Let $I,J\subset [m]$ be two different bases. Prove that for all $i\in I \setminus J$, there is an index $j\in J\setminus I$ such that $(I \backslash \{i\})\cup\{j\}$ is a basis and so is $(J \setminus \{j\})\cup \{I\}$.
Note: $I,J\subset \{1,2,...,m\}$ are index sets.
I am having trouble coming up with a way to prove this. I think I should us the Steinitz exchange lemma in some way but I need help starting this out. Any help would be appreciated!
Let $i \in I$. suppose $(I \setminus \{i\}) \cup \{j\}$ is not linearly independent for any $j \in J$. Then, Since $(I \setminus \{i\})$ is linearly independent it follows that $j$ must be a linear combination of $(I \setminus \{i\})$. This is true for each $j$ so every vector, in particular $i$ must be a linear combination of $(I \setminus \{i\})$ (because it is a linear combination of $J$). This contradiction shows that $(I \setminus \{i\}) \cup \{j\}$ is linearly independent for some $j$. Since it has the same number of elements as $I$ it is a basis. Similar argument for $(J \setminus \{j\}) \cup \{i\}$.