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Let $(u_1, \ldots, u_n)$ and $(v_1, \ldots, v_n)$ be two ordered bases of $\mathbb R^n$. The orientation of the first basis is defined as the sign of the determinant of $[u_1 \cdots u_n]$, and similarly for the second basis. Prove that the first basis can be continuously transformed into the second one, while remaining linearly independent at all times, if and only if the two bases have the same orientation.

The "only if" direction is easy, because the determinant, which must change continuously, cannot change from positive to negative without going through zero. I'm looking for a proof of the "if" part.

More broadly, I'm looking for comments on the issue of defining the determinant in a nice way. The definitions I've seen say something like:

"The determinant is a quantity that has some nice properties. For one, the determinant is zero if and only if the corresponding matrix is singular. Furthermore, its absolute value equals the volume of the parallelepiped spanned by the vectors. And the sign corresponds to the orientation of the vectors. And what is the "orientation" of a tuple of vectors? Well, it's defined as the sign of the determinant!"

The above claim, if correct, might lead to a more natural (and less circular) definition of orientation, and also of the determinant.

Also, is it necessary to define the n-dimensional volume of a parallelepiped as the absolute value of the determinant (as I have seen in some places)? Can't they be shown to be equal via elementary arguments? Consider the "cut-and-paste" proof that the area of a parallelogram equals the area of a rectangle with the same base and height. I think a similar n-dimensional cut-and-paste can show that

$$\mathrm{vol}(u_1, u_2, \cdots , u_n) = \mathrm{vol}(u_1 + k u_2, u_2, \cdots, u_n),$$

and similarly for the other elementary properties of the determinant. But I haven't thought it through.

Thanks in advance!

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    $\begingroup$ Your question is off-topic in this site, where the subject is research-related math. You can ask your question at math.stackexchange.com or some of the other sites listed in the FAQ. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Nov 11, 2012 at 18:33
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    $\begingroup$ @Dmitri: the definition of the determinant in that answer is formulated differently (although equivalently), and there is no definition of orientation. $\endgroup$
    – Margaret Friedland
    Commented Nov 11, 2012 at 19:21
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    $\begingroup$ Hi Gabriel, your question is a great one: well written, with good motivation and comments on your own thoughts. But I agree with the other comments that it is standard material in many linear algebra classes. I think thus question would be excellent at math.stackexchange. Perhaps part of what's causing an objection is the opening paragraph, which is worded like a homework question. I should say, this is a nontrivial result that you are asking about: why should the group of invertible matrices have precisely two components, and not more? $\endgroup$
    – Theo Johnson-Freyd
    Commented Nov 11, 2012 at 22:07
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    $\begingroup$ Speaking as someone who has never had to teach the definition of a determinant, and isn't completely satisfied with the approaches seen in the books I've consulted for teaching: I vote to re-open. It might not be "research mathematics" but it is something where a research-level perspective might illuminate the pedagogical niceties. $\endgroup$
    – Yemon Choi
    Commented Nov 11, 2012 at 22:12
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    $\begingroup$ I agree with Theo that perhaps the opening paragraph creates a misleading impression. $\endgroup$
    – Yemon Choi
    Commented Nov 11, 2012 at 22:13

3 Answers 3

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Basically what you're asking in the first question is whether $SL_n({\mathbb R})$ is pathwise connected. Using the polar decomposition, this follows from the fact that $SO_n({\mathbb R})$ is pathwise connected. That, in turn, follows from the properties of Givens rotations.

