# Topology problem

1. May 7, 2005

### kakarukeys

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

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

2. May 7, 2005

### Galileo

Any fact about projections in general that might help?

3. May 7, 2005

### kakarukeys

The projections are open maps.
But that doesn't help.

multiplication of open maps may not be open map

$$x$$ is open map but
$$x \cdot x = x^2$$ is not

4. May 7, 2005

### kleinwolf

Just take the matrices M=diag(a,1,1,1,1,1....)...then

$$\lim_{a->\pm\infty}det(M)=\pm\infty$$

Is that what you mean ?

5. May 7, 2005

### Hurkyl

Staff Emeritus
Hrm. I wonder if

$$\mathop{det} e^M = e^{\mathop{tr} M}$$

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.

6. May 7, 2005

### Don Aman

let U be a neighborhood of your matrix M. Let $a = \inf \{\det X \in U\}$ and $b=\sup \{ \det X \in U\}$. by the intermediate value theorem, all points in $(a,b)$ are equal to some $\det X$. Thus $\det U \supseteq (a,b)$ and $\det$ is open.

7. May 7, 2005

### Hurkyl

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

8. May 7, 2005

### mathwonk

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?

9. May 8, 2005

### kakarukeys

only certain type of open neighbourhoods of A, not all.

10. May 8, 2005

### kakarukeys

can any matrix be written in exponential form?

11. May 8, 2005

### Hurkyl

Staff Emeritus
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.

12. May 9, 2005

### mathwonk

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.

13. May 9, 2005

### mathwonk

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?

14. May 9, 2005

### Hurkyl

Staff Emeritus
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!

Last edited: May 9, 2005
15. May 9, 2005

### kakarukeys

OK I will think about it carefully.
My mind tends to be rusty these days.

16. May 9, 2005

### kakarukeys

ya, I got it.
every open ball is a union of $$\{(1-\epsilon, 1+\epsilon)A, A\in \bf{the ball}\}$$

and each $$(1-\epsilon, 1+\epsilon)A$$ 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. $$(-1, 0) \cup (0, 2)$$), an open interval union {0} is not open.

17. May 10, 2005

### Doodle Bob

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 $R^{n^2}$. The $n^2$ Cartesian product of open intervals is open in $R^{n^2}$, and in fact forms a basis of the topology of $R^{n^2}$.

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

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

18. May 10, 2005

### Hurkyl

Staff Emeritus
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.

19. May 10, 2005

### Doodle Bob

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 $x_{ij}$ slice, i.e. on $(a_{11},b_{11}) \times ... \times (a_{nn},b_{nn})$, the derivative ${\partial \over {\partial x_{ij}}}det$ is not the constant zero function for any i, j.

20. May 10, 2005

### mathwonk

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.

Last edited: May 10, 2005