Question

Show that the set W = {(x 1 , x 2, x3, x4):
x1 + x2 + x3 + x4 = 0) is a subspace of R4.
Verify that (1, —1, 1, —1) and (1, 0, 0, —1)
are in W. Find a basis of W containing
(1,-1, 1, —1) and (1, 0, 0, —1).

Answer 1

Show that the set

$$W = \{(x _ {1}, x _ {2}, x _ {3}, x _ {4}): x _ {1} + x _ {2} + x _ {3} + x _ {4} = 0 \}$$

is a subspace of $\mathbb{R}^4$ . Verify that $(1, -1, 1, -1)$ and $(1, 0, 0, -1)$ are in $W$ . Find a basis of $W$ containing $(1, -1, 1, -1)$ and $(1, 0, 0, -1)$ .

Solution.

We need to prove that $\forall x, y \in W, \forall \alpha, \beta \in \mathbb{R} : \alpha x + \beta y \in W$ .

$$\begin{array}{l} x = \left(x _ {1}, x _ {2}, x _ {3}, x _ {4}\right), y = \left(y _ {1}, y _ {2}, y _ {3}, y _ {4}\right) \Rightarrow \\ \Rightarrow \alpha x + \beta y = \left(\alpha x _ {1} + \beta y _ {1}, \alpha x _ {2} + \beta y _ {2}, \alpha x _ {3} + \beta y _ {3}, \alpha x _ {4} + \beta y _ {4}\right); \\ \end{array}$$

Assume that $x, y \in W$ . Hence:

$$\begin{array}{l} \alpha x _ {1} + \beta y _ {1} + \alpha x _ {2} + \beta y _ {2} + \alpha x _ {3} + \beta y _ {3} + \alpha x _ {4} + \beta y _ {4} = \\ = \alpha \left(x _ {1} + x _ {2} + x _ {3} + x _ {4}\right) + \beta \left(y _ {1} + y _ {2} + y _ {3} + y _ {4}\right) = \alpha \cdot 0 + \beta \cdot 0 = 0 \Rightarrow \alpha x + \beta y \in W. \\ \end{array}$$ $$\begin{array}{l} \left(x _ {1}, x _ {2}, x _ {3}, x _ {4}\right) = (1, - 1, 1, - 1) \Rightarrow x _ {1} + x _ {2} + x _ {3} + x _ {4} = 1 - 1 + 1 - 1 = 0 \Rightarrow \\ \Rightarrow (1, - 1, 1, - 1) \in W. \\ \end{array}$$ $$\begin{array}{l} \left(x _ {1}, x _ {2}, x _ {3}, x _ {4}\right) = (1, 0, 0, - 1) \Rightarrow x _ {1} + x _ {2} + x _ {3} + x _ {4} = 1 + 0 + 0 - 1 = 0 \Rightarrow \\ \Rightarrow (1, 0, 0, - 1) \in W. \\ \end{array}$$

Consider the following vectors:

$$e _ {1} = (1, - 1, 1, - 1), e _ {2} = (1, 0, 0, - 1), e _ {3} = (0, 1, 0, - 1);$$

Prove that $\{e_1, e_2, e_3\}$ is a basis in $W$ . We need to prove that $\{e_1, e_2, e_3\}$ is linearly independent and $\forall x \in W: \exists a_1, a_2, a_3 \in \mathbb{R}: x = a_1e_1 + a_2e_2 + a_3e_3$ .

$$\begin{array}{l} a _ {1} e _ {1} + a _ {2} e _ {2} + a _ {3} e _ {3} = (0, 0, 0) \Rightarrow \left(a _ {1} + a _ {2}, - a _ {1} + a _ {3}, a _ {1}, - a _ {1} - a _ {2} - a _ {3}\right) = (0, 0, 0, 0) \Rightarrow \\ \Rightarrow \left\{ \begin{array}{c} a _ {1} + a _ {2} = 0 \\ - a _ {1} + a _ {3} = 0 \\ a _ {1} = 0 \\ - a _ {1} - a _ {2} - a _ {3} = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{c} a _ {2} = 0 \\ a _ {3} = 0 \\ a _ {1} = 0 \end{array} \right. \Rightarrow \{e _ {1}, e _ {2}, e _ {3} \} \text {is linearly independent}. \\ \end{array}$$ $$\begin{array}{l} x \in W \Rightarrow x = \left(x _ {1}, x _ {2}, x _ {3}, - x _ {1} - x _ {2} - x _ {3}\right) = x _ {1} (1, 0, 0, - 1) + x _ {2} (0, 1, 0, - 1) + x _ {3} (0, 0, 1, - 1) = \\ = x _ {1} e _ {2} + x _ {2} e _ {3} + x _ {3} \left(e _ {1} - e _ {2} + e _ {3}\right) = x _ {3} e _ {1} + \left(x _ {1} - x _ {3}\right) e _ {2} + \left(x _ {2} + x _ {3}\right) e _ {3}. \\ \end{array}$$

Hence, $\{e_1,e_2,e_3\}$ is a basis in $W$