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Geometrically, the determinant of a matrix is the signed volume of a unit cube after the transformation defined by the matrix is applied. However, I'm have trouble understanding what the "signed" mean here. Since effectively volumes are only positive (or zero but for now let's not worry about it). So what's the geometric meaning of a negative determinant? How should I understand the negative volume produced by applying such transformation?

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    $\begingroup$ It means orientation has been reversed. Start by using examples to see what happens in two and three dimensions. $\endgroup$ – Ted Shifrin Dec 21 '15 at 18:30
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    $\begingroup$ @TedShifrin What does "orientation" mean here? $\endgroup$ – OneZero Dec 21 '15 at 18:40
  • $\begingroup$ Think about the right-hand rule, and see this. $\endgroup$ – Ted Shifrin Dec 21 '15 at 18:45
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    $\begingroup$ Here is an analogy: Given two points on the real line $a,b$ we could define a signed length between them as $b-a$. It carries a little more information (the orientation, which can be positive or negative) than just the length. $\endgroup$ – copper.hat Dec 21 '15 at 19:01
  • $\begingroup$ I beg to disagree with other commenters here. I think the sign of a determinant has no geometric significance. It is an algebraic property, not a geometric one. It's a -- somewhat useful -- artifact of the definition of determinant: we want the determinant to be multilinear in the matrix columns and we want it to be zero when the column vectors are linearly dependent. Consequently, determinants are signed and if we swap two columns, the sign flips. While we can interpret the sign as some sort of orientation in two or three dimensions, this analogue breaks down in higher dimensional cases. $\endgroup$ – user1551 Dec 22 '15 at 2:26
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The idea has to do with orientation, that is, which direction is right and which is left. Ultimately this has to do with which order is given for the basis elements. For example consider $\Bbb R^2$ with basis elements $(1,0)$ and $(0,1)$. We can consider $(1,0)$ to be pointing "to the right", and $(0,1)$ pointing " forward".

Putting $(1,0)$ and $(0,1)$ in a matrix gives the identity matrix whose determinant is $1$ of course.

Now switch the two basis elements. Imagine rotating the plane around the line $y=x$. Now $(1,0)$ is pointing forward but $(0,1)$ is pointing to the left.

Of course if we switch rows in the identity matrix the resulting matrix has determinant $-1$.

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  • $\begingroup$ Of course is not an explanation in my opinion. In particular if of course comes exactly, where you should find a preciser explanation. $\endgroup$ – Bman72 Apr 23 at 12:35

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