Question: I have some questions about determining which subset is a subspace of R^3. Here are the questions:

```
a) {(x,y,z)∈ R^3 :x = 0} 
b) {(x,y,z)∈ R^3 :x + y = 0} 
c) {(x,y,z)∈ R^3 :xz = 0} 
d) {(x,y,z)∈ R^3 :y ≥ 0}
e) {(x,y,z)∈ R^3 :x = y = z}

```

I am familiar with the conditions that must be met in order for a subset to be a subspace:

1. 0 ∈ R^3
2. u+v ∈ R^3
3. ku ∈ R^3

When I tried solving these, I thought i was doing it correctly but I checked the answers and I got them wrong. Can someone walk me through any of these problems? I've tried watching videos but find myself confused. Any help would be great!Thanks.

Answer: 
Do it like an algorithm. I'll do the first, you'll do the rest. I want to analyze $$I = \{(x,y,z) \in \Bbb R^3 \ : \ x = 0\}$$

The first condition is ${\bf 0} \in I$. Is it? Well, ${\bf 0} = (0,0,0)$ has the first coordinate $x = 0$, so yes, ${\bf 0} \in I$. 

The second condition is ${\bf v},{\bf w} \in I \implies {\bf v}+{\bf w} \in I$. Now, I take two elements, ${\bf v}$ and ${\bf w}$ in $I$. I know that their first components are zero, that is, ${\bf v} = (0, v\_2, v\_3)$ and ${\bf w} = (0, w\_2, w\_3)$. Is their sum in $I$? Test it! $${\bf v} + {\bf w} = (0 + 0, v\_2+w\_2,v\_3+w\_3) = (0 , v\_2+w\_2,v\_3+w\_3)$$
Since the first component is zero, then ${\bf v} + {\bf w} \in I$.

The third condition is $k \in \Bbb R$, ${\bf v} \in I \implies k{\bf v} \in I$. Then, I take ${\bf v} \in I$. I know that it's first component is zero, that is, ${\bf v} = (0,v\_2, v\_3)$. Is $k{\bf v} \in I$? Compute it, like this:
$$k{\bf v} = k(0,v\_2,v\_3) = (k0,kv\_2, kv\_3) = (0, kv\_2, kv\_3)$$
Is its first component zero? Yes, it is, then $k{\bf v} \in I$, and hence $I \leq \Bbb R^3$. Think alike for the rest.
a) Take two vectors $u$ and $v$ from that set. $y = u+v$ satisfies $y\_x = u\_x + v\_x = 0 + 0 = 0$. Take $k \in \mathbb{R}$, the vector $k v$ satisfies $(k v)\_x = k v\_x = k 0 = 0$. Finally, the vector $(0,0,0)^T$ has $x$-component equal to $0$ and is therefore also part of the set.

Can you take it from here ?