User denis serre - MathOverflow

Answer by Denis Serre for What can be said about pairs of matrices P,Q that satisfies $(P^{-1})^T \circ P = (Q^{-1})^T \circ Q$ ?

2013-05-22T12:15:09Z

Actually, if $D, E$ are diagonal and $Q=EPD$, then $(P,Q)$ is such a pair. A natural question is whether every pair such that $P^{-T}\circ P=Q^{-T}\circ Q$ is of the form above. But even this is false, because if $P$ is triangular, then $P^{-T}\circ P=I_n$.

I take the occasion to mention an open question: define $\Phi(P):=P^{-T}\circ P$ (the gain array). What are the matrices $P$ such that $\Phi^{(k)}(P)\rightarrow I_n$ as $k\rightarrow+\infty$ ? According to Johnson & Shapiro, this is true at least for

Strictly diagonally dominant matrices
Symmetric positive definite matrices

On the contrary, $\Phi$ has fixed points, for instance the mean $\frac12(P+Q)$ of two permutation matrices. See Exercises 335, 336, 342, 343 of my blog about Exercises on matrix analysis.

Applications of Hilbert's metric

2011-09-27T06:06:39Z

Among the fascinating constructions in mathematics is the Hilbert metric on a bounded convex subset of ${\mathbb R}^n$.

Where, within mathematics, is it used ? I know at least a proof of the Perron--Frobenius Theorem for non-negative matrices.

What are its applications in other sciences ?

Answer by Denis Serre for Which popular games are the most mathematical?

2013-05-17T11:22:14Z

The two-player single suit whist has been analyzed completely in this paper by Johan Wastlund. This was mentionned by Alison Miller in her answer to my MO question Bridge game with only one suit: strategy.

cube + cube + cube = cube

2011-01-24T12:07:20Z

The following identity is a bit isolated in the arithmetics of natural integers $$3^3+4^3+5^3=6^3.$$ Let $K_6$ be a cube whose side has length $6$. We view it as the union of $216$ elementary unit cubes. We wish to cut it into $N$ connected components, each one being a union of elementary unit cubes, such that these components can be assembled so as to form three cubes of sizes $3,4$ and $5$. Of course, the latter are made simultaneously: a component may not be used in two cubes. There is a solution with $9$ pieces.

What is the minimal number $N$ of pieces into which to cut $K_6$ ?

About connectedness: a piece is connected if it is a union of elementary cubes whose centers are the nodes of a connected graph with arrows of unit length parallel to the coordinate axes.

Edit. Several comments ask for a reference for the $8$-pieces puzzle, mentionned at first in the question. Actually, $8$ was a mistake, as the solution I know constists of $9$ pieces. The only one that I have is the photograph in François's answer below. Yet it is not very informative, so let me give you additional informations (I manipulated the puzzle a couple weeks ago). There is a $2$-cube (middle) and a $3$-cube (right). At left, the $4$-cube is not complete, as two elementary cubes are missing at the end of an edge. Of course, one could not have both a $3$-cube and a $4$-cube in a $6$-cube. So you can imagine how the $3$-cube and the imperfect $4$-cube match (two possibilities). Other rather symmetric pieces are a $1\times1\times2$ (it fills the imperfect $4$-cube when you build the $3$-, $4$- and $5$-cubes) and a $1\times2\times3$. Two other pieces have only a planar symmetry, whereas the last one has no symmetry at all.

Here is a photography of the cut mentionned above.

Answer by Denis Serre for Inverse of a totally unimodular matrix

2013-04-19T22:49:12Z

The answer is yes, because if $B=A^{-1}$, then we have an equality between minors: $$B(I,J)=\pm\frac{A(J^c,I^c)}{\det A},$$ for every subsets $I,J\subset[[1,n]]$ of same cardinals. This formula generalizes that giving the entries of $A^{-1}$ in terms of minors of $A$. The $\pm$ sign is not essential to prove the stability of the TU class under inversion.

Answer by Denis Serre for A series question related to solution of Laplace equation

2013-04-12T06:26:01Z

Let $v,w,z$ be the functions obtained from $u$ by composing with a rotation of angle $\frac\pi4,\frac\pi2,\frac{3\pi}4$ about $(\frac12,\frac12)$. The sum $u+v+w+z$ is a harmonic function, whose value at the boundary is constant equal to $1$. Thus $u+v+w+z\equiv1$. This gives you $4u\left(\frac12,\frac12\right)=1$.

Answer by Denis Serre for Fixed point theorems

2013-04-10T05:50:30Z

The infinite dimensional version of Brouwer's FPT is Schauder's FPT. If $K$ is a non-void closed convex subset of a TVS, and $f:K\rightarrow K$ is compact ($f$ is continuous and $f(K)$ is compact), then $f$ has a fixed point.

It has numerous applications in nonlinear analysis. One of the earliest being the existence of a solution to the stationnary Navier-Stokes equations with Dirichlet boundary condition, proven by J. Leray.

Concavity of $\det^{1/n}$ over $HPD_n$.

2010-10-18T08:08:28Z

One of my beloved theorems in matrix analysis is the fact that the map $H\mapsto (\det H)^{1/n}$, defined over the convex cone $HPD_n$ of Hermitian positive definite matrices, is concave. This is accurate, if we think that this map is homogeneous of degree one, thus linear over rays.

it has important applications in many branches of mathematics,
it has many elegant proofs. I know at least three complety different ones.

