Question 7

7

Note that (0,0,0,0) \in W and that if (t_1, t_2, t_3, t_4), (s_1, s_2, s_3, s_4) \in W then:

\lambda t_1 + s_1 + 2(\lambda t_2 + s_2) - (\lambda t_3 + s_3) = \lambda(t_1 + 2t_2 - t_3) + (s_1 - 2s_2 + s_3) = 0, so \lambda v_1 + v_2 \in W for any v_1,v_2 \in W. So W is a subspace of V.

Let (x_1, x_2, x_3, x_4) \in V and \lambda = x_1, \mu = x_2. Then x_3 = \lambda + 2\mu, and x_4 is arbitrary say x_4 = t. So each element of V is expressible in the form:

\begin{pmatrix}\lambda \\ \mu \\ \lambda + 2\mu \\ t\end{pmatrix} = \lambda \begin{pmatrix}1 \\ 0 \\ 1 \\ 0\end{pmatrix} + \mu \begin{pmatrix}0\\ 1 \\ 2\\0\end{pmatrix} + t\begin{pmatrix}0\\0\\0\\1\end{pmatrix}

Noting that these are linearly independent, we have \dim W = 3 and a basis is B_W = \{(1,0,1,0), (0,1,2,0), (0,0,0,1)\}.

The standard basis B_V = \{(1,0,0,0), (0,1,0,0), (0,0,0,1), (0,0,0,1)\}. However only (0,0,0,1) \in W (since 0 + 2 \times 0 - 0 = 0) and a basis of W must be of size 3.

We can add another vector to B_W to get a basis for W. Any vector linearly independent to (1,0,1,0), (0,1,2,0) and (0,0,0,1) works, say (1,0,0,0).

In general, any spanning set of a subspace of a vector space can be extended to a basis of the entire space.

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