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, \(\displaystyle c(x,y) \notin W\)