I was reading Milnor and Stasheff's Characteristic Classes but I got stuck at the beginning of Chapter 9 when a homological interpretation of the orientation of a real vector space is given. More precisely, an orientation of an $n$-dimensional real vector space $V$ can also be viewed as a choice of generator of the infinite cyclic group $H_n(V,V-\{0\})$.
It follows from the text that the claim behind this fact is the following:
Let $V$ be a real vector space of dimension $n$ and $\phi: V \to V$ be a linear isomorphism. Then the induced map $$\phi_*: H_n(V,V-\{0\}) \longrightarrow H_n(V,V-\{0\})$$ is $$\phi_*= \begin{cases} \text{id} \quad & \text{if} \ \det \phi > 0 \\ -\text{id} \quad & \text{if} \ \det \phi < 0 \\ \end{cases} \quad .$$
Of course, $\phi_*$ can only be $\pm \text{id}$ because it is a group automorphism of the infinite cyclic group.
Could someone help me to prove this statement? Any help is very welcome.
Thank you in advance!
Without loss of generality, we may assume $V = \mathbb{R}^n$. Note that any invertible matrix $A\in M_n(\mathbb{R})$ can be written as a finite product of elementary matrices. It follows that $\phi$ is a composition of linear maps representing these elementary matrices. It suffices to show that if $E$ is the elementary matrix of interchanging two rows, then the corresponding linear map induces $-\text{id}$ in $H_n(\mathbb{R}^n,\mathbb{R}^n\setminus0)$, and if $E$ is the matrix representing the addition of a nonzero multiple of a row to another row, then the associated map induces $+\text{id}$ in $H_n(\mathbb{R}^n,\mathbb{R}^n\setminus0)$. This can be achieved by observing that an elementary matrix of the former type is similar to the diagonal matrix $\operatorname{diag}(-1,1,1,\ldots, 1)$, and an elementary matrix of the latter type is similar to the identity matrix.