Determine if the following subsets are subspaces of $\mathbb R^3$ a) (a,b,1)
b) (a,0,c)
c) (a,b,c) where $a-b=c$
d) (a,b,c) where $a\cdot b = 0$
My professor said to check three properties:
1) $0$ is there
2) Closed by addition of vectors
3) Closed by multiplication of real-valued scalar
The answers I have are:
a) No
b) Yes
c) Yes
d) No
How do I check the three properties to determine this myself?
 A: Let’s see which of the four subsets contain $\langle 0,0,0\rangle$, the zero vector.


*

*Is it possible to choose real numbers $a$ and $b$ so that $\langle a,b,1\rangle=\langle 0,0,0\rangle$? Clearly not, so the zero vector is not in this set, and this set is therefore not a subspace of $\Bbb R^3$.

*Is it possible to choose real numbers $a$ and $c$ so that $\langle a,0,c\rangle=\langle 0,0,0\rangle$? Sure: just let $a=c=0$. Thus, the zero vector is in this set, which might still turn out to be a subspace of $\Bbb R^3$.

*Is it possible to choose real numbers $a,b$, and $c$ so that $\langle a,b,c\rangle=\langle 0,0,0\rangle$ and $a-b=c$? Yes: $a=b=c=0$ does the trick. Again, this set might turn out to be a subspace of $\Bbb R^3$.

*Is it possible to choose real numbers $a,b$, and $c$ so that $\langle a,b,c\rangle=\langle 0,0,0\rangle$ and $ab=0$? Yes, $a=b=c=0$ does the trick again.
How about closure under addition?


*

*We already know that this set isn’t a subspace of $\Bbb R^3$, but let’s check closure under addition just for the practice. The vector $\langle 0,0,1\rangle$ is certainly in this set, but when you add it to itself, you get $\langle 0,0,2\rangle$, which is not in the set: this set is not closed under vector addition.

*I’ll leave this one for you; it’s pretty straightforward.

*Suppose that $\langle a,b,c\rangle$ and $\langle x,y,z\rangle$ are in this set; that means that $a-b=c$ and $x-y=z$. Now $$\langle a,b,c\rangle+\langle x,y,z\rangle=\langle a+x,b+y,c+z\rangle\;;$$ is this vector in the set? In other words, is it true that $(a+x)-(b+y)=c+z$? Just do the algebra: $$(a+x)-(b+y)=(a-b)+(x-y)=c+z\;,$$ so the answer is yes, and this set is closed under vector addition.

*HINT: Consider the vectors $\langle 0,1,0\rangle$ and $\langle 1,0,0\rangle$.
That should be enough to get you started; see if you can check for closure under scalar multiplication on your own.
A: Yes, in each case you need to determine if all three properties hold. If any one or more of the properties does not necessarily hold, then the option is not a subspace of $\mathbb R^3$.
Hints: 
For $(a)$, is there any way $(0, 0, 0) \in (a, b, 1)$? Property 1 fails. There can be no zero vector in the set.

For $(d)$, consider the two vectors $(0, b, c), (e, 0, g)$, i.e. $a = 0, f= 0$, $b \neq 0,\; e\neq 0$. 
So $ab = 0\cdot b = 0$, and $e\cdot f = e \cdot 0 = 0$. So each vector is in in the set defined by $(d)$. 
Now, we need closure under vector addition: we need to see if $(0, b, c) + (e, 0, g)$ is also in the set. Now, $(0, b, c) + (e, 0, g) = (e, b, c+g)$.  But since $b\neq 0,\;e\neq 0\;\;eb \neq 0$. Hence vector addition is not closed for vectors of the form defined by $(d)$. Property 2 fails.

Now, review the properties you know must hold for all subspaces of a vector space, and determine why $(b), (c)$ both satisfy all the properties, and hence, define a subspace of $\mathbb R^3$.
