Prove that invertible linear map in $T:\Bbb{R}^n\to \Bbb{R}^n$,that $\det{T}>0$ is isotopic to the identity map.That is find some smooth map $\Bbb{R}^n\times I \to \Bbb{R}^n$ that is diffeomorphism (invertible) at each time $t\in I$.
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1Well, of course there is no hope of such an isotopy passing through only linear maps, because as soon as $\det T<0$ you are guaranteed to send any basis to a lineraly dependent set at some point. Also, the claim fails for $n=1$ and $T=id$ because $H(0,t)H(1,t)$ is $1$ at $t=0$ and $1$ at $t=1$, therefore it will be $0$ for some $0
– Oct 04 '21 at 14:51 
Oh sorry,I didn't realized $T$ needs to preserve orientation,that is $\det{T}>0$ yesterday.I found this question in Milnor's book,but forget one condition. – yi li Oct 05 '21 at 01:56

1$\det(T) > 0$ does not imply $\lambda(T) > 0$. For example $T = \operatorname{diag}\{1/4, 1, 1\}$ fails for $t = 1/2$ in your method. Or am I missing something? – obareey Oct 05 '21 at 11:08

Oh sorry,I realized this point before,I don't know why I write it on the post.....@obareey – yi li Oct 05 '21 at 11:13

See https://math.stackexchange.com/questions/100978/howmanyconnectedcomponentsdoesmathrmglnmathbbrhave . – Thorgott Oct 05 '21 at 11:22

thank you,I didn't realize this is equivalent to set of matrix $GL(\Bbb{R}^n)^+ = \{\det{T}>0\}$ are path connected – yi li Oct 05 '21 at 11:27

there is another nice note talking about it,see http://virtualmath1.stanford.edu/~conrad/diffgeomPage/handouts/gramconnd.pdf – yi li Oct 05 '21 at 11:30