Find a basis for the subspace of R4 consisting of all solutions to the equations $2x_1 −x_2 +x_3 −2x_4 =0$
$2x_1 + x_3 = 0$

Started by just subtracting eq2 from eq1, giving: $-x_2-2x_4$$=0$

set $x_4$=k
so $x_2=-2k$
Basis is: $(0,-2,0,1)$

Is there a mistake somewhere?

  • $\begingroup$ Your matrix is $\begin{pmatrix} 2 & -1 &1&-2 \\ 2&0&1&0 \end{pmatrix}$ and you are looking for the Null space the matrix can be row reduced to $\begin{pmatrix} 2 & -1 &1&-2 \\ 0&1&0&2 \end{pmatrix}$ $\endgroup$ – gbox Feb 5 '18 at 12:05
  • $\begingroup$ So the $rank=2$ and $Null=2$ So you must have $2$ vectors in the null space as showed in the answer $\endgroup$ – gbox Feb 5 '18 at 12:20

Consider that $2x_1+x_3=0$, now set $x_1=t$ and $x_3=-2t$ so $(1,0,-2,0)$ is also a basis. So the solutions are $$Span\{ (0,-2,0,1),(1,0,-2,0)\}$$


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