Find three subspace$ W_{1} ,W_{2}, W_{3}$ such that for a given subspace U, $F^{5} = U\oplus W_{1}\oplus W_{2} \oplus W_{3}$

Question

$$U=\{(x,y,x+y,x-y,2x) \in{F^{5}} : x,y\in{F}\}$$ then find $W_{1},W_{2}$ and $W_{3}$, none of which equals $\{0\}$ such that $$F^{5} = U \oplus W_{1}\oplus W_{2}\oplus W_{3}$$ What I have done is $(a,b,c,d,e) \in{F^{5}}$ Then $$(a,b,c,d,e)=(a,b,a+b,a-b,2a)+(a,b,c-a-b,d-a+b,e-2a)+(-a,b,0, 0,0)+(0,-2b,0,0,0)$$ Hence $$W_{1}=(a,b,c-a-b,d-a+b,e-2a)$$ $$W_{2}=(-a,b,0,0,0)$$ $$W_{3}=(0,-2b,0,0,0)$$ I don't have answer. I don't know what I have done is correct or not. Thanks in advance:)

Answer 1

You already have $(1,0,1,1,2)$ and $(0,1,1,-1,0)$ as linearly independent members of $U$.

You need to find three more (linearly independent) basis vectors.

Any set of linearly independent vectors can be put into row echelon form in a matrix. Let's assume we have the three additional vectors:

$$\left(\begin{array}{ccccc} 1 & 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & a & b & c \\ 0 & 0 & 0 & d & e \\ 0 & 0 & 0 & 0 & f \end{array}\right)$$

These five row-vectors are linearly independent if, and only if, the determinant of the five-by-five matrix is non-zero: i.e. $adf \neq 0$. Personally, I would just choose

$$\left(\begin{array}{ccccc} 1 & 0 & 1 & 1 & 2 \\ 0 & 1 & 1 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right)$$

meaning that $W_1 = \operatorname{span}(0,0,1,0,0)$, $W_2 = \operatorname{span}(0,0,0,1,0)$ and $W_3 = \operatorname{span}(0,0,0,0,1)$.