How to find a subspace that satisfies a direct sum

Question

The problem states:

Suppose

$$U=\{(x,x,y,y)\in F^4:x,y\in F\}$$

Find a subspace $W$ of $F^4$ such that $F^4$ equals the direct sum of $U$ and $ W$.

So I know that we are basically trying to find a $W$ such that when added with you there is only one way to write the sum such that it equals $F^4$.

My thoughts:

My best guess is that $W = \{(0,0,0,0)\}$ because that way your sum is unique. For example, if you get $U = (1, 1, 1, 1)$ the only possible way to obtain this value with the addition of these sets is by doing $(1+0, 1+0, 1+0, 1+0)$ which ends up being $(1, 1, 1, 1)$. If you were to make a change to any of the $x$'s or $y$'s from $U$, it is impossible to obtain the same result from a different way. Can anyone corroborate if I'm going in the right direction?

Answer 1

What we are basically trying to find is a space $W$ such that every $(x_1,x_2,x_3,x_4)\in F^4$ can be written as $w+(x,x,y,y)$ with $w\in W$ and $(x,x,y,y)\in U$. This can be done for example if $W=\{(u,0,v,0):u,v\in F\}$.

Your suggestion that $W=\{(0,0,0,0)\}$ isn't going to work, because if you add $(0,0,0,0)$ to an element of $U$, that element still has the form $(x,x,y,y)$. You thus won't be able to get through that addition vectors in $F^4$ which don't have that form, say $(0,1,0,1)$.

Answer 2

Put \begin{align*} u_1 &= \begin{bmatrix}1\\1\\0\\0\end{bmatrix} & u_2 &= \begin{bmatrix}0\\0\\1\\1\end{bmatrix} \end{align*} and note that $\{u_1,u_2\}$ is a basis for $U$.

If $\{u_1,u_2,v,w\}$ is a basis for $\Bbb F^4$, then $\Bbb F^4=U\oplus\DeclareMathOperator{Span}{Span}\Span\{v,w\}$

Can you prove this boxed statement?

Now, we can find suitable $v$ and $w$ algorithmically by computing the reduced row-echelon form of the matrix $$ \left[\begin{array}{rr|rrrr} 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right] $$ Row-reducing gives $\DeclareMathOperator{rref}{rref}$ $$ \rref \left[\begin{array}{rrrrrr} 1 & 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{rrrrrr} 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 \end{array}\right] $$ The second two pivot columns in the reduced row-echelon form correspond to our desired $v$ and $w$ in the original matrix. That is, choosing \begin{align*} v &= \begin{bmatrix}1\\0\\0\\0\end{bmatrix} & w &= \begin{bmatrix}0\\0\\1\\0\end{bmatrix} \end{align*} allows us to write $\Bbb F^4=U\oplus\Span\{v,w\}$.