I am trying to understand this definition...
Let $V$ be a vector space of dimension $m\in \{0\} \cup \mathbb{N}$. Two bases on $V$, which are descrived by two isomorphisms $\psi,\phi: \mathbb{R}^m \rightarrow V$, are said to have the same orientation iff $det(\psi^{-1} \circ\phi)>0.$ Otherwise, they have the opposite orientation.
I am having a hard time picturing this in a graph and understanding what isomorphism is. I slightly remember group isomorphism from abstract algebra but I am having trouble with this concept in my analysis class. How does the determinent of an inverse function composed with another function play a role here?
Here's an example to make things a little more concrete. Let $V$ be the vector space of all $2 \times 2$ real symmetric matrices. (So $V$ has dimension $3$.)
Let $\phi:\mathbb R^3 \to V$ be defined by $$ \phi(x,y,z) = \begin{bmatrix} x & y \\ y & z \end{bmatrix}. $$ Then $\phi$ is an isomorphism that maps the standard ordered basis $(e_1,e_2,e_3)$ to the basis $\alpha = \left( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}\right).$
Let $\Psi:\mathbb R^3 \to V$ be defined by $$\Psi(x,y,z) = \begin{bmatrix} z & y \\ y & x \end{bmatrix}. $$ Then $\Psi$ is an isomorphism that maps the standard ordered basis $(e_1,e_2,e_3)$ to the basis $\beta = \left( \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\right).$
The function $\Psi^{-1} \circ \phi$ is given by \begin{align} (\Psi^{-1}\circ \phi)(x,y,z) &= \Psi^{-1}\left( \begin{bmatrix} x & y \\ y & z \end{bmatrix} \right) \\ &= (z,y,x). \end{align} You can check that $\Psi^{-1}\circ \phi$ has a negative determinant, which shows that the ordered bases $\alpha$ and $\beta$ have opposite orientation.