Proving Determinant Function Open Mapping: nxn Matrix

[snoble]

But it does it in a predictable way. So if it switches the sign then the small matrix you added to give a negative determinant will actually give a positive determinant after you undo the row reduction. But the small matrix that you added to give a positive determinant fixes that. To formallize a little take P=diag{1,1,1,...,1} and N=diag{1,1,1,...,-1}. Let S be your singular matrix and let A be the invertable matrix such that AS is upper triangular with only non-negatives on the diagonal. Then for t>0,
det(AS+tN)<0<det(AS+tP).
And so $$det(A^{-1})\cdot det(AS+tN)=det(S + tA^{-1}N)\ne 0, det(A^{-1})\cdot det(AS+tP)=det(S + tA^{-1}P)\ne 0$$ and they have opposite signs. But t can be arbitraly small... small enough so that $$S + tA^{-1}N$$ is inside an epsilon ball of S. Ok so not that formal.

Bah I probably should have just kept my nose out of your guys conversation.

Cheers,
Steven

 

[kakarukeys]

reading the posts...
is the question really that difficult?

 

[Doodle Bob]

reading the posts...
is the question really that difficult?

No, it is not.

Simply induct on n: the size of the matrix.

n=1 is trivial.

Then the determinant of an (n+1)x(n+1) matrix is given by $\Sigma_j (-1)^{j+1}(x_{1j} det(A_{1j}))$ where $x_{11},...,x_{1(n+1)}$ is the top row of the matrix, and [itex]A_{1j}[/tex] are the nxn submatrices of M formed by taking out the first row and jth column (sorry, I can't remember the name of these things).

To show the result then, one needs only to argue that the image of the above sum on the open set $(a_{11},b_{11})\times ... \times (a_{(n+1)(n+1)}, b_{(n+1)(n+1)})$ will be open itself. There's a little work here but it's mostly taken care of by induction and the fact that an open set in R times (literally, multiplied by) another open set in R is an open set.

Thus, a clunky but elementary proof that involves no other knowledge about matrices except how to calculate the determinant. Plus no worrying about zero determinants

 

[Hurkyl]

row reduction changes the sign of the determinant.

I was thinking "using just the operation of adding a row to another", I just forgot to say it this time.

But as snoble points out, a particular redction sequence will either always preserve the sign, or always flip the sign, so this restriction isn't vital.

 

[mathwonk]

oh isee, that doesn't matter since you then undo the reduction by the same processes! I think you've done it in an elementary way! nice work as usual.

(I realized you were right as soon as i left the room and lay down, but did not have the will to get up again.)

(the one operation you mention cannot actually reduce a matrix to echelon form, but as noted your argument works anyway.)

 

[kakarukeys]

The proof by the two advisors is really beyond my head...

To show the result then, one needs only to argue that the image of the above sum on the open set $(a_{11},b_{11})\times ... \times (a_{(n+1)(n+1)}, b_{(n+1)(n+1)})$ will be open itself. There's a little work here but it's mostly taken care of by induction and the fact that an open set in R times (literally, multiplied by) another open set in R is an open set.

I don't understand, please elaborate.
I have shown earlier that the product of two open maps may not be open map.

 

[mathwonk]

the map det is open if on every nbhd of the matrix A, detA takes on values both larger and smaller than det(A).

surely this argument is not over your head for the case where detA is non zero. i.e. in every nbhd of A there are matrices of form tA with t both less than 1 and greater than 1.

then det(tA) = t^n detA, takes values less than detA and rgeater than detA.

that was my original argument.


to do the case of detA = 0, hurkyl said choose an invertible matrix E such that EA is upper triangular. this is called row reduction.

then since the det of an upper triangular matrix is merely the product of the entries on the diagonal, some diagonal entry is zero. to get a non zero determinant, change all zero diagonal entries to small positive numbers. to get a det of the opposite sign change one of those entries to a small negative number.


this shows that in every nbhd of EA there are matrices with both positive and negative determinant.

then the same holds for A because multiplying by E^(-1) just changes all the determinants by multiplying by the same number detE.

doodle bobs argument looks short because he does not actually give it.

