For every two subspaces U,W of a vector space V, there exists a subspace S of V such that $V = U \oplus S = W \oplus S$ iff dim(U) = dim(W)

Question

I have found this question in an old exam for linear algebra. Specifically the last question here: http://math.ucdenver.edu/~langou/5718/Spring2015-MATH5718-exam1.pdf . I can however not convince myself of the truth of this result. Would not

$ V = R^3 $, U := {(x,0,0) |$x \in R $}, W := {(0,0,z) | z$\in R$}

be a counterexample of this claim?

The only explanation i have for this is difference in notation where i am used to $\oplus$ referring specifically to a direct sum whereas they may use it as a normal plus sign. Then however would the result not be trivial by choosing S = V?

Answer 1

It is not a counterexample, since you can take $$ S = \{(x,x+z,z) \mid y\in\mathbb{R}\}. $$ Let us show that this subspace satisfies $U\oplus S=\mathbb{R}^{3}$ (the other is similar).

A basis for $U$ consists of the vector $(1,0,0)$. A basis for $S$ consists of $(1,1,0)$ and $(0,1,1)$.

You can easily chech that $A=(1,0,0)$, $B=(1,1,0)$ and $C=(0,1,1)$ forms a basis for $\mathbb{R}^{3}$. This can be done by showing that $$ (0,1,0) = B - A, \qquad (0,0,1) = A+C-B. $$ Therefore, the standard basis is generated by $A,B,C$, proving that any element of $\mathbb{R}^{3}$ is generated by $A$, $B$, $C$, i.e., is a basis.

Answer 2

To prove the statement note that $W\neq U\cup W$ (set theoretic union). To see this we have $U \not \subseteq W$ since they have the same dimension. (assuming they are not equal in which case its easy). Similarly $W\not \subseteq U$ so let $u\in U-W$ and $w\in W-U$ then $u+w\not \in U\cup W$. So chose $s_1\in V-(U\cup W)$. Now repeat the argument for $U\oplus <s_1>$ and $W\oplus <s_1>$.

Answer 3

The assertion in title is true in the more general context of finite dimensional spaces.

$\Rightarrow$ is true because $\dim V= \dim U+\dim S=\dim W+\dim S$.

$\Leftarrow$: if $\dim U=\dim W<\dim V$, we have to find a subspace $S$ such that $\;U\oplus S=W\oplus S=V$.

By the avoidance lemma for vector spaces, one can find a vector $v_1$ not in $U\cup W$. If $\DeclareMathOperator\codim{codim}\codim U=\codim W=1$, setting $S=\langle v_1\rangle$ solves the problem.

If $\codim U=\codim W=r>1$, set $U_1=U\oplus\langle v_1\rangle$, $W_1=W\oplus\langle v_1\rangle$, an easy induction on $r$ shows one can find a subspace $S_1$ such that $\;U_1\oplus S_1=W_1\oplus S_1=V$. Then $S=S_1\oplus\langle v_1\rangle$ solves the problem.