Wikipedia:Reference desk/Archives/Mathematics/2009 October 17

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October 17

Quadratic forms on real matrices

Hi all - I've been given a problem to show that the map $A\mapsto tr(A^{2})$ is a quadratic form on Mat_n($\mathbb {R}$), and find its rank & signature (where tr(A) denotes the trace of A).

The first part is no problem - I'm using the definition of a quadratic form where Q is a q.f. iff $Q(\lambda \,v)=\lambda ^{2}\,Q(v)$ and the map $(u,\,v)\,\mapsto \,{\frac {1}{2}}(Q(u+v)-Q(u)-Q(v))$ is bilinear, so that's fairly simple. However, I don't even understand how to begin finding the trace or the signature - with quadratics like $x^{2}+2xy+4xz+z^{2}$ I know to write them in the form x^TAx for vector x=(x,y,z) and symmetric matrix A, but I'm in a little over my head here, how do I get started when I'm working with n*n matrices instead?

Many thanks, Mathmos6 (talk) 01:08, 17 October 2009 (UTC)

Well you can write it out as a quadratic: If A=[a,b;c,d] then tr(A^2) = a^2 + d^2 + 2bc. The matrix of the quadratic form is just a permutation matrix: [1,0,0,0;0,0,1,0;0,1,0,0;0,0,0,1]. If A=[a,b,c;d,e,f;g,h,i], then tr(A^2) = a^2 + e^2 + i^2 + 2bd + 2cg + 2fh. JackSchmidt (talk) 04:37, 17 October 2009 (UTC)

You can write it out as a quadratic for fixed n, but I think the OP wants it for general n. --Tango (talk) 09:54, 17 October 2009 (UTC)

If A = (a_ij) is an n × n matrix then, as the trace article shows:

${\mbox{tr}}(A^{2})=\sum _{i=1}^{n}\sum _{j=1}^{n}a_{ij}a_{ji}.$

This gives an explicit expression for the quadratic form. The trace and signature should follow from calculations. Find the symmetrix matrix, say M such that vMv^T = tr(A²) where

$v=(a_{11},\ldots ,a_{1n},a_{21},\ldots ,a_{2n},\ldots ,a_{nn}).$

Finding M shouldn't be all that difficult. For example, in the case where n = 2 then tr(A²) = a₁₁² + 2a₁₂a₂₁ + a₂₂² and so

$M=\left({\begin{array}{cccc}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{array}}\right).$

The eigenvalues are given by −1, 1, 1, and 1, the eigenvectors are (0,−1,1,0), (0,0,0,1), (0,1,1,0), and (0,0,0,1) respectively. The rank and signature follow. ~~ Dr Dec (Talk) ~~ 11:32, 17 October 2009 (UTC)

If A is symmetric and B is skew-symmetric then tr(AB)=0. Moreover, Q(A):=tr(A²) is positive definite on the space of symmetric matrices, and it is negative definite on the space of skew-symmetric matrices. So it's a non-degenerate quadratic form with signature ( n(n+1)/2 , n(n-1)/2 ). --pma (talk) 21:26, 17 October 2009 (UTC)

To help the OP "get started when [he's] working with n × n matrices", could you please explain how you came to this conclusion? ~~ Dr Dec (Talk) ~~ 13:03, 18 October 2009 (UTC)

Details: we know that the n²-dimensional vector space Mat_n is the direct sum of the n(n+1)/2 dimensional subspace Sym_n of all symmetric matrices plus the n(n-1)/2 dimensional subspace Skw_n of all skew-symmetric matrices; the decomposition being M=A+B, with A:=(M+M^T)/2 (symmetric) and B:= (M-M^T)/2 (skew-symmetric). The above quadratic form Q splits in this decomposition: Q(M)=Q(A)+Q(B), because tr(AB)=0. Since Q is positive definite, respectively negative definite, on Sym respectively on Skw, the conclusion follows. In terms of the Hilbert-Schmidt norm, we can write Q(M)=|A|²-|B|².

Note that for n=2 Dr Dec's computation gives a signature (3,1) as it has to be. Moreover, if you check the eigenvectors he provides, they correspond to the skew-symmetric 2x2 matrix with 0 on the diagonal (of the eigenvalue -1), and three symmetric 2x2 symmetric matrices (of the eigenvalue 1). This suggests the generalization of the diagonalization of Q for any n. Let E_ij be the matrix with all entries 0 but the ij-entry equal to 1. Consider the base of Mat_n given by the n(n+1)/2 symmetric matrices S_ij:=E_ij+E_ji for i≤j and the n(n-1)/2 skew-symmetric matrices S_ij:= E_ij-E_ji for i>j: clearly tr(S_ij S_i'j')=0 if (i,j)≠(i',j'); Q(S_ij)>0 for i≤j and Q(S_ij)<0 for i>j. (As usual, making computations is a necessary step in order to avoid making computations ;-) --pma (talk) 22:40, 18 October 2009 (UTC)

Good way to remember meet vs join?

I can never remember which of Meet and Join is which (and the symbols $x\land y$ and $x\lor y$). Any suggestions for a good way to remember? — Matt Crypto 17:29, 17 October 2009 (UTC)

Meet looks like intersection which looks like an 'n' and join looks like union which looks like a 'u'. Does that help?--RDBury (talk) 18:10, 17 October 2009 (UTC)

The meet of two sets is where they meet, i.e. the intersection (and has a symbol resembling the intersection symbol). The join of two sets is the two sets joined together, i.e. the union (and again, the symbol is similar). Algebraist 18:15, 17 October 2009 (UTC)

(ec) One way I can think of is that if you have a power set with partial order x ≤ y for $x\subset y$, then $x\land y=x\cap y$ and $x\lor y=x\cup y$.

