This is exercise 3.13 from Roman's Advanced Linear Algebra.
Let $V$ be a vector space with
$V = S_1 \oplus T_1 = S_2 \oplus T_2$.
Prove that if $S_1, S_2$ have finite codimension in $V$, then so does $S_1 \cap S_2$ and
$codim(S_1 \cap S_2) \le \dim(T_1)+\dim(T_2)$.
My approach : Consider the projection mappings $p_1$, $p_2$ onto $T_1,T_2$ respectively.
Then if we consider the product of these two maps $p : V \mapsto T_1 \times T_2$, then the kernel of this map is $S_1 \cap S_2$. So by the isomorphism theorem we have $V/(S_1 \cap S_2) \equiv T_1 \times T_2$. I know that if $T_1 \oplus T_2$ then this is isomorphic to $T_1 \times T_2$. But what can I say about this in the general case? I don't know how to conclude that $\dim(V/(S_1 \cap S_2)) \le \dim(T_1)+\dim(T_2)$ from here. For instance, do we have $\dim(T_1 \times T_2) \le \dim(T_1) + \dim(T_2)$? I would greatly appreciate any help.
Best Answer
You have the diagonal map $V\to T_1\times T_2$, with kernel $S_1\cap S_2$. Thus $V/S_1\cap S_2$ is a subspace of the finite dimensional space $T_1\times T_2$. Hence $\dim V/S_1\cap S_2\leq\dim (T_1\times T_2)=\dim T_1+\dim T_2$.
Note that $T_1\times T_2$ is the Cartesian product, is not isomorphic to $T_1+T_2$, and is not even a subspace of $V$. Also the map $V\to T_1\times T_2$ is not onto.