From Advanced Linear Algebra (Roman):
Let $\dim(V) < \infty$ and suppose that $V = U \oplus S_1 = U \oplus S_2$. What can you say about the relationship between $S_1$ and $S_2$? What can you say if $S_1 \subseteq S_2$?
I couldn't completely answer it, but here are my thoughts:
We know that $U \cap S_1 = \{0\}$ and $U \cap S_2 = \{0\}$.
We also know that all elements of $V = U \oplus S_i$ are of the form $u + s_i$ where $s_i \in S_i$ for i=1,2.
If $s_i = 0$, then $u+s_i=u$, so we don't really need to examine this one since it doesn't really say anything about $S_1$ or $S_2$.
What we want to look at is $u+s_1$ for $u=0$ and $u \not=0$ (where $s_i \not= 0$).
The subspaces $S_1$ and $S_2$ do not need to be equal, since an element $w \in S_1$ doesn't need to be in $S_2$ as long as we have $u_1 + s_1 = w$ for some $u_1 \in U, s_1 \in S_1$.
However, for $S_i$ (where $i = 1,2$), the elements of the form $u+s_i$ where $u\not=0$ must be different from those of the form $u + s_i$ where $u=0$ (we are assuming that $s_i \not= 0$ in these cases).
Suppose that they are equal. Then $u+s_i = s_i'$ for some $s_i, s_i' \in S_i$. But then we have $s_i' – s_i \in S_i$, contradicting the fact that $U$ and $S_i$ are disjoint.
From here, we know that there are $|S_i| + |U|$ elements of the form $u+s_i$ where $u=0$ and where $u \not=0$ (and $s_i \not= 0$ in both cases). In other words, form here we can conclude that both $S_1$ and $S_2$ have the same number of elements; since there are $|S_i| + |U|$ possible elements and we know that they are all different.
So if $S_1 \subseteq S_2$, then $S_1 = S_2$.
So we know that both subspaces must be of the same cardinality. However, the textbook mentions somewhere that such subspaces must be isomorphic. But I wasn't able to prove that.
I'm guessing that I need to define a module homomorphism from $S_1$ to $S_2$ and show that it is injective or surjective. But I'm not really sure how. In other words, how can I define a module homomorphism when we only know that we have two subspaces without any further detail? Or is there another way to prove this without using a module homomorphism?
Thank you in advance
Best Answer
I don't think you want to argue that $S_1$ and $S_2$ have the same cardinality. Rather you want to show they have the same dimension. Then they must be isomorphic since all vector spaces of the same dimension are isomorphic.
To see they have the same dimension notice that a basis for $U$ joined with a basis for $S_i$ is a basis for $V$. So $\dim(S_i)=\dim V-\dim U$.