Finding Subspaces $U, V, W < \mathbb{R}^4$ such that $\mathbb{R}^4 = U \oplus V = U \oplus W = V \oplus W$

Question

So, the dimension of each of the subspaces must be equal to $2$. Now, I kind of brute-forced this problem by trying different values and came up with \begin{gather*} U = \operatorname{Lin}((1, 0, 1, 0), (0, 1, 0, 1)) \,, \\ W = \operatorname{Lin}((1, 1, 1, 0), (0, 0, 1, 0)) \,, \\ V = \operatorname{Lin}((0, 1, 0, 0), (0, 1, 1, 1)) \,. \end{gather*} I do not like this method, it feels very unsatisfying. What would be the way to do it with a more systematic approach (no trial and error)?

Answer 1

By symmetry we may WLOG assume one or more of $U,V,W$ are "nice" subspaces. (That is, this ought to be true if we pick a nice basis anyway). For $\mathbb{R}^4=U\oplus V$ we can pick coordinate planes

$$ U=\mathrm{span}\{e_1,e_2\}=\{(w,x,0,0)\mid w,x\in\mathbb{R}\}, $$ $$ V\,=\,\mathrm{span}\{e_3,e_4\}\,=\,\{(0,0,y,z)\mid y,z\in\mathbb{R}\}. $$

Then it remains to figure out a $W$ which is complementary to both $U$ and $V$. In the case of $\mathbb{R}^2=X\oplus Y$ with the $x$-axis $X$ and the $y$-axis $Y$, we can pick any other line $Z$, like the diagonal line $y=x$ to be a complement. In the 4D case, there is a "diagonal" plane "between" $U$ and $V$:

$$ W=\mathrm{span}\{e_1+e_3,e_2+e_4\}=\{(u,v,u,v)\mid u,v\in\mathbb{R}\}. $$

More generally, the diagonal line in 2D is the graph of the linear function $f(x)=x$ from $\mathbb{R}$ to itself; if we take any linear isomorphism $\phi:\mathbb{R}^2\to\mathbb{R}^2$, given as $\phi(x_1,x_2)=(\phi_1(x_1,x_2),\phi_2(x_1,x_2))$, then the graph of the function $\{(x_1,x_2,\phi_1(x_1,x_2),\phi_2(x_1,x_2))$ is a 2D plane in 4D which is complementary to both $U$ and $V$.