Find the basis vectors of space $$\{x\in R^5: x_1+x_3=x_2+2x_4=x_5\}$$
Well so I got equations $x_1+x_3=x_2+2x_4$ and $x_1+x_3=x_5$
So $x_5=x_1+x_3$ and $x_4=\frac{1}{2} x_1-\frac{1}{2} x_2+\frac{1}{2} x_3$
I used canonical basis and got basis $(1,0,0,\frac{1}{2},1), (0,1,0,-\frac{1}{2},0), (0,0,1,\frac{1}{2},1)$
I'm pretty sure it's wrong result. I have no idea how to solve problems like this one. Any help would be appreciated.


The subspace you are looking for has two independent restrictions associated: $$ x_1-x_2+x_3-2*x_4=0\\ x_1+x_3-x_5 = 0 $$

and thus it has dimension $5-2 = 3$.

It is not too complicated to look for a basis "manually" by substituting.

For example, $\{(1,1,0,0,1), (0,0,2,1,2), (0,1,1,0,1)\}$ is a possible solution.

If you want a mechanical procedure to solve it, then treat the equations as a linear system with three free variables. Solve in terms of your favorite variables and substitute those to get three independent solutions.

  • $\begingroup$ Thanks. I tried your procedure, but I'm still getting dimenson = 3 :/ $\endgroup$
    – Speedding
    Dec 7 '16 at 21:23
  • 1
    $\begingroup$ @Speedding Oops I made a silly mistake. the restrictions I wrote were dependent. Now it should be fixed. $\endgroup$ Dec 7 '16 at 21:29
  • 1
    $\begingroup$ It has two independent restrictions. $\endgroup$
    – Bernard
    Dec 7 '16 at 21:29

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