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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...)

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2 Answers 2

up vote 3 down vote accepted

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.

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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.

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Thanks a lot! As usual the problem that one cannot accept two answers :-( –  Hauke Strasdat Jan 14 '12 at 15:16
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