Exercise 4.4.24

(a) One basis for the subspace $S$ of solutions to $x_1+x_2+x_3-x_4=0$ is the 3 special solutions $v_1=(-1,1,0,0),v_2=(-1,0,1,0),v_3=(1,0,0,1)$

(b) Since $S$ contains solutions to ${(1,1,1,-1)}^{T}x=0$, a basis for $S^{\text{⊥}}$ is $(1,1,1,-1)$

(c) Split $(1,1,1,1)$ into $b_1+b_2$ by projection on $S^{\text{⊥}}$ and $S:b_2=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2}\right)$ and $b_1=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2}\right)$.