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 ?
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.
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.