How to show that the matrix of the following map has positive determinant? Let $(v_1,v_2)$ and $(h_1,h_2)$ and be two basis of a plane $P$ in $\mathbb{R}^3.$ Then I want to show that,
$$\hat{v} = \frac{v_1 \times v_2}{||v_1\times v_2||}= \hat{h}= \frac{h_1 \times h_2}{||h_1\times h_2||}\iff \det(A) > 0$$
where $A$ is the isomorphism that sends $v_i\to h_i$ for $i=1,2.$
I understand geometrically the problem statement but I am not sure how to prove it since 
the expression for the map $A$ is not clear to me. Any hints will be much appreciated.
 A: Assuming $\;T\;$ is the matrix corresponding to the isomorphism you mention with respecto to those given two basis , then it equals $\;\begin{pmatrix}1&0\\0&1\end{pmatrix}\;$, which obviously has positive determinant....and that's all.
If you would want to express your isomorphism with respect to any other two basis of that plane, then you'll get the very same determinant, as determinant is an invariant of linear operators.
A: Define a linear map $B$ on $\mathbb R^3$ by $B=A$ on $P$ and $B=$ the identity map on $V=\operatorname{span}\{\hat{v}\}$. Then
$$
\det(B)\det[v_1,v_2,\hat{v}]=\det[Bv_1,Bv_2,B\hat{v}]=\det[h_1,h_2,\hat{h}].
$$
Note that $\det[v_1,v_2,\hat{v}]=(v_1\times v_2)\cdot\hat{v}=(v_1\times v_2)\cdot\frac{v_1\times v_2}{\|v_1\times v_2\|}=\|v_1\times v_2\|>0$ and similarly, $\det[h_1,h_2,\hat{h}]>0$. Therefore $\det(B)>0$.
However, since $\mathbb R^3=P\oplus V$ and both $P$ and $V$ are invariant subspaces of $B$, we have
$$
\det(B)=\det(B|_P)\det(B|_V)=\det(A).
$$
Hence $\det(A)>0$.
