Suppose

U={(x,y,x+y,x−y,2x)∈F5:x,y∈F}U = \left\{(x, y, x+y, x-y, 2 x) \in \mathbf{F}^{5}: x, y \in \mathbf{F}\right\}

Find a subspace WW of F5\mathbf F^5 such that F5=U⊕W\mathbf F^5 = U \oplus W.

Let

W={(0,0,z1,z2,z3)∈F5:z1,z2,z3∈F}W = \left\{(0,0,z_1,z_2,z_3) \in \mathbf{F}^5 : z_1,z_2,z_3 \in \mathbf{F}\right\}

Clearly U∩W={0}U \cap W = \{0\} since if u=wu = w then

(x,y,x+y,x−y,2x)=(0,0,z1,z2,z3)(x,y,x+y,x-y,2x) = (0,0,z_1,z_2,z_3)

Implies x=y=0x=y=0 which then implies both sides are zero. Thus U⊕WU \oplus W is a direct sum by 1.45.

To see U+W=F5U+W = \mathbf F^5 simply write a general element of F5\mathbf F^5 then solve for x,y,z1,z2,z3x,y,z_1,z_2,z_3.