For a given matrix $X$, find two linearly independent vectors in $C(X)^{\perp}$. Let $$X = \begin{pmatrix} 
1 & 1 & 4 \\ 
1 & 2 & 1 \\ 
1 & 3 & 0 \\ 
1 & 4 & 0 \\ 
1 & 5 & 1 \\ 
1 & 6 & 4 \end{pmatrix}$$ 
Is there an easy way to calculate the form of a vector in $C(X)^{\perp}$, $C(X)$ being the column space of $X$?
What is unfortunate is that $C(X) \neq \mathbb{R}^6$ (of course that would be too easy...). 
I mean, I could start by noticing that 
$$C(X) = \left\{\left.\begin{pmatrix}
a_1 + a_2 + 4a_3 \\
a_1 + 2a_2 + a_3 \\
a_1 + 3a_2 \\
a_1 + 4a_2 \\
a_1 + 5a_2 + a_3 \\
a_1 + 6a_2 + 4a_3
\end{pmatrix}\right\vert a_i \in \mathbb{R}\right\}$$
but this doesn't seem to be at all helpful since I would have to dot this with some fixed vector in $\mathbb{R}^4$, impose some conditions on the $a_i$ since they are arbitrary, and use those conditions to solve for the components of the fixed vector. [Unless, for some reason, this is actually how you are supposed to solve it!]
If helpful at all, I have an orthogonal basis for $X$, found here.
 A: Hint:
Do you know how to find a basis for the null space of a given matrix?  If so, it is a fact that $C(A)^\perp = N(A^T)$ for any matrix $A$.  That is, take the transpose of your matrix and find a basis for its null space, and the result will give you what you are looking for.  
For troubleshooting purposes, I have gone through some of the process below.

So if we have the matrix: $$X = \begin{pmatrix} 
1 & 1 & 4 \\ 
1 & 2 & 1 \\ 
1 & 3 & 0 \\ 
1 & 4 & 0 \\ 
1 & 5 & 1 \\ 
1 & 6 & 4 \end{pmatrix}$$ 
Then: $$X^T = \begin{pmatrix} 
1 & 1 & 1 & 1 & 1  & 1 \\
1 & 2 & 3 & 4 & 5 & 6 \\
4 & 1 & 0 & 0 & 1 & 4\end{pmatrix}$$ 
If we wish to find a basis for the null space of $X^T$, we consider the system $X^T \mathbf{x} = 0$.  The first step to finding a solution set to such homogeneous systems is, of course, to put $X^T$ into reduced row echelon form:
$$\operatorname{rref}(X^T) = \begin{pmatrix} 
1 & 0 & 0 & 1/2 & 3/2 & 7/2 \\
0 & 1 & 0 & -2 & -5 & -10 \\
0 & 0 & 1 & 5/2 & 9/2 & 15/2 \end{pmatrix}$$ 
This tells us that the components of the vector $\mathbf{x}$ satisfy the following system:
\begin{align}
& x_1 = (-1/2)x_4 - (3/2)x_5 - (7/2)x_6 \\
& x_2 = 2x_4 + 5x_5 + 10x_6 \\
& x_3 = -(5/2)x_4 - (9/2)x_5 - (15/2)x_6
\end{align}
And $x_4, x_5, x_6$ are so-called free variables.  Now there's just one more step to finish up.
