Can you please check if my solution is correct? I need to find a direct complement to \begin{align*} \mathbb{U} = \{(x_1,x_2,x_3,x_4) \in \mathbb{R}^4\,|\, x_1+x_2+x_3+x_4 = 0\,,~ x_1+x_2-x_4 = 0\} \end{align*} and describe it by its basis.
What i have done: My basis for $\mathbb{U}$ is $(-1, 1, 0, 0)$, $(-1, 0, 2, 1)$. (I expressed $x_1$ and $x_3$ as they are leading variables in RREF of matrix $ \begin{pmatrix}1&1&1&1\\1&1&0&-1\end{pmatrix} $). Then I extended the basis and got the answer : $(-1, 1, 0, 0), (-1, 0, 2, 1), (1, 0, 0, 0), (0, 0, 1, 0)$.
First of all you have to find a basis of $\mathbb{U}$. To do this, simply solve the following system of linear equations:
\begin{cases} x_1 + x_2 + x_3 + x_4 = 0\\ x_1 + x_2 - x_4 = 0 \end{cases}
I will leave the details to you. The general answer is
\begin{pmatrix} w - t\\ t\\ -2w\\ w\\ \end{pmatrix}
with $t,w \in \mathbb{R}$ arbitrary. A basis of the space where this general vector lives might be $\{(1, 0, -2, 1), \ (-1, 1, 0, 0)\}$.
Now, because $\mathbb{U}$ is a subspace of $\mathbb{R}^4$, we have that $\mathbb{U} \oplus \mathbb{U}^\perp = V$. So, if we find two vectors that are perpendicular to our basis, that are also linearly independent, then we have found what we are looking for. Two vectors that satisfy this are $\{(0, 0, 1, 2), \ (1, 1, 0, -1)\}$
The problem with the two vectors you proposed is that $\mathbb{W} \cap \mathbb{U} \neq \{0\}$, where $\mathbb{W} = \langle (1, 0, 0, 0), (0, 0, 1, 0) \rangle$. That means that it is not a direct complement, whereas it is the case when $\mathbb{W} = \mathbb{U}^\perp$. Probably the reason why you got confused is because it is indeed true that your proposed extended basis spans $\mathbb{R}^4$.