Proof: Tangent space of the general linear group is the set of all squared matrices

Question

Let us assume we have the following definition of a tangent space:

Definition of smooth path

Let $X\subset\mathbb{R}^n$. Let $I$ be a real interval. \begin{equation} P \text{ is a smooth path in } X \quad:\Leftrightarrow\quad P:I\rightarrow X \text{ is a differentiable function}~. \end{equation}

Definition: Tangent vector at the identity

Let $P$ be a smooth path with $P(0) = \mathtt{I}$. \begin{equation} t \text{ is the tangent vector of } P \text{ at the idenity} \quad:\Leftrightarrow\quad t=\frac{\partial}{\partial x}P(x)|_{x=0} \end{equation} Moreover, we call $t$ a tangent vector of a space $X$, if a path $P:I\rightarrow X$ exists such that $t$ is tangent vector of a $P$.

Definition: Tangent space

We call the space of all tangent vectors (at the identity), the tangent space (at the identity).

---

Using these definitions, (how) can we show that the tangent space of all invertible matrices $GL(n)$ is indeed the space of all $n\times n$ matrices?

(I'd like to rely mainly on these definitions and do not use the notion of the exponential map if possible...)

Answer 1

Hint:

• Let $A \in \mathbb R^{n\times n}$ be arbitrary. What is the most simple path $P$ you can think of, which goes through the identity and has derivative $A$?

• Having found such a path $P$ with $P(0) = I$ and $\dot P(0) = A$, show that $P(t) \in GL(n)$ for all small enough $t$ (this is actually true for any such path).

I don't know if this is already enough, but if I say one more thing, then I'm afraid I'll rob you of the joy of figuring it out for yourself.

Answer 2

Recall that a matrix $A$ is invertible if and only if $\det A \ne 0$. Put your favourite norm on the space of all matrices $M_n (\mathbb{R})$. The function $\det : M_n (\mathbb{R}) \to \mathbb{R}$ is continuous, so there is $\epsilon > 0$ such that for all matrices $H$ with $\|{ H }\| < \epsilon$, $\det (I + H) \ne 0$. Hence, for all matrices $X$, there is $\eta > 0$ such that for $-\eta < t < \eta$, $\det (I + t X) \ne 0$. So the path $$t \mapsto I + t X$$ has tangent vector $X$ at $t = 0$, exactly as required.