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Show that $D(v_1,v_2) > 0$ if and only if there exists a rotation $T_α$ such that the vector $T_αv_1$ is parallel to $e_1$ (and looking in the same direction), and $T_αv_2$ is in the upper half-plane $x_2 > 0$ (the same half-plane as e2)

Hint : What is the determinant of a rotation matrix?

I've been studying Treil's Linear Algebra Done Wrong and come across this question. I think intuitively the statement looks obvious.

If $D(v_1,v_2) > 0$ then the linear transformation must not be reversing the orientation. However, I don't know how to write it explicitly. I have also tried:

Since the determinant of a rotational matrix is 1 what I did was: $det(T_α A) = det(T_α) . det(A) = det(A) > 0$ where A is the matrix with column vectors $v_1$ and $v_2$

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  • $\begingroup$ Thank you, I have edited my original post and deleted the comment. $\endgroup$ – Polygebra Nov 20 '17 at 19:56
  • $\begingroup$ That's a much better question. Now it can be answered without wasting a lot of time explaining stuff you already know. (So maybe wasting just a little time explaining stuff you already know ...) $\endgroup$ – David K Nov 20 '17 at 20:05
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I think you pretty much have the solution. You've just written the pieces in a sequence that may be a bit incompletely explained.

Your sequence of equations/inequalities is equivalent to the following set of statements: \begin{align} \det(A) &= \det(T_\alpha) \cdot \det(A) \\ \det(T_\alpha) \cdot \det(A) &= \det(T_\alpha A) \\ \det(T_\alpha A) &> 0. \end{align}

You clearly know how to explain why the two equations are true. If you can just show that the last inequality is true, then you can put it all together to prove that if $\det(T_\alpha A) > 0$ then $\det(A) > 0.$ So consider what you know about the columns of $T_\alpha A$ and see what that tells you about its determinant, and you can show that if a rotation $T_\alpha A$ exists as described in the problem, then $\det(A) > 0.$ (Perhaps you've already done this but thought it was too obvious to state; I would say it's reasonably obvious, but not so obvious that it's not worth mentioning.)

For the other direction, the sequence in which you wrote the equations and inequalities is already a good one for explanation. You just need to establish somehow that if $\det(A) > 0$ then $v_1 \neq 0$ and there is some rotation $T$ that will rotate $v_1$ to a vector in the same direction as $e_1,$ that is, $Tv_1 = ce_1$ for $c\in \mathbb R$ and $c>0.$ Call that rotation $T_\alpha,$ and from your formulas you can show that if $\det(A) > 0$ then $\det(T_\alpha A) > 0.$ Now look at how you compute $\det(T_\alpha A)$ and what you know about the coordinates of $T_\alpha v_1,$ and see what that tells you about the second coordinate of $T_\alpha v_2$ (recalling that all you need to show now is that $T_\alpha v_2$ points somewhere in the upper half-plane).

And you're still right, it's pretty much obvious, the only trick is to convince the reader that you haven't just forgotten about one of the fact that makes it obvious.

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  • $\begingroup$ Doesn't this only prove the one way of the if and only if statement? $\endgroup$ – Polygebra Nov 20 '17 at 20:19
  • $\begingroup$ Yes, good point. Let me try to fit the other way in there too. $\endgroup$ – David K Nov 20 '17 at 20:20

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