Let $S_1$ and $S_2$ complementary subspaces of $U$. What can be said about the relationship of $S_1$ and $S_2$? This is an exercise of Advanced Linear Algebra of Roman

Let $\dim V<\infty$, $U$ a subspace of $V$ and $V=U\oplus S_1=U\oplus S_2$. What can be said about $S_1$ and $S_2$?

This is what I did, but I am not sure that its correct: suppose that $U$ is a proper subspace of $V$, then $S_1,S_2\neq\{0\}$. If $S_1\cap S_2=\{0\}$ then the direct sum $U\oplus S_1\oplus S_2$ is well-defined but then
$$\dim (U\oplus S_1\oplus S_2)=\dim V+\dim S_2>\dim V$$
what cannot be possible, so $S_1\cap S_2\neq\{0\}$. Other thing that can be said is that $S_1+S_2=:S_3$ is also a complementary subspace of $U$.
Are these two statements correct? There is something more important to say?

UPDATE: ok, I saw my mistake. It is not true that the direct sum $U\oplus S_1\oplus S_2$ is well-defined. Then I dont know what can be said about the relationship of $S_1$ and $S_2$. Some idea?
 A: I'm assuming $S_1, S_2$ are also subspaces of $V$, and that $\oplus$ is the inner direct sum, i.e. $X \oplus Y = \operatorname{span}X \cup Y$.
$V = U \oplus S$ immediately implies $\dim S = \dim V - \dim U$, and also $U \cap S = \{0\}$, so $S_1$ and $S_2$ have the same dimension, and intersect $U$ trivially.
In fact, this is also a sufficient condition for a subspace to be a direct complement of $U$.

Let $S \le V$ such that $\dim S = \dim V - \dim U$ and $U \cap S = \{0\}$. Then $S \oplus U = V$.

Proof:
Let $\{u_1,\ldots, u_m\}$ be a basis for $U$ and $\{s_1, \ldots, s_k\}$ be a basis for $S$. Then $$\{u_1, \ldots, u_m, s_1 , \ldots, s_k\}$$ is a basis for $V$.
It suffices to check linear independence since $m + k = \dim V$ by assumption.
Assume
\begin{align}
0 &= \alpha_1u_1 + \cdots + \alpha_mu_m + \beta_1s_1 + \cdots + \beta_ks_k\\
\end{align}
So $$\underbrace{\alpha_1u_1 + \alpha_2u_2 + \cdots + \alpha_mu_m}_{\in U} = \underbrace{- (\beta_1s_1 + \cdots + \beta_ks_k)}_{\in S}$$
Therefore $$\alpha_1u_1 + \alpha_2u_2 + \cdots + \alpha_mu_m = - (\beta_1s_1 + \cdots + \beta_ks_k) = 0$$
so linear independence of $\{u_1, \ldots, u_m\}$ and $\{s_1, \ldots, s_k\}$ implies $$\alpha_1 = \cdots = \alpha_m = \beta_1 = \cdots = \beta_k = 0$$
