Let V = $\Bbb{R}^3$. Find the subspace generated by set $A=\{v_1, v_2\}$, where $v_1=(1, -2, -1)$ and $v_2=(2, 1, 1)$.

My attempt:

$$ \left\{ \begin{array}{c} k_1 + 2k_2=x \\ -2k_1 + k_2=y \\ -k_1+k_2=z \end{array} \right. $$

It follows then that $k_1=x-2k_2$ and $k_1=\frac{y+k_2}{2}$. So, $k_2=\frac{2x+y}{5}$. Substituting the previous results in the third equation, we have $6x+3y-5z=0$ which according to the answer key should actually be $x+3y-5z=0$.

What am I missing?

  • $\begingroup$ It’s always a good idea to check your own work. For this problem, you could check your solution against $v_1$ and $v_2$ by plugging them into the equation that you derived. Neither one satisfies this equation, so you know that your solution is incorrect even without checking the answer key. $\endgroup$
    – amd
    Commented Aug 8, 2018 at 23:37

2 Answers 2


The subspace is a plane $ax+by+cz=0$ therefore

  • $a-2b-c=0$

  • $2a+b+c=0$

and adding up

  • $3a=b$
  • $a=-\frac15 c$



As an alternative by cross product

$$\vec n=\begin{vmatrix}\vec i&\vec j&\vec k\\1&-2&-1\\2&1&1\end{vmatrix}=(-1,-3,5)$$

  • $\begingroup$ @mfl Thanks I update! $\endgroup$
    – user
    Commented Aug 8, 2018 at 22:46
  • $\begingroup$ I was not aware of that approach unit now, so here are some questions. How did you conclude that the subspace is a plane that crosses the origin? Why did you change the sign of the vector when writing the constraint? $\endgroup$ Commented Aug 8, 2018 at 23:04
  • $\begingroup$ @MauricioMendes By the definition given for the subspace as a span of 2 linearly independent vectors we can directly conclude that it is a plane through the origin (indeed the origin is reached by $0v_1+0v_2$). $\endgroup$
    – user
    Commented Aug 8, 2018 at 23:08
  • $\begingroup$ @MauricioMendes Which change of sign are you referring to? $\endgroup$
    – user
    Commented Aug 8, 2018 at 23:08
  • $\begingroup$ I am referring to your alternative solution. $\endgroup$ Commented Aug 8, 2018 at 23:13

You lost a minus sign: should be $k_1=\frac{-y+k_2}2$. Then $x-2k_2=\frac{-y+k_2}2\implies k_2=\frac{2x+y}5\implies k_1=\frac{-2y+x}5\implies 3y+x=5z\implies x+3y-5z=0$...


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