<s>[INST] Find base in F's nullspace Question: 
$F:R^4 \to R^3$  is gotten by 
$$F((x_1,x_2,x_3,x_4)) = (5x_1+11x_2-13x_3+17x_4, 2x_1+8x_2-16x_3+14x_4, 3x_1+6x_2-6x_3+9x_4)$$
Fina a base in F range-space: V(F) and a base in F null-space: N(F). Also find the dimensions of the space. Is it correct to rewrite it into:
$$
F(x)=(e_1,e_2,e_3)
\begin{pmatrix}5 & 11 & -13 & 17 \\ 2 & 8 & -16 & 14 \\ 3 & 6 & -6 & 9 \\\end{pmatrix} 
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}
$$
And then do Gaussian elimination?: 
$$
\begin{pmatrix}5 & 11 & -13 & 17 \\ 2 & 8 & -16 & 14 \\ 3 & 6 & -6 & 9 \\\end{pmatrix}
\sim
\begin{pmatrix}0 & 1 & -3 & 2 \\ 0 & 2 & -6 & 4 \\ 1 & 2 & -2 & 3 \\\end{pmatrix}
\sim
\begin{pmatrix}1 & 2 & -2 & 3 \\0 & 1 & -3 & 2\\0 & 0 & 0 & 0 \end{pmatrix}
\sim
\begin{pmatrix}1 & 0 & 4 & -1 \\0 & 1 & -3 & 2\\0 & 0 & 0 & 0 \end{pmatrix}
$$
What to do next? Thanks.
Edit1: When F(x)=0 is zero then this is the base of null-spece right?: $$(x_1,x_2,x_3,x_4) = r(-4, 3 ,1 ,0) + s(1, -2, 0, 1) $$
 [/INST] The result obtained by Gaussian elimination will let you determine the null space: write the general solution of $F(x)=0$ as a function of the secondary variables (those without pivot, here $x_3,x_4$) and write the expreion given the four components of $x$ as a matrix product: $B\cdot\binom{x_3}{x_4}$ (the bottom two rows of $B$ will form a $2\times2$  identity matrix); then the columns of $B$ form a basis of the null space. But the result of Gaussian elimination will no tell you what the range of $F$ is, since row operations do not preserve the column space. However you know that the range has dimension $2$ (the number of primary variables), so any two linearly independent columns of the original matrix will be a basis of the range.
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