 

[Doodle Bob]

The proof by the two advisors is really beyond my head...



I don't understand, please elaborate.
I have shown earlier that the product of two open maps may not be open map.

What I mean is the following: given two open sets U and V in R, if we define W to be the set W={xy: x in U and y in V}, i.e. the set of the products of elements in U and elements in V. Then W has to be open. Essentially this boils down to showing that the set of products of two open intervals is in fact an open interval, e.g. U=(-1, 1) and V=(0,5) yields W=(-5, 5).

I absolutely admit that my "proof" is not complete. However, the main hook is there: we can write the determinant of an (n+1)x(n+1) matrix as the sum of n variables ($x_{11}, ..., x_{1n}$) times the determinants of nxn matrices. Using the open sets that I described, we can look at each addend and see that the image will be an open set (using the fact that nontrivial linear maps are open and that, by induction, the determinant on nxn matrices is also open).

Let me reiterate that this is the skeleton of the proof. It needs to be fleshed out some.

 

[Hurkyl]

So you're claiming that multiplication is an open map from UxV --> W...

Anyways, there is a problem with your approach: notice that the function:

f(x, y) = xy - xy

is precisely of the form you describe, but it is obviously not an open map!

 

[Doodle Bob]

So you're claiming that multiplication is an open map from UxV --> W...

Anyways, there is a problem with your approach: notice that the function:

f(x, y) = xy - xy

is precisely of the form you describe, but it is obviously not an open map!

Actually, no, it is not. I said, nontrivial linear functions, i.e. not including the zero function, which you will never get on an open set of matrices.

 

[Hurkyl]

That is a nontrivial linear function: I'm composing the function f(x, y) = x - y with the functions g(x, y) = xy and h(x, y) = xy to get f(g(x, y), h(x, y)) = 0.

 

[Doodle Bob]

Well, if a linear function is identically zero, that's pretty trivial to me. But nevertheless let me give you an updated form of my proof. I call this an elementary proof, since all it uses is the definition of the determinant and induction.


First, we start with two lemmas, the proofs of which are super-easy and I will leave to the reader.

Lemma 1: Let $O_1$ and $O_2$ be open sets in R. Then the set $O=\{x+y\in R: x\in O_1, y\in O_2\}$ is an open set of R.

Lemma 2: Let $O_1$ and $O_2$ be open sets in R. Then the set $O=\{xy\in R: x\in O_1, y\in O_2\}$ is an open set of R.

Let $det_n$ be the determinant function on $n\times n$ matrices. We will use induction to prove the following claim.

Claim: $det_n$ is an open mapping for all natural numbers n.

n=1: this is trivial, since $det_1$ is the identity map on R

Suppose the claim is true for $det_1, ..., det_{n-1}.$

Let U be the open set in $R^{n^2}$ given by $U=(a_{11},b_{11})\times … \times (a_{nn},b_{nn}),$ where $a_{ij}<b_{ij}\forall\ i,\ j.$

For each i, j, set $V_{ij}=(a_{ij},b_{ij}),$ an open interval in R, and let $W_{ij}=\pi_{ij}(U)\subset R^{(n-1)^2}$, where $\pi_{ij}:\ M_{n\times n}(R) =R^{n^2} \rightarrow M_{(n-1)\times (n-1)}(R) =R^{(n-1)^2}$ is the projection map that takes out the $i^{th}$ row and $j^{th}$ column of an $n\times n$ matrix. Note that $W_{ij}$ is an open set of $R^{(n-1)^2}$, since it can be written in the form of $(c_{11},d_{11})\times … \times (c_{(n-1) (n-1)},d_{(n-1) (n-1)})$, where $c_{kl}<d_{kl}\forall\ k,\ l$.

