# Proving Determinant Function Open Mapping: nxn Matrix

• kakarukeys
In summary: This follows from the fact that the determinant is a homogeneous polynomial.In summary, the determinant function of an n x n matrix is an open mapping. This can be proved by showing that it is a continuous mapping, which is easy to do since it is a sum of products of projections, which are continuous maps. Using the fact that projections are open maps does not help in this case. It is important to note that not all affine functions are open maps, so the approach of using affine functions to prove the open mapping property must be done carefully. One possible solution is to show that on the open sets as described, the determinant function is never constant along any x_{ij} slice, which
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.

Any fact about projections in general that might help?

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

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 ?

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.

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.

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

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?

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.

Hurkyl said:
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.

can any matrix be written in exponential form?

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.

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.

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?

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.

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.

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.

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.

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 $x_{ij}$ 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.

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.

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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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)

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.

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!

ok i think i see how to show that GL(n,R) has exactly two connected components using row and column operations.

i.e. i think i mentally connected each matrix with det = +1 to the identity and each matrix with det = -1 to the diagonal matrix with a single -1 in the top corner and all other diagonal entries +1.

now it should follow that every nbhd of a matrix with det=0 contains a mtrix of both kinds?

sorry hurkyl i did not see your proof of this.

"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."

is one i thought was true the other day but why is this true? that would do the openness problem of course immediately, right?

but i am worried that the taylor series for y = det(A+xB) might have only positive values for all x near zero. i.e. the curve A+xb might not cross from one component of GLn to the other.

what am i missing?

this is what bothers me: take the diagonal 3 by 3 matrix with -1 in top corner and zeroes elsewhere. then add t times the dientity, the det is (t-1)t^2. this is not positive on any small nbhd of t =0. so your first statement is false.

mathwonk said:
this is what bothers me: take the diagonal 3 by 3 matrix with -1 in top corner and zeroes elsewhere. then add t times the dientity, the det is (t-1)t^2. this is not positive on any small nbhd of t =0. so your first statement is false.

The reason why you're having trouble here is that the set you describe above is not open in the space of 3 by 3 matrices.

mathwonk said:
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?

The above is kosher. But your specific application of it to the situation at hand is a little slippery. Looking at the image of det(tA) for small enough t, works fine if the neighborhood U contains no A such that det(A) is not zero, and if n is odd. It does not work for n even, since that is precisely the case when the map t --> t^n is not open. In particular, the image of det(tA) will be [0,e) or (-e,0] for a small enough e>0.

t --> t^n is open near t = 1, though, and that's all that matters.

Aw phooey. I forgot it could have a nonzero eigenvalue. But, I have it sealed up now.

Use a basis in which A is in Jordan canonical form.

Then, you can choose your diagonal matrix to only have nonzero entries where A's diagonal entries are zero, and that way you can get a positive or negative determinant.

i agree, doodle bob in fact i tried to point that problem out in post 27 above.

i am sure you all agree that i have given a 3 by 3 counter example to the following assertion:

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

and I apologize for tapping your creative powers hurkyl to answer my dumb questions, not all of which are relevant to the current problem.

But it is very interesting to me, to know that G(n,R) has exactly two components, and to see that your proof is similar to my own, as that reassures me greatly as to the accuracy of my own insight.

it now seems clear however that the relevant point for this problem is your other assertion, given without proof, that the discriminant locus: {A:detA = 0} is the common boundary of both components of GL(n,R).

i.e. the assertion that det is an open map on all of matrix space is proved as follows:

if the scalars are complex it is immediate by restriction to any line thorugh A on which det is not constant, by the openness of non constant analytic functions.

if we restrict to real A, with det not zero, i.e. to GL(n,R), it is open by my trivial homogeneity argument.

near an A with detA = 0, it is not as obvious to me by any trivial argument, but provable as follows:

the point is to show that on every nbhd of A with detA = 0, det changes signs.

first consider the "rank filtration" of the discriminant locus.

1) the locus where A has rank = n-1 is open and dense in the discriminant locus ( = locus where rank is less than n). this seems easy by arguments above of A+tB type.

2) the rank n-1 locus is the smooth locus on the hypersurface {det=0}, i.e. at every point where A has rank n-1, the locus {det=0} is a manifold of dimension n-1, with a well defined n-1 dimensional tangent space. this is "well known".

3) at a point A where A has rank n-1, if we restrict det to a line normal to the tangent plane to this manifold, then det has non zero derivative at A, hence changes sign on every nbhd of A.

i claim this doos it. what do you think?

[the problem with naive A+tB arguments near a point where A has rank less than n-1, is to show the line A+tB actually passes from one "side" of the discriminant locus to the other, as indeed my 3 by 3 example shows it need not do.]

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well as usual hurkyl, i failed to see your post before posting my own, because i did not "turn the page". anyway it is fun to see two different arguments.

oh by the way jordan canonical form does not exist over the reals, the only place where the problem is interesting.

i am extremely proud of my solution in the post above but i assume hurkyl is not reading it for the same reason i did not read his, he wants to give his own solution.

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Aw phooey. Guess it shows I really need to actually learn something about Jordan Canonical Form.

Okay, I'll try again:

Row reduce A.
Add the appropriate sufficiently small diagonal matrix to produce a positive or negative determinant.
Invert the row reduction.

Voila, a recipe for generating matrices near A with positive and negative determinants.

row reduction changes the sign of the determinant.

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