Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need to prove that this is a linear subspace:

$$x-y+2z=0$$ $$x+y+4z=0$$

How am I going to do this?

share|cite|improve this question
Did you actually mean to ask how to prove that the set of all solutions in $\,\Bbb R^3\,$ (or in any other vector space) to the given linear system of equations is a subspace? Well, prove that the sum of any two of them is again a solution, and the product of any of them by any scalar is also a solution...oh, and to avoid the case of an empty set, prove the zero vector is a solution, too. – DonAntonio Dec 18 '12 at 11:24

What you’ve written isn’t a linear subspace of anything: it’s a pair of equations. The first step is to state the problem correctly:

Let $V$ be the set of vectors $\langle x,y,z\rangle\in\Bbb R^3$ such that $x-y+2z=0$ and $x+y+4z=0$; then $V$ is a subspace of $\Bbb R^3$.

In order to show this, you must show three things:

  • $V$ is non-empty.
  • $V$ is closed under vector addition: if $\vec u,\vec v\in V$ then $\vec u+\vec v\in V$.
  • $V$ is closed under scalar multiplication: if $\vec v\in V$ and $\alpha\in\Bbb R$, then $\alpha\vec v\in V$.

(It’s possible to combine the last two into a single statement, but at this point it’s probably simplest for you to keep the ideas separate.)

Is $x=0,y=0,z=0$ a solution to both of the equations $x-y+2z=0$ and $x+y+4z=0$? Clearly it is, so $\langle 0,0,0\rangle\in V$, and $V$ is not empty.

To check that $V$ is closed under vector addition, suppose that $\vec u=\langle x_1,x_2,x_3\rangle$ and $\vec v=\langle y_1,y_2,y_3\rangle$ are in $V$; this means that $x_1-x_2+2x_3=0$ and $x_1+x_2+4x_3=0$ (since $\vec u$ is in $V$), and that $y_1-y_2+2y_3=0$ and $y_1+y_2+4y_3=0$ (since $\vec v$ is in $V$. Now

$$\vec u+\vec v=\langle x_1+y_1,x_2+y_2,x_3+y_3\rangle\;,$$

and you need to show that this vector is in $V$. To do so, you must show that

$$(x_1+y_1)-(x_2+y_2)+2(x_3+y_3)=0$$ and $$(x_1+y_1)+(x_2+y_2)+4(x_3+y_3)=0\;;$$

I’ll leave that to you.

To check that $V$ is closed under scalar multiplication, proceed similarly: let $\vec v=\langle x_1,x_2,x_3\rangle$ be any vector in $V$ and $\alpha$ any scalar. You know that $x_1-x_2+2x_3=0$ and $x_1+x_2+4x_3=0$ (why?), and you know that $\alpha\vec v=\langle \alpha x_1,\alpha x_2,\alpha x_3\rangle$; what must you prove about $\alpha x_1,\alpha x_2$, and $\alpha x_3$ in order to show that $\alpha\vec v\in V$?

share|cite|improve this answer
  • Verify that the null vector $\overrightarrow 0 = (0,0,0)$ is a solution of the system.
  • Supposing $\overrightarrow x =(x_1, x_2, x_3)$ and $\overrightarrow y =(y_1, y_2, y_3)$ are solutions to the system, verify that the linear combination of $\overrightarrow x$ and $\overrightarrow y$ is also a solution, i.e. $$ \forall \lambda \in \mathbb R, \ \ \overrightarrow x + \lambda \overrightarrow y \text{ is a solution to the system}$$
  • If both properties are verified, then you have a vector subspace in $\mathbb R^3$, if not then it is not a vector subspace.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.