Then, for any $A=(A_{ij})\in U$, $det_n(A)=\Sigma_{j=1}^n (-1)^{j+1}A_{1j} det_{n-1}(\pi_{1j}(A))$. So, the image of the determinant on U can be written as
$det_n(U)=\Sigma_{j=1}^n (-1)^{j+1}det_1(V_{1j}) det_{n-1}(W_{1j})$.

By induction, both maps, $det_1$ and $det_{n-1}$, are open maps. So, that each of the terms $det_1(V_{1j})$ and $det_{n-1}(W_{1j})$ for any $j=1, ..., n,$ are open sets in R. Thus, by the two lemmas, $det_n(U)$ is an open set in R.

That pretty much does it.

 

[mathwonk]

thats a lot better, but you still seem to be missing the key point of hurkyls example.

i.e. in the definition of det(U) you have to use the same matrix in every term of the formula for det.

But your application of your lemma requires you to be able to vary the matrices independently over the different terms.

so although i am beginning to believe you, i do not think you have proved your claims, especially the key one in the formula for the image of det(U).

i.e. by hurkyl's example, although the sum of two open sets is always open, the sum of two open maps is not always open, and you are apparently using that false statement in your proof.

i.e. although 0 = xy-yx, it is not true that the image of an open interval under the map 0, equals the difference of its images under xy and yx.

so you seem to be assuming that the image of an open rectangle under the map det, equals the sum of its images under the terms of your sum.

this requires proof since it can fail.

but you have noit used any hypotheses that distinguish your situation from hurkyl's.

i.e. you have to use somehow that the terms in the sum do not cancel each other in any such unfortunate way.

and even this inadequate proof is longer than hurkyls correct one.

i admit the mistake is subtle. you had me convinced for a minute there.

 

[Doodle Bob]

thats a lot better, but you still seem to be missing the key point of hurkyls example.

i.e. in the definition of det(U) you have to use the same matrix in every term of the formula for det.

But your application of your lemma requires you to be able to vary the matrices independently over the different terms.

It's bit more subtle than that. The proof uses the fact that there isn't really just one determinant function. There is a different determinant function defined for each size of a square matrix. By using induction, we can already assume that the determinant functions defined on the smaller sized matrices (ii.e. of sizes 1x1, 2x2, ..., (n-1)x(n-1)) are open mappings. In particular, if W is any open set of $R^{k^2}$ for k=1, ..., n-1, then det(W) is open by the induction hyphothesis.


Using this induction hypothesis, the lemmas, and the fact that the determinant on nxn matrices is actually defined recursively as the sum of determinants of smaller matrices, we get that the determinant on nxn matrices is open as well.

Quoth mathwonk:

"so although i am beginning to believe you, i do not think you have proved your claims, especially the key one in the formula for the image of det(U).

"i.e. by hurkyl's example, although the sum of two open sets is always open, the sum of two open maps is not always open, and you are apparently using that false statement in your proof.

"i.e. although 0 = xy-yx, it is not true that the image of an open interval under the map 0, equals the difference of its images under xy and yx.

"so you seem to be assuming that the image of an open rectangle under the map det, equals the sum of its images under the terms of your sum.

"this requires proof since it can fail."


Be careful here. Nowhere do I use that the addition or multiplication of open *mappings* is open; I say that the "addition" and "multiplication" of open *sets* (as defined in the lemmas) are open sets.

This boils down to: let (a,b), (c,d) be non-empty intervals in R. Then the set $O_+$ defined by $O_+=\{x+y\in R: x\in(a,b), y\in (c,d)\}$ is the open interval (a+c, b+d), the set $O_\times$ defined by $O_\times=\{xy\in R: x\in(a,b), y\in (c,d)\}$ is actually the open interval $(min\{ac, ad, bc, bd\},max\{ac, ad, bc, bd\})$.

This needs some proving and one then has to show that it's true for a general open set. But that's easy enough.

Nevertheless, by writing det(U) as the sum of products of open sets in R (using again the induction hyphothesis), we have that det(U) is open.


"but you have noit used any hypotheses that distinguish your situation from hurkyl's.

