Verify if the basis $E=(e_1,e_2,e_3)$ and $F=(f_1,f_2,f_3)$ are positive or negative with:
$$f_1=e_1\quad \quad\quad\quad\quad f_2=e_2+e_3\quad \quad \quad\quad \quad f_3=e_1+e_2 $$
I did the following:
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I assumed that $e_1=(1,0,0),\;e_2=(0,1,0),\; e_3=(0,0,1)$.
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Then $f_1=(1,0,0),\; f_2=(0,1,1), \; f_3=(1,1,0)$.
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Then I took the determinant $[e_1,e_2,e_3]=1$ and $[f_1,f_2,f_3]=-1$.
Is this procedure enough to verify if they are positive or negative? I'm not sure if assuming that first step is legitimate.
Best Answer
Oh Jesus! If you take two bases $E$ and $F$, the matrix of change from $E$ to $F$ is given by: $$M_{EF} = [{\rm id}]_{FE}.$$Then $E$ and $F$ have the same orientation if $\det M_{EF} > 0$. Note that this determinant won't ever be zero (why?). What you did is valid, but your notation was sloppy: ${\bf e}_1 = (1,0,0)$ is being used to mean $[{\bf e}_1]_E = (1,0,0)$, etc - the vector ${\bf e}_1$ is not necessarily the vector $(1,0,0) = {\bf i}$.
A sure-fire way to do this is to compute $[{\rm id}]_{EF}$. $$\begin{cases} {\rm id}({\bf f}_1) = {\bf f}_1 = 1\cdot {\bf e}_1 + 0 \cdot {\bf e}_2 + 0\cdot {\bf e}_3 \\ {\rm id}({\bf f}_2) = {\bf f}_2 = 0\cdot {\bf e}_1 + 1 \cdot {\bf e}_2 + 1\cdot {\bf e}_3 \\ {\rm id}({\bf f}_3) = {\bf f}_3 = 1\cdot {\bf e}_1 + 1 \cdot {\bf e}_2 + 0\cdot {\bf e}_3 \end{cases}$$
Building the matrix: $$M_{EF} = [{\rm id}]_{FE} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{bmatrix} \implies \det M_{EF} = -1 <0,$$ so they have opposite orientations. Note, though, that "positive" and "negative" orientations only make sense w.r.t. a fixed basis in space - usually the canonical basis. If you define $E \sim F$ if and only if $\det M_{EF} > 0$, you get an equivalence relation. Since the determinant is always positive or negative, we only get two equivalence classes. Pick the class of the canonical basis and call it "positive".