Given $W_1,W_2$ subspaces of $V$, then exists $U$ such that $V=W_1\oplus U=W_2\oplus U$.

Question

I am struggling with this problem

Let $V$ be a vector space with $\dim(V) = n < \infty$, and let $W_1, W_2 \subset V$ be subspaces of $V$ with $\dim(W_1) = k =\dim(W_2)$. Prove that there exists a subspace $U \subset V$ that satisfies $W_1 \bigoplus U = W_2 \bigoplus U = V$.

I managed to prove it for the case $W_1 = W_2$, but for the general case, I have no good idea where to start. Can you point me in the right direction? Thanks in advance

Answer 1

Hint: First, take the basis $(e_1, e_2,\ldots, e_i)$ of $W_1\cap W_2$.

It can be extended to the basis $(e_1,\ldots,e_i,f_1,f_2,\ldots,f_{k-i})$ of $W_1$ and to the basis $(e_1,\ldots,e_i,g_1,g_2,\ldots,g_{k-i})$ of $W_2$.

It can further be proven that $e_1,\ldots,e_i,f_1,\ldots,f_{k-i},g_1,\ldots,g_{k-i}$ will then all be linearly independent, so you can extend this set to the basis $(e_1,\ldots,e_i,f_1,\ldots,f_{k-i},g_1,\ldots,g_{k-i}, h_1,h_2,\ldots,h_{n-2k+i})$ of $V$.

Now, prove that $U=\text{span}(f_1+g_1,f_2+g_2\ldots,f_{k-i}+g_{k-i},h_1,\ldots,h_{n-2k+i})$ satisfies the desired property.

Answer 2

An idea for you by hints:

Let us try to use Zorn's Lemma: let $\;T:=\{ K\le V\;|K\cap W_i=\{0\}\;,\;\;i=1,2\;\}\;$ . Observe that $\;T\neq\emptyset\;$ since $\;\{0\}\in T\;$. We partial order $\;T\;$ by set inclusions and let $\;C\subset T\;$ be any chain. It's easy to see that

$$\;D:=\bigcup_{F\in C}F\;\;\;\text{is an upper bound of}\;\;C\;\;\text{(Why? Be sure you can prove this clearly)}$$

Thus, there exists a maximal element $\;U\;$ in $\;T\;$, and certainly $\;U\cap W_1=U\cap W_2=\{0\}\;$ (Why?) , and thus $\;U+W_i=U\oplus W_i\,,\,\,i=1,2\;$.

We now must prove that $\;U\oplus W_i=V\;$ . Suppose this isn't true. Then there exists a vector $\;v\in V\;,\;\;v\notin U\oplus W_i\;$. But then $\;Z:=\text{Span}\left(U\cup\{v\}\right)\;$ still fulfills $\;Z\cap W_i=\{0\}\;$ (why?) and thus $\;Z\in T\;$ , and we get a contradiction to maximality of $\;U\;$.

Now just fill in little detials in the proof above and end the argument.

Answer 3

Choose $\varphi\in\text{Aut}(V),\psi\in\text{End}(V)$ such that $$\varphi[W_1]=W_2\;\;\text{and}\;\;\psi[V]=\psi[W_2]=W_1$$ Consequently, $U:=\ker(\psi)=\ker(\varphi\circ\psi)$ and by the Rank Theorem we must have $$\dim(U)+\dim(W_1)=\dim(U)+\dim(W_2)=\dim(V)=n<\infty$$ Most importantly $U\cap W_1=U\cap W_2=\{\mathbf 0\}$ because both $\psi|_{W_2}$ and $(\varphi\circ\psi)|_{W_1}$ are injections. $$\therefore V=U\oplus W_1=U\oplus W_2$$