"i.e. you have to use somehow that the terms in the sum do not cancel each other in any such unfortunate way.

"and even this inadequate proof is longer than hurkyls correct one.

"i admit the mistake is subtle. you had me convinced for a minute there."

The terms may cancel each other out at individual points of the open set of U, but that doesn't matter. What matters is that the image of U is an open set of R. And the proof shows that.

Incidentally, hurkyl's example never pops up. The variables x_11, ..., x_1n show up exactly once in the definition of the determinant I'm using, since the other smaller matrices do not use the first column at all. They are the submatrices formed by taking out the 1st column and the jth row. And, we don't have to worry about the determinants of the smaller matrices because of induction.

As for the length, I wasn't concerned so much about that as what machinery was involved. As is usually true in math, the more machinery, the shorter the proof. I just knew that there was a proof that did not involve Jordan canonical forms.

 

[mathwonk]

Doodle Bob, the following claim of yours:

"Nevertheless, by writing det(U) as the sum of products of open sets in R (using again the induction hyphothesis), we have that det(U) is open."

is exactly what you have not proved. I.e. in your 5th and 6th sentences from the end, you essentially say: "because the function det(n) is a sum of prodicts of determinants of lower degree, it follows that the image det(n)(U) is the analogous sum of products of the images of U under those lower degree determinants."

this is equivalent to claiming that the sum of products of those open maps is itself an open map. that may be true but you have not proved it nor even argued it.

since it is not true that the sum of products of arbitrary open maps is open, you still need to prove it is true for your case.

and hurkyls proof does not use jordan forms, only the elementary fact that for any matrix A there exists an invertible matrix E such that EA is upper diagonal. this follows from row reduction, the first technique taught in matrix theory.

and my proof for the special case that detA>0 say, uses only that if 0<t<1, then t^n detA < detA, while if t>1 then t^n detA) > detA.

it doesn't get much shorter or more elementary than that.



let me make my objection to your proof as simple as possible:

just because h = f+g, it does not follow that h(U) = f(U)+g(U), which is what you are apparently claiming in your argument. this is what hurkyl's counter example shows.

i.e. h(U) consists only of all points of form f(x)+g(x) for all x in U, while f(U)+g(U) consits of the much larger collection of points of form f(x)+g(y) for all x,y in U.

I.e. f(U) +g(U) is the image of the product set UxU under the map f+g, while h(U) is the image only of the diagonal.

If I am misunderstanding you, and I may be, please tell me how.

 

[Doodle Bob]

let me make my objection to your proof as simple as possible:

just because h = f+g, it does not follow that h(U) = f(U)+g(U), which is what you are apparently claiming in your argument. this is what hurkyl's counter example shows.

i.e. h(U) consists only of all points of form f(x)+g(x) for all x in U, while f(U)+g(U) consits of the much larger collection of points of form f(x)+g(y) for all x,y in U.

I.e. f(U) +g(U) is the image of the product set UxU under the map f+g, while h(U) is the image only of the diagonal.

Well, now that you put it that way, I see the problem. I tried to eliminate that problem by writing the addends in the form of h(x,y)=f(x)g(y), but now I see that the sum of them does not quite conform to the way I wanted it to.

 

[mathwonk]

here is simpler conundrum for you: your type of argument would also prove that det^2 is an open mapping whereas clearly it is not, even for n=1.

 

[mathwonk]

ok here's a quicky solution, inspired by hurkyl's:

we must show that in every nbhd of any matrix A with det = 0, there are matrices whose determinants have both signs.

first of all, note that most polynomials of degree n have n distinct roots (those that do not lie on the hypersurface where the discriminant equals zero).

hence in every open set around A there are matrices which are diagonalizable. now it is trivial that on every nbhd of a diaonalizable matrix of det zero, there are matrices with det of boths signs. thus we are done.

i.e. if A is diagonalizable, the for some invertible P, P^-1 AP = D is diagonal. Hence there are matrices near D with determinants of both signs, and thus applying
P(.)P^-1 to them yields matrices near A with determinants of both signs.