Let $\ K = SP \{ (-5,8,14,0),(-1,4,2,4) \} $
So based on the solutions given here and here it should be same as finding orthogonal vectors so $\ (x,y,z,t) \cdot (-5,8,14,0) = 0 \\ (x,y,z,t) \cdot (-1,4,2,4) = 0 $
and then I find those four vectors $\ \{ (\frac{10}{3},0,1,0),(-\frac{8}{3},0,0,1),(0,\frac{1}{3},1,0),(0,-\frac{5}{3},0,1) \} $
setting those vectors in a matrix isn't the solution though... Im not sure what am I missing?
A simple way to do this: use elementary column operations to transform the matrix $$\begin{bmatrix} -5 & -1 \\ 8 & 4\\ 14 & 2\\ 0&4 \end{bmatrix}$$ into 'reverse' echelon form: $$\begin{bmatrix}-\frac{5}{14} & -\frac{1}{14}\\\frac{4}{7} & \frac{5}{7}\\1&0\\0&1\end{bmatrix}.$$ Notice that this matrix is in the form $$\begin{bmatrix} -F \\ I \end{bmatrix},$$ and so if we multiply it by $$\begin{bmatrix} I & F \\ 0 & 0 \end{bmatrix},$$ (where the 0's in this case can be any amount of rows with zero entries) the result will be the zero matrix, which is what we want. So depending on how many rows we want in our resulting matrix we can select as many zero rows at the bottom as we want. It seems as if you were aiming for a $4 \times 4$ matrix so let's add two rows of zero...so therefore $$\begin{bmatrix} 1 & 0 & \frac{5}{14} & \frac{1}{14}\\0&1&-\frac{4}{7} & -\frac{5}{7}\\0&0&0&0\\0&0&0&0 \end{bmatrix}$$ would work. Notics that this matrix is in row reduced echelon form, and so you can left multiply it by any elementary matrix to get other matrices that would have the same null space. The theory for the above, you can find in the following article: Row reduction of a matrix and $A=CaB$ by Steven L Lee and Gilbert Strang.