Proving linear transformation theorem with two vector spaces I am currently proving a theorem where $L: \mathbb R^n \to \mathbb R^n$ and $ M:\mathbb R^n\to \mathbb R^n$ are linear transformations if:


*

*$L+M:\mathbb R^n\to \mathbb R^n$ defined by $(L+M)(v)=L(v)+M(v)$ is a linear transformation.

*$-L:\mathbb R^n \to \mathbb R^n$ defined by $(-L)(v)=-(L(v))$ is a linear transformation

*$cL:\mathbb R^n \to \mathbb R^n$ defined by $(cL)(\vec v)=c(L(\vec v))$ is a linear transformation (Where c is some scalar in the reals)

*$Z:\mathbb R^n \to \mathbb R^n$ defined by $ Z(\vec v)= \vec 0$ (where $Z=\vec 0$) is a linear transformation.
I believe I have proved the first two but am tripping up on the simple proofs for numbers 3 and 4.
3: $L+M(-\vec v) = L(-\vec v) + M (-\vec v) = -L + -M(\vec v) = -(L+M)(\vec v)$ 
4: $L+M (Z(\vec v)) = L(Z(\vec v)) + M(Z(\vec v)) = Z*L(\vec v) + Z*M(\vec v) = Z((L+M)(\vec v))$
 and then since $Z=\vec 0$ we have $\vec 0*((L+M)(\vec v))= \vec 0$? I'm not sure this gives us the zero vector though...
 A: Well if you are trying to prove that $Z$ is a linear transformation. That should be easy. 
Step 1: Show $Z$ is closed under addition. That is to say that if we have $\vec u,\; \vec v \in \mathbb R^n$ then 
$$Z( \vec u+\vec v)= Z(\vec u)+ Z(\vec v)$$
Step 2: Show $Z$ has the homogeneity property. That for $\vec v \in \mathbb R^n$ and $a \in \mathbb R$ we have,
$$Z(a\vec v)= a Z(\vec v) $$
However, I have a feeling this isn't what you're trying to do. By what it sounds you are trying to prove that linear transformations have a unique property:
that the $\vec 0$ of one space maps to $\vec 0$ of the other. 
To do this we're going to denote two different $\vec 0$'s $\vec 0_1$ for the first $\mathbb R^n$, and $\vec 0_2$ for the second $\mathbb R^n$. Even though they are the same we need to do this so it won't be so confusing.
We're already given that $L$ is a linear transformation so it has the homogeneity property. We need to make one further note here. That is $a*\vec 0= \vec 0,\; \forall a \in \mathbb R$, and namely if $a=0$.
Let look at $L(\vec 0_1)=\vec u$ for $\vec u \in \mathbb R^n$,
$$\begin{array}{ccc}
L(\vec 0_1)&=& L(0*\vec 0_1)\\
&=& 0*L(\vec 0_1)\\
&=& 0*\vec u
\end{array}$$
Now from here lets look at what $0*\vec u$ means. As
$$\vec u = \begin {bmatrix}
u_1\\
u_2\\
\vdots\\
u_n
\end{bmatrix},$$
this implies that,
$$L(\vec 0)=0*\vec u = \begin {bmatrix}
0*u_1\\
0*u_2\\
\vdots\\
0*u_n
\end{bmatrix}=\begin {bmatrix}
0\\
0\\
\vdots\\
0
\end{bmatrix}=\vec 0_2$$
as desired.
