Suppose that $S_1$ and $S_2$ are subspaces of $\mathbb{R}^n$, with $\dim(S_1) = m_1$ and $\dim(S_2) = m_2$. If $S_1$ and $S_2$ have only $0$ in common, then what is the maximum value of $m_1 + m_2$?
So I know the theorem where if $S_1 \subseteq S_2 \subseteq \mathbb{R}^n$, then $\dim(S_1) \le \dim(S_2)$. My question has to do with the "have $0$ in common." How does this change the problem and what is the importance of including this information? I was thinking originally the maximum value of $m_1 + m_2$ is just $m_1 + m_2$ going by the theorem but it is not working. Any help is appreciated!
Best Answer
The sum is $\le n$, because it's in $\Bbb R^n$. $m_1+m_2$ is the dimension of $S_1+S_2$, which "lives" in $\Bbb R^n$.
Use that $\dim (S_1+ S_2)=\dim S_1+\dim S_2-\dim (S_1\cap S_2)=m_1+m_2-0=m_1+m_2$.
(Notice, without the condition on the intersection, we can always say that $m_1+m_2\le2n$.)