How might you identify which set/function is a vector space? I understand that for a function or a set to be considered a vector space, there are the 10 axioms or rules that it must be able to pass. My problem is that I am unable to discern how exactly we prove these things given that my book lists some weird general examples.
For instance: the set of all third- degree polynomials is not a vector space but the set of all fourth degree or less polynomials is? Is this because I can have $$f(x) = x^3$$ $$g(x) = 1 + x - x^3$$ $$f(x) + g(x) = 1 + x$$ 
which isn't in 3rd degree where as the fourth degree or less means I can have it not have to be in 4th degree?
Other curious sets I can't seem to discern or wrap my head around include
The set $${(x, y)}$$ where $$x>=0$$ and y is a real number. 
A 4x4 matrix with symmetrical ordering except the diagonal is 0, 0, 0, 1 in descending order.
And the set of all 2x2 singular matrices.
The most confusing of all is what I like to call the modified set which changes an operation or two into something like this:
Let V be a set of all positive real numbers; determine whether V is a vector space with the operations shown below
$$x + y = xy$$
$$cx = x^c$$
If anyone could help explain why these sets break whatever axiom (because I just feel like these sets fit as a vector space but my book says otherwise) it would really help me out. I just started the vector space unit in my class and I gotta say being this clueless is scary.
 A: A (real) vector space has an addition(like operation) with a zero element and additive inverses, and also have a multiplication(like operation) by scalar (=real numbers).
The first thing to check is the zero element. If the addition is the usual one (either among numbers, $n$-tuples, matrices, polynomials or functions), then the zero must be indeed (constant) $0$. 
This already rules out the set of symmetric $4\times 4$ matrices with diagonal $(1,0,0,0)$ (unless the addition is modified).
Next thing to verify is closure under the given addition/subtraction operation. 
The set of polynomials of degree exactly $n$, is not closed under addition, as you correctly pointed out.
The set of $(x, y)$ with $x\ge0$ is not closed under subtraction. 
If the given operations are the usual addition and multiplication by scalar, we don't have to verify the axioms about their properties (associativity distributivity, etc.), since those are already known to be satisfied.
So, for example, the set of polynomials of degree $\le n$ is closed under the usual addition and scalar multiplication, and that's enough to state that's a vector space. In fact, in such a situation it is indeed a subspace of an already known vector space, now the ambient space consists of all polynomials.
Finally, if the operations are not usual, but the given set will be a vector space w.r.t. them, then there must be an isomorphism to an already known vector space. 
The given example $U$ with $x\hat+y:=xy$ and $c\hat\cdot x:=x^c$ is isomorphic to $\Bbb R$ with the usual operations, by $x\mapsto e^x, \ \Bbb R\to U$. 
[The exponential map takes addition to multiplication, its inverse is the logarithm.]
If you can't spot such an isomorphism, then you have to check the axioms one by one (which is easy in this example). 
A: Without too much precision, a real vector space $V$ is just a set endowed with two operations. The first operation is often denoted by $+$ and just turns $V,+$ into a commutative group. The second operation, the scalar multiplication $\cdot$, allows you to multiply real numbers with elements in $V$. Obviously you want some silly properties to hold, that is $1\cdot v=v, (\lambda+\mu)\cdot (v+w)=\lambda\cdot v+\lambda\cdot w+\mu\cdot v+\mu \cdot w$ and so on.
The important thing of a vector space is that it is a structure on a set which allows you to take linear combinations of elements, i.e. let $v_i\in V$ for each $1\leq i\leq n$ and let $\lambda_i\in \mathbb{R}$ for each $1\leq i\leq n$, then the element $\sum_{i=1}^n\lambda_i\cdot v_i$ of $V$ is defined.
Now consider the set $W=\left\{(x,y)\in \mathbb{R}^2\mid x\geq 0\right\}$. Obviously this is a subset of the vector space $\mathbb{R}^2$ (and when I say vector space I imply a sum and scalar product structure on it). The question is whether $W$ is itself a vector space for the same operations. Note that $(1,0)\in W$ but $(-1)\cdot (1,0)=(-1,0)\notin W$. Hence $W$ is not closed under linear combinations and hence it's not a vector space!
Try to figure out whether the given examples survive the test.