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  • $\begingroup$ Robert: Do you mean that any rotation in $\mathbb R^n$ can be decomposed into Givens rotations? I see such a statement in the Wikipedia for dimension 3. Is it true in general? $\endgroup$
    – Gabriel Nivasch
    Commented Nov 11, 2012 at 20:59
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    $\begingroup$ If $R$ is a rotation matrix in ${\mathbb R}^n$, we can multiply it on the left by Givens matrices $G(1,2,\theta_2), G(1,3,\theta_3), \ldots, G(1,n,\theta_n)$ successively making the $(2,1), (3,1), \ldots, (n,1)$ entries $0$. Then the $(1,1)$ entry will be $\pm 1$ and we can make it $+1$ by adding $\pi$ to $\theta_n$ if necessary; the $(1,2), \ldots, (1,n)$ entries will be $0$. Thus now we have a block matrix $\pmatrix{1 & 0\cr 0 & S\cr}$ where $S$ is an $(n-1) \times (n-1)$ rotation matrix. Proceed by induction. $\endgroup$
    – Robert Israel
    Commented Nov 12, 2012 at 18:12
  • $\begingroup$ Let me add for the record that there is a simpler way of doing the continuous transformation in the same spirit as the suggestion of Robert Israel. We wish to transform $(u_1,...,u_n)$ into $(e_1,...,e_{n-1},\pm e_n)$. We do Gauss-Jordan elimination continuously, with the restriction that a row cannot be multiplied by a negative number (because the multiplier, which starts at 1 and changes continuously, would have to go through $0$). We get $(\pm e_1,...,\pm e_n)$. Then we eliminate the minus signs in pairs with Givens rotations, as Israel suggested. We might be left with one minus at the end. $\endgroup$
    – Gabriel Nivasch
    Commented Dec 21, 2012 at 8:27
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About volume: first, I think it is reasonable, especially when addressing beginning students, to admit that we do have a prior notion of "volume", even if it is imprecise. Probably most people would agree that rigid rotations (=$K=SO(n,\mathbb R)$) preserve volume, and that dilation (by positive amounts) of coordinates (=$A^+$, the positive diagonal matrices) multiply by the product of the diagonal entries, by looking at boxes with sides parallels to the axes. Then by a Cartan decomposition (or some more elementary-sounding name, if desired) $K\cdot A^+ \cdot K$ is close to being $GL(n,\mathbb R)$: it is all positive-determinant matrices. Thus, we've given an "accommodating" argument that all matrices with positive determinant change volume by the determinant.

Robert Israel's point about path-connectedness can be re-used in this context.

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  • $\begingroup$ In a first course in linear algebra, I admit that we have a prior notion of volume (and a prior notion of orientation) in dimensions up to 3. I do not claim to have, nor do I suggest that they should have, such prior notions in higher dimensions. I develop the basic properties of determinants in low dimensions by treating them as oriented volumes (pointing out that including the orientation makes the algebra cleaner). Then I extend the algebra to higher dimensions and tell them that this extension can be used to define volume and orientation in spaces where we lack intuition. $\endgroup$
    – Andreas Blass
    Commented Nov 12, 2012 at 2:04
  • $\begingroup$ I might disagree in a very mild way to @AndreasBlass' comment about, in effect, legitimizing ways of working with new situations. Of course I understand the methodological point, that, on one hand, one should mistrust an untutored intuition. However, there is not a unique way to extend intuition/sensibilities from the familiar to the novel. I truly do think that it is defensible to claim that, whatever else may be going on, rigid rotations preserve volume, for example, even if we admit that we are beyond 3 dimensions, etc. ... (cont'd) $\endgroup$
    – paul garrett
    Commented Nov 12, 2012 at 3:20
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    $\begingroup$ ... so, although many students do need "checks" on reckless and thoughtless facile conclusions, my experience also indicates that many have too-well "learned" to distrust even qualitative conclusions based on prior experience. My specific methodological point is that, although, of course, one must always be cautious, there is nothing else upon which to base speculation than... the past, and prior experience. So, to distrust that, or, worse, to be taught to distrust prior experience too strongly, is simply defeatist. $\endgroup$
    – paul garrett
    Commented Nov 12, 2012 at 3:22
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    $\begingroup$ Calculus already has a definition of volume, which is by Riemann sums, which boils down to dividing space into a grid of tiny cubes and counting how many cubes are inside your shape. So why should Linear Algebra come and introduce a different notion of volume?? At the very least one needs to prove that the two definitions are equivalent! $\endgroup$
    – Gabriel Nivasch
    Commented Nov 12, 2012 at 5:52
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    $\begingroup$ Calculus only defines volumes given a coordinate system... At the very least, we'd presume that changing coordinates in the most trivial possible, but real-life, fashion, namely, "rigid rotations" (and translations) oughtn't change "volume". $\endgroup$
    – paul garrett
    Commented Nov 12, 2012 at 13:35
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Re your general questions: in multilinear algebra, the determinant (of $n$-tuples of vectors with $n$ coordinates in some previously chosen ordered basis) is often defined as the (unique) $n$-linear alternating form which takes value $1$ on the basis. From this the properties of determinant follow almost tautologically: the alternating property causes determinant to be zero on linearly dependent $n$-tuples, and changing the basis requires scaling the determinant by a factor. Also, the definition of the volume of a parallelepiped becomes natural.

