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January 13

Cube root function on the space Mn of matrices

Hi all:

I've been asked by my supervisor to investigate the possibility of the existence of a continuous square root function in some open ball about the identity matrix in $M_{n}$ ; that is, a continuous function g, with $[g(A)]^{2}=A$ for all matrices A in this open ball about I. Now I've used the inverse function theorem and the fact that $f(A)=A^{2}$ is continuously differentiable about I to prove that the inverse of f exists and is continuous in some open ball about I (an I-ball, as it were - oh dear...), but I also managed to prove that you can't extend such a function to the whole space $M_{n}$ , because a matrix like, for example,

[0,1

0,0]

has no square root (I don't know how to LaTeX a matrix, sorry!). However, I couldn't find any such matrix which has no cube root: does there exist then a continuous 'cube root' function which extends to the whole of $M_{n}$ ? (I presume that every matrix does have a cube root, and so a discontinuous function would be fairly trivial by defining it pointwise, right?) My intuition tells me that no such function should exist - but how would I go about proving it? If not, why am I wrong?

Thanks in advance for any responses, Typeships17 (talk) 13:57, 13 January 2010 (UTC)[reply]

The claim that every matrix has a cube root sounds highly unlikely, and in fact I think that

A={\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}}

is a counterexample. Let us assume that B³ = A, and write B = P⁻¹CP, where P is invertible, and C is in Jordan normal form. Note that A = P⁻¹C³P. If J is a Jordan block of C with eigenvalue λ, then J³ appears as a block of C³, and it is a triangular matrix with λ³ on the diagonal, hence λ³ is an eigenvalue of A. However, the only eigenvalue of A is 0, hence λ = 0. Thus C is a triangular matrix with 0 on the diagonal, hence C³ is a triangular matrix whose main diagonal as well as two adjacent shifted diagonals are zero. As it is a 3-by-3 matrix, it is in fact the zero matrix, thus A = P⁻¹C³P = 0, a contradiction. — Emil J. 14:48, 13 January 2010 (UTC)[reply]

Actually, now that I think about it, your matrix

{\begin{pmatrix}0&1\\0&0\end{pmatrix}}

should have no cube root either, by pretty much the same argument. — Emil J. 14:55, 13 January 2010 (UTC)[reply]

The way I see it is: if N is any nilpotent nxn matrix, then Nⁿ=0 (because of the normal form of nilpotent matrices, which is just their Jordan form. Check also PST's hint to a close question below). This fact implies a necessary condition on the existence of a p-th root of N. Indeed if X^p=N then X is nilpotent too, hence Xⁿ=0, therefore N is more nilpotent than X: N^k=0 as soon as kp≥n. In particular a nilpotent N such that N^n-1≠0 (for instance the above mentioned shift matrix) has no root of any order p>1. (not even if you allow complex coefficients) --pm a 15:34, 13 January 2010 (UTC)[reply]

Also recall that an analytic p-th root in the unit ball around the identity is provided by the binomial series, that works in any Banach algebra A. You may also like to investigate the set of all square roots of the identity: it is an analytic submanifold of A (in general not connected); the tangent space at x is given by the closed linear subspace V of all v that anti-commute with x, that is vx=-xv. Note that A splits as A=V⊕W where W is the closed linear subspace of all w that commute with x. --pm a 16:16, 13 January 2010 (UTC)[reply]

If ever you forget how to LaTeX a particular mathematical notation, go to the corresponding Wikipedia article, click the "edit page" tab, and observe how the LaTeX is done. By repeatedly following this procedure, you will quickly learn how to LaTeX the given notation. For instance, to LaTeX a matrix, go to Matrix (mathematics), and notice that some examples of matrices are given at the beginning of the article. Now, click the "edit page" tab at the top, and you can observe the exact LaTeX for the particular matrix shown. At this point, it should not be difficult to duplicate the LaTeX for other matrices. Of course, you can also see how the above posters have written matrices in LaTeX, in this instance, by simply editing this section and viewing the underlying <math></math>. --PS T 03:36, 14 January 2010 (UTC)[reply]

In fact, in this case, I can give you an example of how to LaTeX a particular matrix:

Visible LaTeX:

<math> \begin{bmatrix} 1 & 9 & 13 \\ 20 & 55 & 4 \end{bmatrix}. </math>

Appearance:

${\begin{bmatrix}1&9&13\\20&55&4\end{bmatrix}}.$

For the matrix you mentioned, the following LaTeX works:

Visible LaTeX:

<math> \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}. </math>

Appearance:

${\begin{bmatrix}0&1\\0&0\end{bmatrix}}.$

It is simple if you try it a couple of times, and become accustomed to the LaTeX, but you are of course not obliged to do so (but just in case you wish to learn how to do so, I have included these tips). Hope this helps. --PS T 03:43, 14 January 2010 (UTC)[reply]

And there's also Help:Formula. -- Meni Rosenfeld (talk) 08:37, 14 January 2010 (UTC)[reply]

finding the centre of a circle

Firstly thanks for any help and if wiki has an article on this topic already then can you please point me in the right direction?

