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January 20

Eigenvalues and eigenvectors

I am working on some project which in part involves derivatives of eigenvalues and eigenvectors and associated problems. The following requires additional insight from someone else (or from some classic textbook, I don't know). Consider a matrix $A(t)$ , linear with respect to complex argument $t\in \mathbb {C}$ , such that $A(0)$ has an eigenvalue of multiplicity two. I could identify three completely different scenarios illustrated by examples below

A=\left({\begin{array}{cc}1+t&0\\0&1-t\end{array}}\right)

where $A(0)$ is in fact diagonalizable.

A=\left({\begin{array}{cc}1+t&1\\0&1-t\end{array}}\right)

where $A(0)$ is not diagonalizable, and even though eigenvalues $1\pm t$ are analytic functions, eigenvectors have a singularity (a pole) at $t=0$

A=\left({\begin{array}{cc}1+t&1-t\\t&1-t\end{array}}\right)

where $A(0)$ is not diagonalizable, and eigenvalues $1\pm {\sqrt {t}}$ as well as eigenvectors have a branch point at $t=0$

There may be other scenarios as well (which I do not see at the moment). In a more general setting, $A(t)=X+tY$ (which later will even become multidimensional), how can I determine which scenario I should expect just by looking at matrices X and Y (their SVD or other decompositions are all at my disposal). (Igny (talk) 04:41, 20 January 2010 (UTC))[reply]

You can determine it by taking the root of the function and integrating wrt a constant wher the value of F(x) does not change and remains constant while the determinant of the diagonalizable form may come in handy--123.237.193.11 (talk) 14:54, 21 January 2010 (UTC)[reply]

Riemann integral

Resolved

When, if ever, is $\int _{a}^{b}|f(x)|dx=0\Rightarrow f(x)=0\forall x\in [a,b]$ true? Thanks-Shahab (talk) 08:09, 20 January 2010 (UTC)[reply]

For starters, this obviously holds if f is continuous. -- Meni Rosenfeld (talk) 08:27, 20 January 2010 (UTC)[reply]

Why? Can you explain.-Shahab (talk) 09:21, 20 January 2010 (UTC)[reply]

Think of it in terms of area-under-the-curve. That area can only be zero if either the function is identically zero or if the area above the x-axis equals the area below it (or the function is discontinuous). |f(x)| is, by definition, positive, so the area below the x-axis must be zero. That leaves you with only f(x)=0 as a possibility. --Tango (talk) 10:14, 20 January 2010 (UTC)[reply]

[ec] If

f(x_{0})\neq 0

, then

|f(x_{0})|=\epsilon >0

. Since f is continuous, so is

|f|

, and there is a neighborhood around

x_{0}

where

|f(x)|>\epsilon /2

. The integral of

|f|

is positive over that region and non-negative elsewhere. -- Meni Rosenfeld (talk) 10:20, 20 January 2010 (UTC)[reply]

I think that's if and only if. If f isn't continuous, then it can't be identically zero. (Obviously, it holds for discontinuous functions where the LHS isn't true, but that isn't interesting.) --Tango (talk) 08:44, 20 January 2010 (UTC)[reply]

The real result is of course that, if

\int _{a}^{b}|f(x)|dx=0

, then f is almost everywhere zero. Algebraist 13:19, 20 January 2010 (UTC)[reply]

Thanks all-Shahab (talk) 17:11, 20 January 2010 (UTC)[reply]

A few more geometry problems!

Hi all,

Sorry to pop up again, I asked a question yesterday (thankyou very much for the help with that, incidentally) and I'm really having some trouble getting my head around some of this geometry stuff, my lecturer is a brilliant mathematician but not so competent when it comes to explanation and I don't want to hand in no work whatsoever because I've not been able to do any of it, but the only book which is recommended on my course (Curved Spaces by P.M.H Wilson) hasn't been a great deal of help - in fact, most of these problems come from the exercises at the end of the chapters, which have no hints or solutions.

I don't expect you to do all these for me, but if I could just get some general guidance on the following points that'd be an amazing help!

1) I'm looking at finite subgroups G of the isometries of $\mathbb {R} ^{n}$ , and I've been told to show G fixes some point of $\mathbb {R} ^{n}$ by considering the barycentre (average) of the orbit of the origin under G. Now if we let G act on some point x, and note that gG=G (fairly simple to prove), then for $c={\frac {\sum _{g\in G}gx}{|G|}}$ , we have gc=c $\forall g\in G$ , right? Since every element in G can be written as $g^{-1}j$ , for some $j\in G$ , so we end up with exactly the same sum and c is a fixed point. My question is, unless my argument is wrong, why do we need to consider specifically the origin to find this fixed point? Why does it not work with an arbitrary point x? I also want to classify all finite subgroups of isometries in $\mathbb {R} ^{2}$ into cyclic and dihedral groups. Could anyone give me any suggestions to get started?

