Topology problem

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Prove that the determinant function of an n x n matrix is an open mapping (from [tex]R^{n^2}[/tex] space to [tex]R[/tex])

proving it to be a continuous mapping is easy, determinant function is a sum of products of projections, which are continuous maps.
 

Galileo

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Any fact about projections in general that might help?
 
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The projections are open maps.
But that doesn't help.

multiplication of open maps may not be open map

[tex]x[/tex] is open map but
[tex]x \cdot x = x^2[/tex] is not
 
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Just take the matrices M=diag(a,1,1,1,1,1....)...then

[tex] \lim_{a->\pm\infty}det(M)=\pm\infty [/tex]

Is that what you mean ?
 

Hurkyl

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Hrm. I wonder if

[tex]\mathop{det} e^M = e^{\mathop{tr} M}[/tex]

would be helpful... I think it might be easier to show the trace is an open mapping, and then you might be able to use this to wrap it up.
 
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let U be a neighborhood of your matrix M. Let [itex]a = \inf \{\det X \in U\}[/itex] and [itex]b=\sup \{ \det X \in U\}[/itex]. by the intermediate value theorem, all points in [itex](a,b)[/itex] are equal to some [itex]\det X[/itex]. Thus [itex]\det U \supseteq (a,b)[/itex] and [itex]\det[/itex] is open.
 

Hurkyl

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You've skipped a key step -- you would need to show that a and b are not in det U.
 

mathwonk

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i would try to use the fact that the determinant is a homogeneous polynomial. i.e.that det(tA) = t^n det(A).

that should show that in every nbhd of A there are enough points to get an interval around det(A) in the image?
 
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mathwonk said:
i would try to use the fact that the determinant is a homogeneous polynomial. i.e.that det(tA) = t^n det(A).

that should show that in every nbhd of A there are enough points to get an interval around det(A) in the image?
only certain type of open neighbourhoods of A, not all.
 
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Hurkyl said:
Hrm. I wonder if

[tex]\mathop{det} e^M = e^{\mathop{tr} M}[/tex]

would be helpful... I think it might be easier to show the trace is an open mapping, and then you might be able to use this to wrap it up.
can any matrix be written in exponential form?
 

Hurkyl

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I don't think so -- but I do think it can be written as a product of matrices that can be written in exponential form. I was hoping you'd remember better. :smile:
 

mathwonk

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i think my suggestion does work, kakarukeys. think about it a little more.

i.e. it is rather obvious that in every open nbhd N of A, tA belongs to N for t near 1.

Hence if det(A) is non zero, this means the numbers t^n det(A) do fill up an interval around det(A). Hence det is an open map near every non singular A.

am i missing something?

Now you try it when det(A) = 0.
 

mathwonk

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is something amiss here? just think what open means: i.e. when you wiggle A a little, then det(A) also wiggles.

so wiggle A to tA, and then det(A) wiggles to t^n det(A).

i.e. when t is less than one so is t^n, so det(A) gets smaller, and when t is greater than one, so is t^n so det(A) gets bigger.

since you can make det(A) get both bigger and smaller in every nbhd of A, you are done.

Unless det(A) = 0, then t^n det(A) does not change. so what then?

but really guys isn't this problem rather elementary? or have i totally lost it somewhere?
 

Hurkyl

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Now that I've thought it over more, I see how it works.

You've sliced your open set into sets of the form {tA | a < t < b} and proved that each of those sets get mapped to an open interval in R (assuming det A != 0). Then, the image of your whole open set is simply a union of open intervals.

I never got that last step from your argument. You might have said it, but I just missed it!
 
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OK I will think about it carefully.
My mind tends to be rusty these days.
 
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ya, I got it.
every open ball is a union of [tex]\{(1-\epsilon, 1+\epsilon)A, A\in \bf{the ball}\}[/tex]

and each [tex](1-\epsilon, 1+\epsilon)A[/tex] is mapped into an open interval except singular matrix.

union of open intervals is open.

But unless zero is already in an open interval, or zero is exactly singled out (e.g. [tex](-1, 0) \cup (0, 2)[/tex]), an open interval union {0} is not open.
 
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Just a suggestion

The problem with math wonk's approach is that you need to show that every element of a ball maps into an open interval, and then you need to make a special case for balls that contain singular matrices.

I think you're giving yourself too much undue trouble by looking at open balls in [itex]R^{n^2}[/itex]. The [itex]n^2[/itex] Cartesian product of open intervals is open in [itex]R^{n^2}[/itex], and in fact forms a basis of the topology of [itex]R^{n^2}[/itex].

Also, if we focus on one of the variables, say [itex]x_{ij}[/itex] and imagine all of the other variables fixed, the determinant function is an affine function in [itex]x_{ij}[/itex], i.e. [itex]det(x_{ij})=A x_{ij} +B[/itex], where A and B are of course functions of all of the other variables in [itex]R^{n^2}[/itex]. The great thing about affine functions is that they're open maps.

So, you should be able to show then that the image of [itex] (a_{11}, b_{11}) \times ... \times (a_{nn}, b_{nn})[/itex] is a (very large) union of open intervals, and thus is open. And thus det maps every open set to an open set.
 

Hurkyl

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The great thing about affine functions is that they're open maps.
Not all of them! f(x) = B is not an open map, and it is certainly an affine map that could arise from your method of proof.
 
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Hurkyl said:
Not all of them! f(x) = B is not an open map, and it is certainly an affine map that could arise from your method of proof.

You are correct, sir. I realized this problematic part of my solution as I was walking the dog this morning.