More generally, $\land ,\cap$ are like 'A' for "and" and $\lor ,\cup$ are like 'U' for "union". Rckrone (talk) 18:28, 17 October 2009 (UTC)

Let's add $\scriptstyle \prod$ and $\scriptstyle \coprod$. Note that cup-like symbols denote inductive limits while cap-like symbols, like the product, denote projective limits. Also, it may help the mnemonic to recall that (like $\cup$ is for "Union"), $\lor$ stands for the Latin conjunction vel, "or" (indeed, x is in $A\cup B$ iff x is in A, or in B). --pma (talk) 21:47, 17 October 2009 (UTC)

Flip the symbol upside-down and that's what the Hasse diagram would look like. Or, remember that "AND" can only make truth values go down, and "OR" can only make them go up. — Carl (CBM · talk) 01:01, 19 October 2009 (UTC)

I always remembered it by $\cap$ looks like the 'n' in AND, whereas $\cup$ looks like the 'u' sound in OR. --Dakuton (talk) 18:17, 21 October 2009 (UTC)

Certain conformal map projections

At Talk:Dymaxion map, I posted a hastily scrawled question, which I then amended a few minutes later. This seems like a better place to find an answer than that page.

The Dymaxion map is a world map on the surface of an icosohedron (or maybe in some cases some other polyhedron?). It can unfold to make a flat map, perhaps with less distortion (except maybe along the edges and certainly at the vertices) than some other projections of the whole world, or nearly the whole world, onto a page.

The simplest way to do that would be 20 separate stereographic projections, I would think. Stereographic projections are conformal.

So that would make the Dymaxion map conformal except at the vertices and possibly along the edges.

My guess is it's not conformal along the edges.

And obviously it could not be conformal at the vertices.

Two questions:

• Is my guess about the edges right? (Maybe I'll fiddle with this and see if I can find the answer before anyone answers here; it doesn't seem like a hard question.)
• If so, a possibly harder question to answer: is there some other way to do it, i.e. not stereographic, that makes it conformal everywhere except at the 12 vertices?

Michael Hardy (talk) 23:04, 17 October 2009 (UTC)

If the map consists of portions of 20 stereographic projections, is the centre of projection in each case the centre of the globe? Is so, the edge of each triangular face will, I think, correspond to a great circle on the globe, in which case conformality will be maintained on each side of that boundary line and, I think, across it. But intuition can only go so far, and my 3D trigonometry isn't up to a formal proof.→217.43.210.13 (talk) 12:37, 18 October 2009 (UTC)

Oh: I was being stupid. I don't think it's a stereographic projection if the center of projection is the center of the sphere. But the center of the sphere seems like the obvious place from which to project. Is projection from the center of the sphere conformal? Michael Hardy (talk) 18:03, 18 October 2009 (UTC)

I think the radial projection (from the south hemisphere S₊ to C, seen as the tangent plane at the south pole) is not conformal. Without making the computation, the reason is: if it were conformal, we could compose the inverse of the radial projection, C → S₊, with the standard stereographic projection S → C; the result would be a radial holomorphic map C →C, that is, of the form f(re^it)=φ(r)e^it, which can be holomorphic only if f(z)=az, which is not the case here (alternatively, it would be a bounded holomorphic map on C, hence constant by Liouville theorem, again not the case here). --pma (talk) 21:37, 18 October 2009 (UTC)

Just to fix some data of the problem. An icosahedron I minus the set of its vertices V is a Riemann surface with the structure given by quotient of an open set U of C (this corresponds exactly to the concrete construction: U is the union of 20 open equilateral triangles of the plane expansion, together with the tabs to be glued to the adjecent faces). Otherwise, we can define the holomorphic structure by an atlas, the charts being the maps that take the rhombus made by 2 equilateral triangles in C, onto a pair of adjacent faces (the transition maps are then of the form az+b). This Riemann surface is of course homeomorphic to the Riemann sphere S minus a set of 12 points (that we may identify as V, the set of vertices of I, if we think I as inscribed in S). The piecewise stereographic projection is certainly a homeomorphism of S onto I, but it is certainly not holomorphic on S\V : it is a stereographic projection on each face, but exactly for this reason it can't be globally holomorphic (by unique continuation it would be a unique stereographic projection, read on the charts. In particular, without all the icosahedral symmetry of the piecewise sereographic map). However, it seems at least $C^{1}$, even at the edges. So I agree with you, as to the first question (however I couldn't find out there the details of the construction of the Dymaxion map, and we are not sure if the Dymaxion is actually the piecewise stereographic projection). The second question is very interesting: is there a homeomorphism h between I and S which is bi-holomorphic between I\V and S\V? I think the answer is yes (and if so, I guess this one is the true Dymaxion map). I suspect the point is that, even if the holomorphic structure of I\V can't be extended to I, there is a 6m-fold branched covering of I that admits a global holomorphic structure, and that is bi-holomorphic to the analogous branched covering of S. This should give the map h. In any case, if we could make this thing clear and at elemenary level, it should be a very beautiful addition to the article. --pma (talk) 21:11, 18 October 2009 (UTC)

Maybe it would also be publishable elsewhere—not as math research but as a folding and unfolding map of the earth. So we need to do that, then copyright it, then sell it at every B. Dalton's, etc. Michael Hardy (talk) 22:51, 22 October 2009 (UTC)

Well, I don't think so; the whole thing is certainly well-known and written in some classical book, and, as usual, understanding the thing by oneself is quicker and funnier. In any case the question you posed is very interesting and deserves a search. --pma (talk) 06:50, 23 October 2009 (UTC)