2) Here I'm looking at isometries of the unit sphere $S^{2}$ in $\mathbb {R} ^{3}$ , and I want to show that any matrix A $\in O(3,\mathbb {R} )$ can be written as the product of at most 3 reflections in planes through the origin, so I can then deduce that an isometry of $S^{2}$ can be written as the product of at most 3 reflections in spherical lines. Then, I need to work out which isometries are obtained as the product of 2, and which are obtained as exactly 3 reflections. Should I look at which isometries are orientation preserving or does this not come in to play with spherical reflections? How else would I go about it, in that case?

3) I want to show that for every spherical line l and point P on $S^{2}$ there is a spherical line l' such that P $\in$ l' and l and l' are perpendicular at their intersection(s). I always find that with problems like this I have to rotate the sphere so that l is in the x-y plane and P has y-coordinate 0 and use some 'WLOG' argument before I can actually make any headway, and then I don't really seem to be proving anything at all. Can anyone get me going on this with any sort of more elegant proof? I want to show also that the distance from P to any point Q on the line l is minimized when Q is one of the 2 intersections of l and l', and whilst this seems intuitively obvious (as with much of geometry) I'm getting stuck actually proving it rigorously. I also want to prove that l' is unique if this minimum distance is < ${\frac {\pi }{2}}$ , but I'll probably be able to give that a go myself once I've made a dent in the earlier parts of the question.

I appreciate that I'm asking a lot and of course any answers you could give me will probably be quite time consuming so don't feel like you have to help me with everything in one go (or at all!) - I did spend most of yesterday trying and failing to make much progress with these problems and knowing geometry isn't my forté will probably just make me want to learn it more, so please please do lend a hand!

Many thanks in advance, 82.6.96.22 (talk) 15:16, 20 January 2010 (UTC)[reply]

Ad 1: your argument is correct, and it does work with an arbitrary point x, there's nothing special about the origin. — Emil J. 15:23, 20 January 2010 (UTC)[reply]

Thanks appreciated for previous problem. I haven't time at the moment for this, in fact I shouldn't be even looking here at the moment, but can I suggest that saying your lecturer sucks might not be the best approach with some people here ;-) Dmcq (talk) 17:09, 20 January 2010 (UTC)[reply]

Noted ;) (same person, temporarily different I.P. address!) 131.111.185.75 (talk) 18:50, 20 January 2010 (UTC)[reply]

For (2), if you choose an orthonormal basis, the isometries are characterized by where they send those basis elements. The reflection that sends a to b is the one through the axis a-b. You can send the basis elements to their new spots with reflections one at a time and show that the later reflections don't mess with the basis elements that were already set. You're right that reflections switch the orientation, so isometries that preserve orientation will be made of an even number of reflections and the orientation reversing ones will be odd. Then it's just a matter of deciding when an isometry requires 1 or 3 reflections. There might be something more clever, but that's how I'd do it. Rckrone (talk) 20:08, 20 January 2010 (UTC)[reply]

For (3), a spherical line is characterized by the axis it goes around. In this way, try rephrasing the problem in terms of finding perpendicular vectors. Rckrone (talk) 20:20, 20 January 2010 (UTC)[reply]

For the second part of (3), the distance between two points a, b on the sphere is the same as their angle, so a.b = cos(dist(a,b)), so minimizing the distance is maximizing the dot product. The basis formed by the axis of l, the axis of l' and the direction orthogonal to them is convenient to work in. Rckrone (talk) 21:03, 20 January 2010 (UTC)[reply]

You guys have been awesome, thanks so much! I don't know how I ever managed without you :-) 82.6.96.22 (talk) 08:33, 22 January 2010 (UTC)[reply]

Geometry question

Is there a way of finding the radius of a circle whose area is equal to the area of a rectangle of given dimensions? (Kind of like the geometric mean describes the size of a square whose area is equal to the area of a rectangle of given dimensions.) Edgeweyes (talk) 18:50, 20 January 2010 (UTC)[reply]

An a by b rectangle has area ab, corresponding to a circle of radius

{\sqrt {\frac {ab}{\pi }}}

, so it's just a (constant multiple of a) geometric mean. Algebraist 18:55, 20 January 2010 (UTC)[reply]

I see now. I should have thought about it a little more... πr² = ab is not tough to solve. Thanks! Edgeweyes (talk) 19:28, 20 January 2010 (UTC)[reply]