Using this definition of determinant, the orientation can be defined as follows (cutting and pasting from Wikipedia): ``Let $V$ be a finite-dimensional real vector space and let $b_1$ and $b_2$ be two ordered bases for $V$. It is a standard result in linear algebra that there exists a unique linear transformation $A : V \to V$ that takes $b_1$ to $b_2$. The bases $b_1$ and $b_2$ are said to have the same orientation (or be consistently oriented) if $A$ has positive determinant; otherwise they have opposite orientations. The property of having the same orientation defines an equivalence relation on the set of all ordered bases for $V$. If $V$ is non-zero, there are precisely two equivalence classes determined by this relation. An orientation on $V$ is an assignment of $+1$ to one equivalence class and $−1$ to the other."

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    $\begingroup$ Yes, I know those are the definitions, and that's exactly my question: Must the volume of the parallelepiped be defined in terms of the determinant? Can't they be shown to be equal? And must the orientation be defined in terms of the sign of the determinant? Can't it be defined more naturally in terms of continuous motion, and then show that there are really only 2 equivalence classes which are given by the sign of the determinant? $\endgroup$
    – Gabriel Nivasch
    Commented Nov 11, 2012 at 19:01
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    $\begingroup$ Of course I'm not going to define a topology in my course, and I'm not going to define volume either. But I want to know whether to tell my students "The determinant can be shown to equal the volume" or "The volume is the determinant by definition". And regarding the "orientation" thing, I can start by arguing that in the plane one cannot continuously transform $(e_1,e_2)$ into $(e_2,e_1)$ without making the vectors parallel or anti-parallel at some point; then go to 3-space and discuss the same for right-handed and left-handed bases; then discuss the general notion of orientation. $\endgroup$
    – Gabriel Nivasch
    Commented Nov 11, 2012 at 21:10
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    $\begingroup$ @Gabriel Nivasch: regarding the volume question, if you check (standard) books on multivariate analysis you will find transformation formulas for the higher-dim integrals that involve a determinant. The proofs (or possibly some result) will show that you could define the volume differently (via integrating the shape) and then show it matches a differently defined determiannt. $\endgroup$
    – user9072
    Commented Nov 11, 2012 at 21:21
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    $\begingroup$ @quid: transformation-of-variable formula in integral calculus says the same thing as the above, just at the particular context of cotangent spaces. $\endgroup$
    – Margaret Friedland
    Commented Nov 11, 2012 at 22:25
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    $\begingroup$ @Gabriel Nivasch: Did I say that? I don't think so. I stressed that at least the proofs but possibly also a result will show the equality of the two. Typically there will be a definition of some notion of volume via Riemann integral say, as you mention yourself. And there will be a transformation formula with some determinant inside. Apply this formula for the unit cube under a linear map, shows that the volume of the parallepiped (the image of the unit cube under this linear map) and the volume of the unit cube are linked via the determinant. My point is [...] $\endgroup$
    – user9072
    Commented Nov 12, 2012 at 19:36

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