Question: ![enter image description here](https://i.stack.imgur.com/PiEJY.png)

![enter image description here](https://i.stack.imgur.com/Azlqy.png)

Here is the following question and answer for the question.
I don't seem to quite grasp the answers or how the answers are what they are.

Requirements for subspace:

* the zero vector is in the subset
* if you add 2 elements of the subset they remain in the subset
* if you multiply any element in the subset by a constant than this is in the subset

a) Since b1 = b2, doesn't that mean that there are effectively only two vectors that define the subspace -> meaning it creates a plane in R3. Since there are no limitations all three conditions are met. Anything I missed?

b) The plane of vectors? What does that even mean. Since there is only one vector does this not only span a line?

c) All three vectors, b1, b2 and b3 must be 0 if they are to result in 0. That fulfills the zero vector requirement. b1+b2 = 0 and b1b3 = 0 (zero vector), is this not in R3? 

d) Scalar multiplier of 0 fulfills the zero vector requirement. Then I'm lost.

e) Isn't this an example of linearly independent set of vectors?

f) Lost completely.

Anyone give a light note of what they all mean! I'm lost in what they mean basically.

Answer: 
I show the steps for the (a). 
Given the plane $(b\_1,b\_2,b\_3)$ with $b\_1=b\_2$

First, zero vector is contained in the set since $b\_1=b\_2=0$

Next, suppose there are two vectors fulfilling the equation of plane given by
$$u=(x\_1,x\_2,x\_3) \text{ and } v=(y\_1,y\_2,y\_3) $$
such that $x\_1=x\_2$ and $y\_1=y\_2$

Now, 
$$u+v=(x\_1+y\_1,x\_2+y\_2,x\_3+y\_3)$$
$u+v$ is contained in the set since $x\_1+y\_1=x\_2+y\_2$

Next,
suppose $r\in F$
$$ru=(rx\_1,rx\_2,rx\_3)$$
$ru$ is contained in the set since $rx\_1=rx\_2$

Since these three conditions is satisfied, so the plane is subspace.
(a) Consider the plane "z = 1" in $R^3$. It's not a subspace, but your argument ("there are no limitations") applies to it just as well as the one in the problem. So your argument cannot be sufficient. 

(b) The set of all vectors of the form $(1, p, q)$, where $p$ and $q$ are real numbers, is a plane in 3-space. (You might call it the "x = 1" plane). They're asking if it's a subspace of $R^3$. It's not, because the vector $(0,0,0)$ is not in it. 

(c) Once again your reasoning is wrong: any ONE of the three coordinates can be zero and still have $b\_1 b\_2 b\_3 = 0$. 

I worry that you may be in the wrong course, since the problems you're having are all with understanding definitions clearly, and with elementary algebra, not with linear algebra.