# Subspace

#### cdlegendary

V= W= Why isn't the given subset W of the vector space V a subspace of V? (check all that apply)

A. W is not closed under addition
B. W does not contain a zero vector
C. Not applicable: W is a subspace
D. W is not closed under multiplication by a nonzero scalar

If I'm not wrong...I think it's because

B. W does not contain a zero vector

and

D. W is not closed under multiplication by a nonzero scalar

Any help is appreciated!

#### Haven

It's A,B and D;

A) Let $$\displaystyle (x,y), (u,v) \in W$$
Assume $$\displaystyle (x,y)+(u,v) \in W$$
The $$\displaystyle (x,y) + (u,v) = (x+u,y+v)$$
Yet $$\displaystyle (x+u)^2 + (y+v)^2 = x^2 + 2xu + u^2 + y^2 + 2yv + v^2 =$$$$\displaystyle (x^2 + u^2) + (y^2 + v^2) + 2xu + 2yv = 1 +1 + 2xu + 2yv = 2 + 2xu + 2yv = 1$$
$$\displaystyle \rightarrow 1 = 2(xu + yv)$$
$$\displaystyle \rightarrow \frac{1}{2} = xu+yv$$
Which isn't satisfied by all elements of W

Example: $$\displaystyle (1,0)$$ and $$\displaystyle (0,1)$$
$$\displaystyle (1,0) + (0,1) = (1,1)$$
but $$\displaystyle 1^2 + 1^2 = 2$$ so $$\displaystyle (1,0) + (0,1) \notin W$$

B) Since W is a supposed subspace of V, the zero vector of $$\displaystyle \mathbb{R}^2$$ must be the zero vector of W. But the zero vector of $$\displaystyle \mathbb{R}^2$$, $$\displaystyle (0,0)$$ is not in W.

D) Let $$\displaystyle c \in \mathbb{R}$$ and $$\displaystyle (x,y), (u,v) \in W$$
Then $$\displaystyle c(x,y) = (cx,cy)$$ but $$\displaystyle (cx)^2 + (cy)^2 = c^2x^2 + c^2y^2 = c^2(x^2+y^2)= c^2$$
This is only true when $$\displaystyle c=\pm 1$$
So for the majority of scalars, $$\displaystyle c(x,y) \notin W$$

• HallsofIvy