Prove that $λ(A)^+ : S^n → S^n$ (the map induced on one-point compactifications) is homotopic to identity if $\det(A) > 0$ 
Let $A ∈ GL(n,\Bbb R)$.
a) Prove that the induced map $λ(A)$ from $\Bbb R^n$ to $\Bbb R^n$ corresponding to the associated linear transformation is a proper map.
b) Prove that $λ(A)^+ : S^n → S^n$ (the map induced on one-point compactifications) is homotopic to identity if $\det(A) > 0$

My attempt:
(a) Any compact set in $\Bbb R^n$ is closed and bounded. Since, $\Bbb R^n$ is finite dimensional, the induced map $\lambda(A)$ ( i.e. $x \mapsto Ax$) is continuous, thus pre-image of a closed set is closed. $\dots (*)$
We also know that  a linear operator between normed spaces is bounded if and only if it is a continuous linear operator. Thus, $||A|| < + \infty$ .
Hence, using $$||x|| =||A^{-1}Ax|| \le ||A^{-1}||||Ax||$$ we get that pre-image of bounded sets under $\lambda(A)$ is bounded.$\dots (**)$
Thus, $(*),(**)$ combined implies that pre-image of compact sets under $\lambda(A)$ is compact. So $\lambda(A)$ is proper.
(b) Couldn't come up with anything!
Please point out mistakes (if any) in part (a) and how to proceed for part(b) ?
Thanks in Advance for help!
 A: As Mike Miller suggests, it is enough to show that $GL_n^{+}(\mathbb R)$ is connected.
However, I'll show this directly.
Let $A:\mathbb R^n \to \mathbb R^n$ be an automorphism. Then 
$\Lambda(A):S^n \to S^n$ is given by $(a_1, \dots ,a_n,a_{n+1}) \mapsto (A(a_1), \dots A(a_n),a_{n+1})/\|\Lambda (A) (a_1, \dots, a_{n+1})\|$. In a matrix, that is letting the function act by identity on the $(n+1)^{th}$ coordinate, and dividing by the norm of the vector it acts on.
We Supposing first that we have a homotopy of $$B:(a_1, \dots ,a_n,a_{n+1}) \mapsto (A(a_1), \dots A(a_n),a_{n+1})$$
to the identity map that is nonsingular at each step, we can obtain a homotopy on the sphere by dividing by the norm. That is to say, a homotopy $F:\mathbb R^{n+1} \times I \to \mathbb R^{n+1}$ will descend to a homotopy on the sphere granted that $\|x\|$ is not zero at any point which is guaranteed if $F_t$ is a nonsingular linear map at each step.
(This can basically be thought of as a map that regards the "positive span" of each vector as the same point, which intersects the sphere in a unique place.)
Actually, to be less wishy washy, a path in $GL_n^+(\mathbb R)$ is precisely a map that meets those criteria (after a choice of path component.) In the most extremely rigorous sense, there is a homeomorphism $Map((I,X),X) \cong Map(I \times X,X)$ given by $F(x(t))=F(X,t)$.
But anyhow, you can convince yourself in an easier way that it is enough to see that GL_n^{+}$ is path connected.  A proof of this statement is essentially Gaussian elimination, which gives a finite sequence of steps to get to the identity. Create a piecewise linear path using these steps.