Anyway my problem is this i have a set of (x,y) co-ordinates randomly distributed inside a circle of radius R and i'd like to find out where the centre of the circle is. At best if we assume all points are evenly distributed over the circle then i guess calculating the mean of the (x,y)'s would yield the centre, but at worst, say all points clustered in one quadrant taking the straight mean would give a wrong answer. Is there a solution to this problem?--86.27.192.94 (talk) 17:02, 13 January 2010 (UTC)[reply]

Without knowing the distribution, you cannot definitively state where the center is. By omitting points in one area of the circle, you have lost information and cannot recreate it. -- kainaw ™ 17:24, 13 January 2010 (UTC)[reply]

No. If they are randomly distributed then the best you can do if find an expected value for the centre (whether that is the mean of the coordinates or not will depend on the details of the distribution). It would be unlikely that the points will be significantly skewed, so that expected centre should be close to the actual centre, but it is entirely possible for it to be a long way off. --Tango (talk) 17:27, 13 January 2010 (UTC)[reply]

If you want something ~~non-random~~ defined as a function of the set of points, then there is a unique closed disk of minimum radius containing your bounded set of points; it depends continuously on the set wrto the Hausdorff distance; it's indeed Lipschitz continuous of constant 1 --this is true in a Hilbert space too. --pm a 17:48, 13 January 2010 (UTC)[reply]

Yet again Wikipedia has the answer - Smallest circle problem. Dmcq (talk) 18:44, 13 January 2010 (UTC)[reply]

No, that's wrong. The "smallest circle problem" doesn't help here at all. Your comment seems very rash. Michael Hardy (talk) 05:07, 15 January 2010 (UTC)[reply]

That is really an answer to the OP's question, although it may be the best the we can do. The OP's circle has a fixed radius, though, so finding the smallest circle won't necessarily help. --Tango (talk) 18:54, 13 January 2010 (UTC)[reply]

Sorry, I'm not sure I got what you meant: did you mean "That isn't "? Anyway. the OP question is interesting but clearly ill-posed. It's not even clear if R is known or not. So something has to be clarified and possibly changed. If the point is, to have a unique ~~non-random~~ solution, then I suggested to modify the question as I wrote. --pm a 22:26, 13 January 2010 (UTC)[reply]

Assuming the distribution is uniformly random, i.e. each point in the circle has an equal chance of being sampled, can we give good estimators of center and radius? I guess in that case the best estimator for the center would indeed be the mean (and it should be unbiased, I think). If we have the center, the distance to the farthest point is a lower bound for the radius. As the sample size goes towards infinity, it should also converge against the true radius. However, for any finite set it's bound to underestimate the true radius. I suspect we can do better... --Stephan Schulz (talk) 18:55, 13 January 2010 (UTC)[reply]

The OP says the radius is R, which I interpreted as meaning R is known, so we only need to worry about the centre. Assuming a uniform distribution (which actually means any region of the circle has a probability of being sampled equal to the ratio of its area to the area of the whole circle - if you look at individual points you just get the probability for every point is zero, which doesn't tell you much) then I agree that the mean should be unbiased. --Tango (talk) 19:29, 13 January 2010 (UTC)[reply]

I'd still like to know if we can estimate the radius as well...

I can imagine some sets of points where the mean is not going to be the best estimate. For example if I have the points (0,0), (1,0), (2,0) and (100,0), I don't think this provides any different information than if I had only the points (0,0) and (100,0). ~~Wouldn't you also need some prior distribution on what the radii of the circles tend to be?~~ 67.100.146.151 (talk) 21:14, 13 January 2010 (UTC)[reply]

Sure, in non-trivial cases you can always find a sample that will mis-estimate the distribution. The charm of an unbiased estimator is that for a random sequence it will stolastically converge to the true value. --Stephan Schulz (talk) 21:40, 13 January 2010 (UTC)[reply]

(the IP 67... is me.) ~~Actually I was wrong. The two cases give different information by merit of the fact that they have a different number of points.~~ But the idea still holds. A simple average is not the best you can do.

