Verify whether the following set are subspaces of the vector spaces taken into consideration

Question

a){(x, y, z) | z^2 = 1} in R^3 over R b){f(x) ∈ R^R| f(2) + f(3) = 0} in R^R over R

I dont totally get those 2 :( Can somebody explain me, how to deal with it?

Answer 1

Given a set $V$ with binary operation $+$ and scalar field $K$ over $V$ with scalar multiplication $\cdot$ which satisfy associativity, commutativity, and distributivity...

$V$ is a vector space over the scalar field $K$ if and only if the following condition is satisfied:

$$u,v\in V,~\text{and}~ \alpha,\beta\in K\implies \alpha\cdot u+\beta\cdot v\in V$$

This is often instead worded as the following three conditions: $(0\in V),~~(u,v\in V\implies u+v\in V),~~(v\in V~\text{and}~\alpha\in K\implies \alpha\cdot v\in V)$

For the first question, the set of all points in $\Bbb R^3$ such that $z^2=1$

Check the first condition: is the "zero" element (in this case $(0,0,0)$) in our set? Is $(0,0,0)\in V$? I.e. does $(x,y,z)=(0,0,0)$ satisfy $z^2=1$?

No, since $(0,0,0)$ has the square of the third term equal to zero, not one, it doesn't satisfy the defining property to be in the set.

Alternatively, check the second condition. If $(x,y,z)\in V$ and $(x',y',z')\in V$ does that imply that $(x,y,z)+(x',y',z')\in V$? Try to find a counter example or give an arbitrary argument why it has to always be true.

No again, since $(0,0,1)\in V$ and so is $(0,0,1)$, but $(0,0,1)+(0,0,1)=(0,0,2)$ is not in $V$ since the square of the third entry is four, not one, and therefore it doesn't satisfy the defining property to be in the set.

Alternatively again, check the third condition. If $(x,y,z)\in V$ and $\alpha\in \Bbb R$, is $\alpha(x,y,z)=(\alpha x,\alpha y,\alpha z)\in V$? Again, find a counterexample or come up with a general argument why it has to always be true.

Again, No, since $(0,0,1)\in V$ and $5\in \Bbb R$ but $5(0,0,1)=(0,0,5)$ is not in $V$ since the third entry squared is equal to 25, not one and therefore it doesn't satisfy the defining property to be in the set.

If the answer to any of the above is no, then it is not a vector space. If you find even one no, you don't need to check any of the other properties about it.

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For the second approach similarly. Check each individual property or the combined property I already mentioned.

If $f$ and $g$ are both functions satisfying $f(2)+f(3)=0$ and $g(2)+g(3)=0$, will $\alpha\cdot f+\beta\cdot g$ be a function with the same property?

Is $(\alpha\cdot f+\beta\cdot g)(2)+(\alpha\cdot f+\beta\cdot g)(3)=0$?

$(\alpha\cdot f+\beta\cdot g)(2)+(\alpha\cdot f+\beta\cdot g)(3)=\alpha (\underbrace{f(2)+f(3)}_{\to 0})+\beta(\underbrace{g(2)+g(3)}_{\to 0})=\dots$ where we can replace those two underbraced expressions with zero because of our assumption that $f$ and $g$ are both elements in our set $V$.