If $U_1, U_2, W$ are subspaces of $V$ such that $U_1 + W = U_2 + W$ then $U_1 = U_2$?

Question

1) Prove or give a counterexample : if $U_1, U_2, W$ are subspaces of $V$ such that $U_1 + W = U_2 + W$ then $U_1 = U_2$ .

2) Prove or give a counterexample : if $U_1, U_2 , W$ are subspaces of $V$ such that $V = U_1 \oplus W$ and $V = U_2 \oplus W,$ then $U_1 = U_2$

attempt 1): Let $U_1 = V, U_2 = {\{0}\}, W = V$ . THen $U_1 + W = V + V = {\{0}\} + V = U_2 + W$, so $V = V$and so we have $U_1 + W = U_2 + W$ but $U_1 \neq U_2$.

attempt 2). Let $U_1, U_2, W$ be subspaces of $V$.

Can someone please help me? I am stuck on part 2). Thank you for any help.

Answer 1

Let $V = \mathbb{R}^2$, and $U_1 = \overline{\{(0,1)\}}$, $W = \overline{\{(1,0)\}}$,$U_2 = \overline{\{(1,1)\}}$. Note that $U_1 \cap W = U_2 \cap W = \{ 0\}$, and $U_1 + W = U_2 + W = \mathbb{R}^2$. From here, it follows that $U_2 \oplus W = U_2 \oplus W = \mathbb{R}^2$, but $U_1 \neq U_2$.

Answer 2

For $(1)$ you have provided a counterexample on your post.

For $(2)$, given that we are dealing with finite dimensional vector spaces, you have to have: $$U_1 \cong U_2$$ and not necessarily $U_1 = U_2$. Intuitively, you can think of that as follows: Consider $\mathbb{R}^2$ (i.e. the $2d$-plane) which can be written either as the direct sum of two $1$-dimensional subspaces consisting of two perpendicular straight lines or as the direct sum of two distinct, non-perpendicular straight lines.