I expect one could readily prove that on average the "simple average" performs considerably better than the center of the smallest disk enclosing the data. Michael Hardy (talk) 03:20, 14 January 2010 (UTC)[reply]

Edit: Sorry, didn't notice that the OP specified that the radius is already known. Rckrone (talk) 22:42, 13 January 2010 (UTC)[reply]

We can find the region S_R formed by the centers of all possible circles of radius R that enclose the points in the initial set (it'll be the intersection of the disks of radius R centered at each point in the set). I'm pretty sure there's an even distribution of where the center is likely to be in S_R. We can think about if we pick a circle and then choose points at random, the chance of the points being clustered in one region of the circle is the same as it is in any other. So the average location of the center over all those possible circles should be the centroid of S_R. I don't know if there's a straight forward way to calculate that though. Rckrone (talk) 22:50, 13 January 2010 (UTC)[reply]

If I understand what you said, S_R is the disk with the same center as the minimal disk, and radius R-r (r being the radius of the minimal disk). Correct?--pm a 23:14, 13 January 2010 (UTC)[reply]

S_R contains that disk of radius R-r, but I think it will generally be larger since a disk of radius R that contains the points in the set doesn't have to contain the minimal disk. Rckrone (talk) 23:48, 13 January 2010 (UTC)[reply]

right :) --pm a 00:04, 14 January 2010 (UTC)[reply]

A few points about the discussion above:

Tango writes of an expected value of the center. The average location of observed points is the expected value only of the discrete distribution of the set of points actually observed; it's not the expected value of the center. Only if there's a prior distribution, and hence a posterior distribution, can you speak of an expected value of the center.
The observed average is indeed unbiased.
pma proposed unique smallest disk enclosing the observed points is not "non-random" if those observed points are random.
One should be able to find confidence regions for the center; these would be large if the number of points is small. I'd have to work out details though.

Michael Hardy (talk) 03:16, 14 January 2010 (UTC)[reply]

I think people (at least I was) were working under the assumption that the prior distribution of the center is an even distribution on R² (or on some really large closed disk if that makes more sense. As long as the points in the set aren't near the edge it doesn't matter). I'm not really sure confidence regions are really useful here, unless I'm misunderstanding the problem. There's a well defined region where the center could be, and the posterior probability is evenly distributed across it. Outside that region the probability is zero. Rckrone (talk) 05:14, 14 January 2010 (UTC)[reply]

"posterior probability is evenly distributed across it" That is WRONG. I didn't even notice this comment the first time I commented. The posterior probability, with any reasonable prior, would be highly concentrated near the mean of the data points, and get less concentrated as you moved further away, and the distribution would have unbounded support unless the prior (i.e. pre-data) distribution had bounded support. Your comments are really seriously confused. Michael Hardy (talk) 05:04, 15 January 2010 (UTC)[reply]

That's why I said "(or on some really large closed disk if that makes more sense. As long as the points in the set aren't near the edge it doesn't matter)". Obviously you can't really have an even distribution on the whole plane, but an even distribution on any large bounded region doesn't effect the outcome of the problem as long as the points are not within 2R of the edge of the region. We can essentially ignore its specific properties. Rckrone (talk) 05:33, 15 January 2010 (UTC)[reply]

You seem confused. The uniform distribution on the whole plane is not a proper probability distribution; it assigns infinite measure to the whole space. I doubt that's what people assumed. I think they were assuming the DATA are uniformly distributed, not within the whole plane, but within the interior of the circle. That is NOT a prior distribution of the center of the circle. Given the data, the center of the circle COULD be anywhere in the plane, even quite remote from a disk-like cloud of data point, but that's improbable---it means by some accident all the points landed close together, far from the center. At any rate, "prior" and "posterior" refer to the probability distribution of the center, not of the data. Using frequentist methods, one wouldn't usually have prior and posterior distributions, but only a distribution of the data. Michael Hardy (talk) 04:54, 15 January 2010 (UTC)[reply]

The only explanation I have for what you're saying is that you're thinking about a quite different problem than I am. Here is the problem as I understood it: You have some large large bounded region of the plane and you choose a point P at random according to an even distribution on the region. Then you draw a circle of radius R centered at P and pick some number of points at random from inside the circle according to a even distribution on that disk. Given that set of points and the radius R, describe the probability of where P lies.

