prove if it is a subspace of a space

Question

So I have two questions about this two examples

1. a) Let $a$ be fixed vector of real vector space $E$, and $b$ fixed integer.

   is $L = \lbrace (x\in E:\langle x,a \rangle = b$ subspace of $E$

   b) Is it $W\lbrace(a,b,c,d):a+b=c+d\rbrace$ subspace of $R^4$

................................................................................................................................................................. I know that by definition i need to prove this:

1.$W \neq \emptyset$

2.$\forall x,y \in W \rightarrow x+y\in W$

3.$\forall x \in W, \forall c \in R \rightarrow cx \in W$

First example(a) i dont know even how to start.

For second example(b) im not sure if this is a correct way

1.trivial

2. $x=(a_{1},b_{1},c_{1},d_{1}) :a_{1}+b_{1}=c_{1}+d_{1}$

   $y=(a_{2},b_{2},c_{2},d_{2}) :a_{2}+b_{2}=c_{2}+d_{2}$

$x+y = (a_{1},b_{1},c_{1},d_{1}) + (a_{2},b_{2},c_{2},d_{2}) = (a_{1}+a_{2}, b_{1}+b_{2}, c_{1}+c_{2}, d_{1}+d_{2})$

$= (a_{1}+a_{2})+(b_{1}+b_{2})+(c_{1}+c_{2}) + (d_{1}+d_{2})$

My question for this now is can i do next and what after this:

$ = (a_{1}+b_{1}) + (a_{2}+b_{2})+(c_{1}+c_{2})+(d_{1}+d_{2})$

and by replacing $a_{1}+b_{1}=c_{1}+d_{1}$ do i get something what i dont see?

Answer 1

For the first let check property "$1$", that is $\vec 0\in W$

• for $x=0$ then $\langle 0,a \rangle = 0 \implies b=0$

then check for the others properties

• $\langle x_1,a \rangle=0,\langle x_2,a \rangle=0 \implies \langle x_1+x_2,a \rangle=?$

• $\langle x,a \rangle=0 \implies \langle cx,a \rangle=?$