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?
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.
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 $\mathbb R^3$.
