Encyclopedia of Mathematics says:
The Lie algebra of an intersection of Lie subgroups coincides with the intersection of their Lie algebras.
I want to prove this statement. Namely, $T_I (A\cap B)=T_I A \cap T_I B$, where $A$ and $B$ are closed subgroups of $GL_n$.
From definitions, it seems trivial. But how can I write a rigorous proof for it? Any hint will be appreciated.
Let $G$ be a Lie group, let $\mathfrak g$ and let $H$ be a Lie subgroup of $G$. Then$$\mathfrak{h}=\left\{X\in\mathfrak{g}\,\middle|\,(\forall t\in\mathbb{R}):\exp(tx)\in H\right\}$$is the Lie algebra of $H$. Using this, it's quite easy. If $X\in\mathfrak g$, then asserting that $X$ belongs to the Lie algebra of $A\cap B$ means that$$(\forall t\in\mathbb{R}):\exp(tX)\in A\cap B,$$which, of course, is equivalent to$$(\forall t\in\mathbb{R}):\exp(tX)\in A\wedge(\forall t\in\mathbb{R}):\exp(tX)\in B.$$And clearly this means that $X\in\mathfrak{a}\cap\mathfrak b$.