I am new with vector space and subspace. The question asks "is the set of all vectors (x,y) where $\displaystyle 0 \leq x \leq 1$, $\displaystyle 0 \leq y \leq 1$ is in subspace $\displaystyle \mathbb{R}^2$ So there's 10 properties a "candidate" must satisfy to be classified as a subspace. These properties are listed alpha a-d beta e-h. I don't know if alpha and beta have some sort of special significance. In the book I'm reading there are two knew operators: vector addition and scalar multiplication. These confuse me because they appear to act in the same way as regular addition and multiplication. Here is my attempt at the question: $\displaystyle \alpha)$ $\displaystyle \mathbf{u}+\mathbf{v}=(a_1+a_2,b_1+b_2)$

this doesn't hold because (1,1)+(1,1)=(2,2) which is outside of the vector space

a) $\displaystyle \mathbf{u}+\mathbf{v}=(a_1,b_1)+(a_2,b_2)=(a_1+a_2 ,b_1+b_2)=(a_2+a_1,b_2+b_1)=\mathbf{v}+\mathbf{u}$ so this holds

b) $\displaystyle \mathbf{u}+(\mathbf{v}+\mathbf{w}) = (a_1,b_1)+[(a_2,b_2)+(a_3+b_3)]=(a_1+a_2,b_1+b_2)+(a_3,b_3)=(\mathbf{u}+\mathbf{v })+\mathbf{w}$ this holds

c) $\displaystyle \mathbf{0}+\mathbf{u}=\mathbf{u}$ I guess this holds, I don't know how it could not.

d) *property: for each ***u** in V, there is an element **-u** in V such that **u** (Thinking) **-u** = **0**where (Thinking) denotes vector addition

so I think this property does not hold true take for example u = 1. How do I say this in a more formal way?

$\displaystyle \beta)$ If **u** is any element of V and c is any real number then c scalar multiplication with **u** is in v.

I said this does not hold if c is outside of [0,1]. Or is the idea that c has to be in [0,1] to start with?

e)

$\displaystyle c \cdot (\mathbf{u}+\mathbf{v}=c \cdot [(a_1,b_2)+(a_2,b_2)]=c \cdot (a_1+a_2,b_1+b_2)$

I have a simillar problem, is c>1 or c<0 then it doesn't hold.

f) g) and h) I have similar problem

So since one property doesn't hold the answer to the question is no.

I'm trying really hard here, can someone help me correct anymistakes I have?