let there be $U$ a subspace of $ V$ vector space such that $U \ne\{0\}$. suppose $W$ is the only subspace such that $W\oplus U=V$ prove that $U=V$

Question

let $U$ be subset of $V$ vector space such that $U\ne\{0\}$. suppose $W$ is the only subspace such that $W\oplus U=V$ prove that $U=V$

Hi everyone, i encountered this question and i was not able to prove it, i tried proof by contradiction:

let $V$ be the only subspace of $V$ such that $U \ne \{0\}$ and $W$ is the only subspace such that $W\oplus U=V$. Lets assume that $U\ne V$ hence $U\subset V$ , from here i got stuck.

Answer 1

Assume that $U \subsetneq V$. Let $B$ be a basis for $U$ and extend it to a basis $B_1$ for $V$.

Since $U \subsetneq V$, clearly $B \subsetneq B_1$. Define $W_1 = \operatorname{span}(B_1 \setminus B)$. Clearly $W_1 \oplus U = V$.

On the other hand, pick $b \in B$ and $b' \in B_1\setminus B$. Define $B_2 = \{b + b'\} \cup (B_1\setminus \{b'\})$. Verify that $B_2$ is still a basis for $V$ which extends $B$.

Define $W_2 = \operatorname{span}(B_2 \setminus B)$ so we also have $W_2 \oplus U = V$. However, $W_1 \ne W_2$ because $b+b' \in W_2 \setminus W_1$.

This is a contradiction so it must be $U = V$.

Answer 2

Take a basis $\{v_1,..,v_m\}$ for $U$ and complete it to a basis $\{v_1,..,v_m, ..v_n\}$ for $V$, where $n > m$.

Then necessarily $W = span(v_{m+1},..,v_n$) and $ U \oplus W = V$.

Also, $\{v_1,..,v_{m+1} - v_1, ..v_n\}$ is a basis for $V$:

$\forall v \in V$, $v = \sum_{i = 1}^{n} \alpha_iv_i = (\alpha_1 + \alpha_{m+1})v_1 + \sum_{i = 2}^{m} \alpha_iv_i + \alpha_{m+1}(v_{m+1} - v_1) + \sum_{m+2}^{n}\alpha_iv_i$.

Therefore $W = span(v_{m+1} - v_1,..,v_n)$ as well.

To conclude the above we must also show that $span(v_{m+1} - v_1,..,v_n) \cap U = \emptyset$. This holds because if there was $z \in span(v_{m+1} - v_1,..,v_n) \cap U$ then $z = \sum_{i=1}^{m}\gamma_iv_i = \delta_1(v_{m+1} - v_1) + \delta_2v_{m+2}+...+\delta_nv_n$ and from here $W \cap U \neq \emptyset$, a contradiction.

Then write $v_{m+1} - v_1 = \sum_{m+i}^{n}\beta_{m+i}v_{m+i}$ where $i \geq 1$, and get that $v_1 \in W$ which is a contradiction. Hence $n = m$ and we're done.