Question

Let V ={(a,b,c,d) ϵ R^4!a+b+2c+2d = 0} and W = { (a;b;c;d) ϵ R^4!a = -b;c = -d}.
Check that V and W are vector spaces.
Further, check that W is a subspace of V.

Answer 1

Answer on Question #44927 – Math – Linear Algebra

Question. Let $V = \{(a, b, c, d) \in \mathbb{R}^4 : a + b + 2c + 2d = 0\}$ and $W = \{(a, b, c, d) \in \mathbb{R}^4 : a = -b, c = -d\}$. Check that $V$ and $W$ are the vector spaces. Further, check that $W$ is a subspace of $V$.

Solution. We shall prove that $V$ and $W$ are the subspaces of $\mathbb{R}^4$. We shall use the next criterion of subspace: the subset $W$ of linear space $V$ is a subspace of $V \Leftrightarrow \left\{ \begin{array}{l} \left( \bar{a} + \bar{b} \right) \in W \ \forall \bar{a}, \bar{b} \in W \\ \lambda \bar{a} \in W \ \forall \lambda \in \mathbb{R}, \forall \bar{a} \in W \end{array} \right.$. Let $\bar{a} = (a_1, b_1, c_1, d_1) \in V, \bar{b} = (a_2, b_2, c_2, d_2) \in V$. Then $\bar{a} + \bar{b} = (a_1 + a_2, b_1 + b_2, c_1 + c_2, d_1 + d_2)$.

$$\begin{array}{l} a_1 + a_2 + b_1 + b_2 + 2(c_1 + c_2) + 2(d_1 + d_2) = (a_1 + b_1 + 2c_1 + 2d_1) + (a_2 + b_2 + 2c_2 + \\ + 2d_2) = 0 + 0 = 0 \Rightarrow (\bar{a} + \bar{b}) \in V. \end{array}$$ $$\begin{array}{l} \lambda \bar{a} = (\lambda a_1, \lambda b_1, \lambda c_1, \lambda d_1). \text{ Then } \lambda a_1 + \lambda b_1 + 2\lambda c_1 + 2\lambda d_1 = \lambda (a_1 + b_1 + 2c_1 + 2d_1) = \lambda \cdot 0 = \\ = 0 \Rightarrow \lambda \bar{a} \in V. \text{ So } V \text{ is a subspace of } \mathbb{R}^4 \Rightarrow V \text{ is a vector space}. \end{array}$$

Let $\bar{a} = (a_1, -a_1, b_1, -b_1) \in W, \bar{b} = (a_2, -a_2, b_2, -b_2) \in W$. Then $\bar{a} + \bar{b} = (a_1 + a_2, -a_1 - a_2, b_1 + b_2, -b_1 - b_2)$. Obviously $(\bar{a} + \bar{b}) \in W$.

$\lambda \bar{a} = (\lambda a_1, -\lambda a_1, \lambda b_1, -\lambda b_1)$. Obviously $\lambda \bar{a} \in W$. So $W$ is a subspace of $\mathbb{R}^4 \Rightarrow W$ is a vector space.

Let $\bar{x} = (a, -a, c, -c) \in W$. Since $a + (-a) + 2c + (-2c) = 0 \Rightarrow \bar{x} \in V \Rightarrow W$ is a subspace of $V$.

Answer. $V$ and $W$ are the vector spaces, $W$ is a subspace of $V$.

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