# Linear Subspace/vector subspace

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?

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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 $\alpha\vec v\in V$?

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• 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.
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