In my lecture today, we were covering matrix groups and Lie algebras. My professor made the statement that given any matrix group $G$, the tangent space of the group at the identity $T_{e}G$ is a Lie subalgebra of $gl_{n}(\mathbb{R})$, though he offered no proof for this (though it may be coming later in the course I guess). But this got me thinking and leads me to the following question:
Let $\gamma:I\subset\mathbb{R}\longrightarrow G$ denote a differentiable curve in $G$ with $\gamma(0)=I_{n}$, where $I_{n}$ denotes the $n\times n$ identity matrix. With this definition, if $T_{e}G$ is a Lie subalgebra of $gl_{n}(\mathbb{R})$, then all $\gamma'(0)\in T_{e}G$ must be invertible. That is, $\det(\gamma'(0)) \neq 0$. Is this obvious or am I missing something? Does $\gamma(0)=I_{n} \;\Longrightarrow\;\det(\gamma'(0)) \neq 0$.
Thanks for any help/hints/clarifications.
You are confusing the definition of $gl_n(R)$. This lie algebra consists of all matrices, not just invertible ones.
In order to prove the claim, use the following general fact: Let $f : M \to N$ be a smooth map between manifolds and let $p \in M$ be a regular point. Then $M' = f^{-1}(f(p))$ is a submanifold of $M$ and $T_p M'$ is the kernel of $T_p M \to T_{f(p)} N$. Conclude that if $G$ is a Lie subgroup of $H$, then $Lie(G)$ is a Lie subalgebra of $Lie(H)$.