely that any linear mapping maps the zero vector to the zero vector because of scalar multiplication by 0. $(u-v,u+v,1)$ is never equal to the zero vector so this is not a linear mapping.
