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Verify whether the following set is a subspace of the vector space taken into consideration:

$\{(x,y,z) \mid x=y+2z\}$, in $\mathbb{R}^3$ over $\mathbb{R}$.

Is my solution ok?

I'm checking if it can be zero vector: $x-y-2z=0$ so if $x=0$, $y=0$ and $z=0$ it is ok.

Vector addition: It is ok for $(6,2,2)$. Is it enough?

Now I'm checking scalar multiplication: For example $-1(6,2,2)$ it's still ok? $-6=-2-2\cdot2$

Do I need to prove it in some other way?

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    Any time you try to prove that it is okay, it must be okay **FOR ALL** of the **infinitely many** possible vector additions and scalar multiplications. Just checking finitely many is not good enough, you would need to check all infinitely many of them. To do so, don't use specific numbers, use a *general* argument. – 2017-01-26
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    You need to check that the set is closed under addition and scalar multiplication, so you need to check that for **all** vectors – 2017-01-26
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    So I need to create something like this: (y+2z,y,z) or I need to create some function? – 2017-01-26
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    Is there any way of proving it for every subspace? Like doing some common steps. or I have to find some rule, which is different for every subspace? @JMoravitz – 2017-01-26
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    Suppose that $u\in V$ and $v\in V$ where $V$ is the set we wish to prove is a vector space, and suppose that $\alpha\in K$ and $\beta\in K$ where $K$ is our scalar field over our set $V$. Assuming all of the standard commutativity, associativity, and distributivity rules hold to begin with (*usually standard to be given all of these for free*), all you need to check is whether or not $\alpha\cdot u+\beta\cdot v\in V$ using a *general* argument that doesn't specify anything more about $u$ and $v$ other than the fact that they are elements in your set $V$. – 2017-01-26
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    So, suppose that $(x,y,z)\in V$ and that $(x',y',z')\in V$. That means in your specific case that $x=y+2z$ and that $x'=y'+2z'$. Is $\alpha(x,y,z)+\beta(x',y',z')\in V$? I.e. is $(\alpha x+ \beta x', \alpha y+\beta y', \alpha z + \beta z')\in V$? I.e. is $(\alpha x + \beta x') = (\alpha y + \beta y') + 2(\alpha z + \beta z')$? – 2017-01-26
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    It seems to be true, for all 3 conditions, but Is it enough to write what you wrote in last comment. And how can I know if it's ok? – 2017-01-26
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    The single condition $((u,v,\in V,~\alpha,\beta\in K)\implies (\alpha u+\beta v\in V))$ is equivalent to the three conditions $(0\in V),(u,v\in V\implies u+v\in V),(u\in V,~\alpha \in K\implies \alpha u\in V)$. Instead of checking all three individually, you may check just the one since all three are true if and only if the one is true and vice versa. If you have any doubts about that, then I encourage you to try to prove that they are indeed equivalent yourself as extra exercise. – 2017-01-26
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    I'm not sure. If I write ((u,v,∈V, α,β∈K)⟹(αu+βv∈V)) and than α(x,y,z)+β(x',y',z')∈V, now I have to get my subspace: x=y+2z. How can I check above condition for my subspace? – 2017-01-26
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    This (αx+βx′,αy+βy′,αz+βz′)∈V is in V since V∈R^3 and what about (αx+βx′)=(αy+βy′)+2(αz+βz′), how can I prove it and say it's ok? Is building such equation enough? – 2017-01-26

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I am mostly just repeating what JMoravitz has said in the comments, but I hope that the extra length allowed in a full answer will help clarify the issue:

First, let's put a label on that set, so we can reference it more easily:

Let $V = \{(x, y, z) \in \Bbb R^3 \mid x = y + 2 z\}$. To show that $V$ is a subspace of $\Bbb R^3$, we need to show that three things are true:

  1. $\mathbf 0 \ = (0,0,0) \in V$

  2. For every $\mathbf v_1, \mathbf v_2 \in V$ we also have $\mathbf v_1 + \mathbf v_2 \in V$

  3. For every $\mathbf v \in V$ and $\alpha \in \Bbb R$, we also have $\alpha\mathbf v \in V$.

As JMoravitz as pointed out, these three conditions together are equivalent to showing that for all $\mathbf v_1, \mathbf v_2 \in V$ and $\alpha, \beta \in \Bbb R$, we also have $\alpha v_1 + \beta v_2 \in V$. But since you are a novice at this, I think it might be wiser to stick with the three invididual conditions for now.

So, let us start with (1). How do you show that $\mathbf 0 \in V$? Well, by definition $(x, y, z) \in V$ if and only if $x = y + 2z$. So to show that $\mathbf 0 = (0,0,0) \in V$, we just have to note that $(0) = (0) + 2(0)$.

For (2), I am not sure what you mean by "it is okay for $(6,2,2)$". Vector addition is about the sum of two vectors, but you have only given one. So you cannot have shown that "it is okay" for that vector. But in any case, to show the $V$ is a vector space, you must show that for EVERY pair of vectors in $V$ that their sum is also in $V$. One case does not prove this.

So instead, we start with two arbitrary vectors $\mathbf v_1 = (x_1, y_1, z_1)$ and $\mathbf v_2 = (x_2, y_2, z_2)$ in $V$. We must show that $\mathbf v_1 + \mathbf v_2 \in V$. Since $\mathbf v_1, \mathbf v_2 \in V$, we know that $$x_1 = y_1 + 2z_1\\x_2 = y_2 + 2z_2$$ Now, $\mathbf v_1 + \mathbf v_2 = (x_1 + x_2, y_1 + y_2, z_1 + z_2)$. To show that it is in $V$, we must show that $$(x_1+x_2) = (y_1+y_2) + 2(z_1+z_2)$$

Can you figure out how to use the facts that $$x_1 = y_1 + 2z_1$$ and $$x_2 = y_2 + 2z_2$$ to prove $$(x_1+x_2) = (y_1+y_2) + 2(z_1+z_2)\text{?}$$

For (3), the approach is similar. Let $\mathbf v = (x, y, z) \in V$. Then we know that $x = y + 2z$ Now for $\alpha \in \Bbb R$, by definition, $\alpha \mathbf v = (\alpha x, \alpha y, \alpha z)$. To prove (3), you must use the fact that $$x = y + 2z$$ to prove that $$(\alpha x) = (\alpha y) + 2(\alpha z)$$