However, there's a way around it by proving that, on the open sets as I described, the determinant function is never constant along any [itex]x_{ij}[/itex] slice, i.e. on [itex](a_{11},b_{11}) \times ... \times (a_{nn},b_{nn})[/itex], the derivative [itex] {\partial \over {\partial x_{ij}}}det[/itex] is not the constant zero function for any i, j.
 

mathwonk

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the point is that the map is open if every nbhd of a domain point A maps to a nbhd of the range point det(A).

so i do not have to slice up the nbhd of A, I only need to observe that every nbhd contains a little interval of matrices of form tA when t is near 1. then these already give me enough values to obtain a nbhd of the range point det(A) by the intermediate value theorem. i.e. i just need to get one value bigger than detA and one smaller.

so i am done, at least when detA is not zero. really this is trivial.

[it would be a little harder to show say that inversion is an open map from non singular matrices to non singular ones, since the range space is bigger dimensional, but fortunately inversion has an inverse, itself.


try showing say that e^A is an open map. or is that what you did? using a local inverse i guess.]


the case where det(A) = 0 is also easy.

i.e. all you have to do is show that every nbhd of a matrix A with zero determinant contains matrices with positive determinants and matrices with negative determinants.

hint: look at matrices of form A + tB, for some appropriate B, and t near zero.
 
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mathwonk said:
the point is that the map is open if every nbhd of a domain point A maps to a nbhd of the range point det(A).

so i do not have to slice up the nbhd of A, I only need to observe that every nbhd contains a little interval of matrices of form tA when t is near 1. then these already give me enough values to obtain a nbhd of the range point det(A) by the intermediate value theorem. i.e. i just need to get one value bigger than detA and one smaller.

The problem with this approach that all it shows is that the image of a nbd. of A contains an open interval containing the number det(A). It doesn't necessarily tell you that the entirety of that image is open, i.e. what about the image of the elements in the nbd. which are not of the form tA?


mathwonk said:
i.e. all you have to do is show that every nbhd of a matrix A with zero determinant contains matrices with positive determinants and matrices with negative determinants.

hint: look at matrices of form A + tB, for some appropriate B, and t near zero.
I thought of this as well, but as far as I know, det(A+tB) does not have a nice form with respect to det(A) and det(B).

Incidentally, my previous "fix" on my approach doesn't quite work either. However, I was able to come up with a proof using induction on n, which is, now that I think about it, the way I solved it in grad school.
 

mathwonk

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let me try again doodlebob: suppose a map f has the property that for every point A, and every nbhd N of A, the image f(N) contains a nbhd of f(A).

then f is an open map.

i.e. let U be any open set in the domain, and f(U) its image. we wish to show that f(U) is open.

so let f(A) be any point of f(U), where A is any point of U, and we will show that f(U) contains a nbhd of f(A).

but this is trivial. i.e. U is itself a nbhd of A, so f(U) contains a nbhd of f(A) by hypothesis. done.

this is what I have almost completely done for the det map. do you agree?
 
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Hurkyl

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I think I see the discreptancy in our interpretations of your proof.

We're looking at it as you've proven certain one-dimensional sets map to an open set, but you're looking at it as you've proven f(A) is in the interior of f(U). (And thus f(U) is open, because all of its points are interior points)
 

mathwonk

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yep. let me be honest here though. i do not now see how to prove that if detA = 0, then det takes both positive and negative values on every nbhd of A. i can prove easily that it takes non zero values simply by joining A to any matrix with non zero determinant by a line.

i would like to know something deep, like perhaps the zero locus of det separates matrix space into two components and is the boundary of both, sort of like in the jordan curve theorem.

i.e. for the result to even be true, you need to show that every A with detA = 0, is an accumulation point of both matrices with det>0 and amtrices with det<0.

of course this is obviously true since there is no reason it to be false.
 

Hurkyl

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For det A = 0, you simply need to add a sufficiently small multiple of diag{1, 1, ..., 1} to get a positive determinant, and diag{-1, 1, 1, ..., 1} to get a negative determinant.


i would like to know something deep, like perhaps the zero locus of det separates matrix space into two components and is the boundary of both, sort of like in the jordan curve theorem.
Hrm. Sounds true. I claim that there is a path connecting any nonsingular matrix A to either the identity = diag{1, 1, ..., 1}, or the matrix diag{-1, 1, ..., 1}, along which the determinant is never zero.

Step 1: there exists a path between any nonsingular matrix A and a diagonal matrix with the same determinant.

The proof is through gaussian elimination -- using only steps of the form "Add a multiple of row m to row n", form a diagonal matrix D.

Each step "Add k times row m to row n" can be turned into an arc via the 1-parameter family "Add tk times row m to row n" (0 <= t <= 1).

Note that the determinant remains constant along this path.

Step 2: reduction to a matrix of 1's and -1's

Now, we have a nonsingular diagonal matrix D. Let B be the diagonal matrix formed by replacing each positive entry of D with 1, and each negative entry of D with -1.

Then, we can form a path connecting D and B as:

(1-t) D + t B

This does not change the sign of the determinant, because the sign of each entry remains the same.

Step 3: reduction to a diagonal matrix with at most one -1.

Suppose the matrix B has more than one -1 in it. In the corresponding plane, form a path by applying a rotation through t radians (0 <= t <= pi). This changes the -1's into +1's, and since the determinant of a rotation is 1, this again leaves the sign unchanged.


To restate: for any matrix A, there is a path connecting A to either the identity, or diag{-1, 1, 1, ..., 1}, and the determimant is never zero along the path.

Therefore, the set of nonsingular matrices has two connected components -- the matrices with positive determinant, and the matrices with negative determinant. The boundary between them is, of course, the singular matrices.


I hope I interpreted the question right, or else all that work for nothing! :smile:
 

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