Linear Mapping Proof for Combination of Zero and Identity Maps So I am mostly finished a proof on mapping some basis vectors and have created my own linear map, $L$, which takes vectors $v_1 \ldots v_k$ and maps them to 0 and also takes vectors $v_{k+1} \ldots  v_n$ and maps them back to themselves. I just need to be sure that the map I have created is linear.
I know that the zero map (which is vectors 1 to k) is linear and I know that the identity mapping (which is the vectors k+1 to n) is linear
I'm just not sure how to go about proving this for the combination of the two.
If I say L(sx + ty) for x, y in the set how do I know if L(x) is mapping to x or zero in my proof?
Thank you!
 A: I don't think you'll be able to prove the linearity of this particular linear map $L$ as the composition of the zero transformation $Z$ and the identity transformation. Since if $L = Z \cdot F$ or $L = F \cdot Z$, then $L$ will be the zero transformation for any linear transformation $F$
Also note that you've only defined $L$ on the basis set of your vector space. The linearity of the map $L$ depends on how you define $L$ on the rest of your vector space. Now for an arbitrary element $x$ in your vector space
$$
{\bf x} = \sum_{i = 0}^{n}a_iv_i
$$
Since,
$$ 
L = 
\begin{cases}
    \begin{align}
        v_1 & \to 0 \\
        \vdots\\
        v_k & \to 0 \\
        v_{k+1} & \to v_{k+1}\\
        \vdots \\
        v_n & \to v_n
    \end{align}
\end{cases}
$$
if you define
$$L(x) = \sum_{i = 0}^{n}a_iL(v_i) = \sum_{i = k+1 }^{n}a_iv_i$$
Then the linearity of the map will follow.
To Prove this, if
$${\bf x} = \sum_{i = 0}^{n}a_iv_i \quad \text{and} \quad {\bf y} = \sum_{i = 0}^{n}b_iv_i$$
then $$p{\bf x} + q{\bf y} = \sum_{i = 0}^{n}(pa_i + qb_i)v_i$$
now, $$L(p{\bf x} + q{\bf y}) = \sum_{i = 0}^{n}(pa_i + qb_i)L(v_i) = p\sum_{i = 0}^{n}a_iL(v_i) + q\sum_{i = 0}^{n}b_iL(v_i) = pL({\bf x}) + qL({\bf y})$$
Hence $L$ will be linear