In the context of that problem, P cannot lie farther than R from any of the points in the set. It CAN'T be anywhere in the plane, rather it's restricted to the region S_R that I described above. Moreover, the posterior distribution is an even distribution on S_R.

Maybe you are assuming that R is not given? In that case (a) you need to know a prior distribution of the values the radius can take in order to answer the question (as I said and then crossed out when I modified my interpretation of the question) and (b) the prior distribution for the location of P that I described makes less sense since the edge effects will always matter for any set of data points, unless the prior for the radius is bounded. Rckrone (talk) 05:33, 15 January 2010 (UTC)[reply]

OK, I think I did neglect that fact that R is known. That apparently does simplify things a lot. Michael Hardy (talk) 07:18, 15 January 2010 (UTC)[reply]

yea, this seems a pretty good description (what else could be said?) --pm a 06:26, 14 January 2010 (UTC)[reply]

Study a simplified problem first. Consider the one-dimensional case of points inside the interval a<x<a+1, and estimate a. Bo Jacoby (talk) 08:12, 14 January 2010 (UTC).[reply]

OP here, thanks for the help so far if it helps i'll describe the actual situation, i initially said "taking point (A,B) get all data within a circle of radius R, now some time later (after losing track of the individual (A,B)), i have all the data points and would like to reconstruct the original (A,B), or even just the best guess at what it would be--86.27.192.94 (talk) 08:37, 14 January 2010 (UTC)[reply]

Well that gives a whole lot more information. You have a whole lot of points which weren't within radius R, so this circle of radius R must be somewhere in the gap between these two sets of points. For instance if you were choosing all towns of more than 10000 inhabitants within 10 miles of a point you might only get three towns and the towns you didn't choose might be as good or better at determining the point.

Where the problem given the points inside determines a convex polygon with curved sides of radius R within which the center must lie, excluding points means you have sides that bend the other way and the figure isn't convex. The corners where the curves intersect still form a convex figure though as far as I can see so it should be a fairly reasonable problem for a computer program. Dmcq (talk) 09:31, 14 January 2010 (UTC)[reply]

Sigh. OK, I wrote some comments neglecting the fact that R is known. But some of what's said about makes me think some people are confusing the terms "prior" and "posterior" with the distribution of the DATA given the location of the center. Michael Hardy (talk) 05:20, 15 January 2010 (UTC)[reply]

I think I'll sigh even deeper. If you consider the points which aren't included the possible positions of the centre with a fixed R needn't even be connected never mind anywhere near convex. If you have three points excluded at the corners of an equilateral triangle and one included point at the centre then with R = half the length of a side or a bit more you get four completely separate areas where the centre can be. Dmcq (talk) 09:29, 15 January 2010 (UTC)[reply]

By the way using Bo Jacoby's very simple version of an interval of length 1 the midpoint of the interval containing the samples is a better estimate of the center of the interval than the mean - see Uniform_distribution_(continuous)#Estimation of midpoint. Generalizing to the circle case and ignoring any points outside I'd be pretty certain the center of the minimum circle would be a better estimate of the center than the mean though of course working out S_R as above would give a better estimate. In fact just getting the midpoint of the minimum and maximum x and y values would probably be a good quick and dirty way of estimating the center. Dmcq (talk) 13:20, 15 January 2010 (UTC)[reply]

Concerning the interval: that point is certainly correct even if the length were not known. I suspect that difference between the problem with known radius and that in which one is left to estimate the radius based on the data is more extreme than that. But maybe you're right in your guess about the circle. Michael Hardy (talk) 22:09, 15 January 2010 (UTC)[reply]

Isn't the center of the minimum circle enclosing S_R the same as the center of the minimum circle for dataset? If so, this center is still the best estimate (in terms of expected value) even without given R or it's distribution (as far as center location doesn't affect this distribution).--95.84.241.53 (talk) 13:25, 17 January 2010 (UTC)[reply]

PS I was wrong:) At least the center of minimum circle of dataset is a good estimate (and the right one when R is close to r of the minimal circle). The centroid of S_R is an unbiased estimate, and should be pretty close to the former.

