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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?: $$(x_1,x_2,x_3,x_4) = r(-4, 3 ,1 ,0) + s(1, -2, 0, 1) $$

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  • $\begingroup$ "valueroom"?.... $\endgroup$ – DonAntonio Dec 17 '13 at 17:06
  • $\begingroup$ Do you mean find a basis for the range and null space of $F$? $\endgroup$ – copper.hat Dec 17 '13 at 17:09
  • $\begingroup$ @Jossi: Your GE result does not look correct. $\endgroup$ – Amzoti Dec 17 '13 at 17:09
  • $\begingroup$ valuerooom = range $\endgroup$ – Jossi Dec 17 '13 at 17:10
  • $\begingroup$ Yes. I corrected the GE. Can you please take a look? $\endgroup$ – Jossi Dec 17 '13 at 17:21
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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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