ASSIGNMENT 03
SOLUTION
Problem 7. Determine whether the sets U are V are subspace of R? defined by
U = \{(x, y, z, u) \in \mathbb{R}^4: x^2 - yz = u\} and
V = \{(x, y, z, v) \in \mathbb{R}^4: x = 2z \text{ and } x - 3y = v\}.
[10 marks]
Solution: (1) As u = (1, 1, 0, 1) and v = (1, 0, 1, 1) belong to U.
$\implies u + v = (2, 1, 1, 2) \notin U$ as $2^2 - (1)(1) \neq 2$.
As $u, v \in U \implies u + v \notin U$, so U is not a subspace of R?.
(2) Take v = ($x_1, y_1, z_1, v_1$) and w = ($x_2, y_2, z_2, v_2$) belong to V.
$\implies u + v = (x_1 + x_2, y_1 + y_2, z_1 + z_2, v_1 + v_2) \in V$ as $(x_1 + x_2) = 2(z_1 + z_2)$ and $(x_1 + x_2) - 3(y_1 + y_2) = (v_1 + v_2)$ (show).
Also take v = ($x_1, y_1, z_1, v_1$) belongs to U and for any scalar k,
$kv = (kx_1, ky_1, kz_1, kv_1) \in U$ since $kx_1 = 2(kz_1)$ and $kx_1 - 3(ky_1) = kv_1$.
Hence, V is a subspace of R?.
