How to decompose $X_p = X_r +X_n$? From an original matrix $A$, I have solved for $X_p$, $C(A^T)$, and $N(A)$. The last step is the problem below. 
How should I approach this? Thank you for taking the time. 
Decompose $X_p = X_r + X_n$, where $X_r \in C(A^T)$ and $X_n \in N(A)$.
$$
X_p =    \begin{pmatrix}
    3 \\
    4 \\
    -2 \\
    0 \\
    \end{pmatrix}, \quad
C(A^T) = \text{span}\left[\begin{pmatrix}16\\2\\3\\13\end{pmatrix},\begin{pmatrix}5\\11\\10\\8\end{pmatrix},\begin{pmatrix}9\\7\\6\\12\end{pmatrix}\right], \quad
N(A) = \begin{bmatrix}-1\\-3\\3\\1\end{bmatrix}$$
 A: I'm not sure where is your doubt. Why one can always do that decomposition, or how can one do it?
First. Given a real matrix $A^{n\times m}$, let $S \subset \mathbb{R}^m $ be its row space (equivalently, in your notation, $S=C(A^t)$) and let $T \subset \mathbb{R}^m $ be the null space of $A$.
By the  Rank–nullity theorm, $\dim(S)+\dim(T)=m$. 
Also, it's easy to see that the spaces are orthogonal: ${\bf s} \in S, {\bf t} \in T \implies {\bf s} \perp {\bf t} $.
Then, $S$ and $T$ together span $\mathbb{R}^m$.
Second. Given two orthogonal spaces $S,T$ that together span $\mathbb{R}^m$ , and given some ${\bf u} \in \mathbb{R}^m$, how can we decompose ${\bf u}={\bf s}+{\bf t}$ for some ${\bf s}\in S, {\bf t}\in T$ ?
This should be easy. Find some base for $S$ : $\{{\bf s_1},{\bf s_2} \cdots\}$ and one for $T$ $\{{\bf t_1},{\bf t_2} \cdots\}$; together they form a base for $\mathbb{R}^m$. 
Then decompose ${\bf u}$ in that base
${\bf u}=\alpha_1 {\bf s_1} + \alpha_2 {\bf s_2} + \cdots + \beta_1 {\bf t_1} + \beta_2 {\bf t_2} + \cdots$. 
Then simply take ${\bf s}=\alpha_1 {\bf s_1} + \alpha_2 {\bf s_2} + \cdots $ , ${\bf t}=\beta_1 {\bf t_1} + \beta_2 {\bf t_2} + \cdots$ and you're done.
A: Since I am breaking my Xp into orthogonal projections, I would just need to solve for my scalars at this point (X below).AX=B solving for X. So my final solution would be my C(AT) and N(A) multiplied by my scalar (x).
A = \begin{pmatrix}
    16,5,9,-1 \\
    2,11,7,-3 \\
    3,10,6,3 \\
    13,8,12,1 \\
    \end{pmatrix}
B = \begin{pmatrix}
    3 \\
    4 \\
   -2 \\
    0 \\
    \end{pmatrix}
X = \begin{pmatrix}
    159/85 \\
    1909/1360 \\
   -3853/1360 \\
    -31/20 \\
    \end{pmatrix}