The average value of the dataset, for a very skewed dataset, can give an answer that is not just improbable but impossible. Say the dataset is {0,100,100,100} and R = 60. The mean is 75, but that would leave one point completely outside the acceptable range. Black Carrot (talk) 23:11, 17 January 2010 (UTC)[reply]

Nilpotent homomorphisms

Let A∈Hom_K(Kⁿ,Kⁿ) and n∈ℕ. Which consequences has the fact that A²=0? --84.61.151.145 (talk) 17:50, 13 January 2010 (UTC)[reply]

For instance, it has a normal form with 2x2 blocks (plus some 1x1 0 blocks, at least one if n is odd), hence few 1's (at most n/2). --pm a 17:55, 13 January 2010 (UTC)[reply]

What about Im(A) and Ker(A)? --84.61.151.145 (talk) 18:03, 13 January 2010 (UTC)[reply]

Of course Im(A) is included in ker(A) (so e.g. the quotient H=ker(A)/im(A) is well defined); further information may be seen via the normal form above. Now that you say it, I have a vague memory that somebody in some book defines a "module with differentiation" to be a module M with an endomorphism A such that A²=0, as an elementary example of chain complex. I think I saw it in Jacobson's "Basic Algebra II" (typical of it). --pm a 18:08, 13 January 2010 (UTC)[reply]

The way in which these questions were phrased suggests that they constitute homework. As pma has already confirmed that Im(A) is contained in Ker(A), the OP might like to consider the following exercise. If V is an n-dimensional vector space over a field K, and A is a linear transformation of V over K with the property that A^p = 0 for some p, show that Aⁿ = 0 (n is the dimension of V over K). A hint is hidden below, but make a solid attempt at the problem before viewing the hint to attain a satisfactory intuition of nilpotent transformations.

(Click the "show" button at the right to see the hint or the "hide" button to hide it.)

Prove that $\{0\}\subset \ker A\subset \ker A^{2}\subset \ldots \subset \ker A^{p-1}\subset \ker A^{p}=V$ is an ascending chain of subspaces (that each inclusion is indeed correct).

Hope this helps but please note that in order to learn anything, you must show us that you have made an attempt at the problem. Simply firing questions at us in quick succession will not aid your intuition of nilpotent transformations. --PS T 03:29, 14 January 2010 (UTC)[reply]

How to prove that Kⁿ has a basis B with _BA_B= ${\begin{pmatrix}0&0\\I_{r}&0\end{pmatrix}}$ , where r is the rank of A? --84.61.165.65 (talk) 15:18, 14 January 2010 (UTC)[reply]

If you trust the above normal form, note that the number of 2x2 blocks in it is r, and that that matrix differs from the one you wrote only by a permutation of the base. If you want an independent proof, start with PST's hint. --pm a 20:30, 14 January 2010 (UTC)[reply]

Your most recent question suggests that you have not yet attempted to visualize the linear operator A. Visualization can prove an important tool in determining canonical forms, and thus it may increase your chances of solving the questions you pose. Personally, I feel that your most recent question should not require a formal proof in that it is very intuitive; you need only formally pinpoint the intuition (which is done in my hint above). Since you are unable to solve it, either you do not have any intuition of A at all, or you do not know how to construct a mathematical proof. If the former, please read my hint and pma's responses carefully. If the latter, we cannot help you. If neither, it is very difficult to decide exactly that which you are asking, especially with the wording "How to prove...". I do not intend to discourage you from asking questions here but please provide context (rather than merely fire questions at us). --PS T 03:26, 15 January 2010 (UTC)[reply]

35 degrees east

Since Indo-Australian Plate said 35 degrees east and the total plate is 67 mm then how many meters north. If is 35 degrees east would east be 10 mm and north be 50 mm?--209.129.85.4 (talk) 21:27, 13 January 2010 (UTC)[reply]

Since the plate is moving 35 degrees east of north (that is, it's moving NNE), and moving at a rate of 67 mm per year, you can find how fast the plate is moving in both the East-West and North-South direction by drawing a right triangle (with the legs lined up along the North-South and East-West directions, and a Hypotenuse of 67 mm/yr) and using trigonometry to determine the lengths of the legs. The plate is moving somewhat faster to the north than you guessed, and significantly faster to the east. Does someone know a good way to get a diagram in here? Buddy431 (talk) 02:00, 14 January 2010 (UTC)[reply]

Oh, you're talking about movement! I couldn't get any sense at all out of the question. --ColinFine (talk) 08:18, 14 January 2010 (UTC)[reply]

Simplistic solution is that northwrds component of velocity is 67 cos(35 degrees) = 54.9 mm/yr and eastwards component is 67 sin(35 degrees) = 38.4 mm/yr. However, as the figure given for the plate's velocity is an average of small movements over a large geographic area and over a long period time, this notional calculation of northwards and eastwards components is probably over-simplifying a much more complex pattern of movement. Different places on the plate boundary may be moving at different rates. Gandalf61 (talk) 10:58, 14 January 2010 (UTC)[reply]