Let UU be the subspace of R5\mathbf R^5 defined by

U={(x1,x2,x3,x4,x5)∈R5:x1=3x2 and x3=7x4}U = \{(x_1,x_2,x_3,x_4,x_5) \in \mathbf R^5 : x_1 = 3x_2 \text{ and } x_3 = 7x_4\}
  1. Find a basis of UU.
  2. Extend the basis to a basis of R5\mathbf R^5.
  3. Find a subspace WW of R5\mathbf R^5 such that R5=U⊕W\mathbf R^5 = U \oplus W.
  1. A basis is
(3,1,0,0,0),(0,0,7,1,0),(0,0,0,0,1)(3,1,0,0,0),(0,0,7,1,0),(0,0,0,0,1)

To prove this is a basis note we can rewrite an element of UU as

(3x2,x2,7x4,x4,x5)=x2(3,1,0,0,0)+x4(0,0,7,1,0)+x5(0,0,0,0,1)\begin{aligned} (3x_2, x_2, 7x_4,x_4,x_5) &= x_2(3,1,0,0,0) \\ &+ x_4(0,0,7,1,0) \\ &+ x_5(0,0,0,0,1) \end{aligned}

And our list is clearly independent, thus it is a basis.
2. Add (1,0,0,0,0),(0,0,1,0,0)(1,0,0,0,0),(0,0,1,0,0) to our list so we can violate the constraints, It's straightforward to see our new list is spanning (since we can get the standard basis via a linear combination of our basis). To see it's independent note that no dependent list of length five can span R5\mathbf R^5 since then we could remove a vector without changing the span (by 2.21) giving a spanning list of length less then five, which is impossible as the standard basis is of length five, meaning (by 2.23) all spanning lists have to be of length at least five.
4. Let W=span((1,0,0,0,0),(0,0,1,0,0))W = \text{span}((1,0,0,0,0),(0,0,1,0,0)), to see U⊕W=R5U \oplus W = \mathbf{R}^5 note that U∩W={0}U \cap W = \{0\} (no vector in WW satisfies the constraints) and clearly U+W=R5U+W = \mathbf{R}^5.

My solution to (2) is overly long, I could have replaced the stuff about independance with a reference to 2.42 but we haven't read that yet so it would be cheating :/