I am interested to learn in both aspects. Which is your prefered proof of the concavity ? Is it useful in your own speciality ? In order to avoid influencing the answers, I decide not to give any example. But those who have visited my page may know my taste.

Answer by Denis Serre for Concavity of $\det^{1/n}$ over $HPD_n$.

2013-04-04T07:03:33Z

Here is a interesting calculus proof. Let $f:A\mapsto(\det A)^{1/n}$, defined over $SPD_n$. Differentiating twice, we find the Hessian $${\rm D}^2f_A(X,X)=\frac1{n^2}f(A)\left(({\rm Tr} M)^2-n{\rm Tr}(M^2)\right),$$ where $M=A^{-1}X$. This matrix, being the product of two symmetric matrices with one of them positive definite, is diagonalisable with real eigenvalues $m_1,\ldots,m_n$. The parenthesis above is now $$\left(\sum_jm_j\right)^2-n\sum_jm_j^2,$$ a non-positive quantity, according to Cauchy-Schwarz. We infer that ${\rm D}^2f_A\le0$ and that $f$ is concave.

Answer by Denis Serre for Examples of interesting false proofs

2013-03-19T16:58:35Z

I have known the following for 45 years: in the Euclidian plane, every triangle is isosceles.

The false proof needs a handmade picture; take your pen, it's easy. Start from a triangle $ABC$. Draw the perpendicular bisector of $BC$, and the angle bisector from $A$. Let $I$ be their intersection (if it is not unique, you are done). Let $J$ be the projection of $I$ over $AB$, $K$ that over $AC$. Considering the right triangles $AIJ$ and $AIK$, we see that (lengths) $AJ=AK$, and that $IJ=IK$. Then looking at righttriangles $BIJ$ and $CIK$, we obtain that $BJ=CK$. We conclude that $$AB=AJ+JB=AK+KC=AC.$$

The falsity is that one of $J$ or $K$ is in the triangle, and the other one is out. Therefore one of the sums above (and only one) should be a difference.

Answer by Denis Serre for On the convexity of element-wise norm 1 of the inverse

2013-03-19T16:01:38Z

The answer is Yes when $n=2$,but No when $n\ge3$. Here is the analysis.

The differential $L_A$ of $A\mapsto A^{-1}$ is $L_A=-A^{-1}BA^{-1}$. Likewise, the Hessian is $$H_A[B]=2A^{-1}BA^{-1}BA^{-1}=\frac2{(\det A)^3}\hat A B\hat AB\hat A,$$ where $\hat A$ is the adjugate matrix (mind that $A$ being symmetric, $(\det A)A^{-1}=\hat A$). Finally, the Hessian of the norm of $A^{-1}$ is $$\phi_A[B]=\frac2{(\det A)^3}\sum_{i,j}{\rm sgn}(\hat a_{ij})(\hat A B\hat AB\hat A)_{ij}.$$ For the function to be convex over $S_n^{++}$, it is therefore necessary and sufficient (the singular part of the Hessian, located at matrices such that some entry of $A^{-1}$ vanishes, is positive) to have, for every $S\in S_n^{++}$ and $B\in Sym_n$ $$\sum_{i,j}{\rm sgn}(s_{ij})(SBSBS)_{ij}\ge0.$$

When $B$ runs over $Sym_n$, $S^{1/2}BS^{1/2}$ covers $Sym_n$, and its square covers $S_n^+$. Thus $SBSBS$ runs over $S_n^+$. We infer that the convexity is equivalent to the property that for every $S\in S_n^{++}$ and $K\in S_n^+$, there holds $$\sum_{i,j}\epsilon_{ij}k_{ij}\ge0,$$ where $\epsilon(S):=(({\rm sgn}(s_{ij}))_{i,j}$. This amounts to saying that the matrix $\epsilon(S)$ is positive semi-definite.

This inequality turns out to be true if $n=2$, because $$\epsilon(S)=\begin{pmatrix} 1 & \pm1 \\ \pm1 & 1 \end{pmatrix}.$$ But this is false if $n\ge3$. Take for instance a matrix $A$ such that $S$, thus $(\det A)A^{-1}$, be a small disturbance of $I_3$, with negative off-diagonal entries. Then $$\epsilon(S)=\begin{pmatrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix}$$ is indefinite.

Answer by Denis Serre for Motivating the Laplace transform definition

2013-03-18T14:44:28Z

Even if we don'y use Laplace Transform as often as Fourier Transform, it is definitely a more subtle tool. The reason is that the FT analyses only functions $f:{\mathbb R}\rightarrow X$ that decay at $\pm\infty$. At best, $f$ can be a temperate distribution, and this means that $f$ and its derivatives grow slowly at infty; but let's think about functions. The LT instead deals with functions $f:(0,+\infty)\rightarrow X$ (the domain may be $(-\infty,0)$ as well, but not their union), whose growth at infinity is moderate (at most exponential). Then the transformed function $\hat f$ is defined and holomorphic on a half-space $H$. This gives us the possibility of employing the tools of complex analysis. Also, $\hat f$ contains a lot of redundancy. Let me give an example (which applies to Dirichlet series as well): it may happen that $\hat f$ has a pole at some $z_0$, a point boundary of $H$. I mean that $\hat f$ has a meromorphic extension in a slightly bigger half-space than $H$. Then the residue calculus gives us an information about the asymptotics of $f$ at $+\infty$. This applies to the Prime Number Theorem and to the Theorem of Arithmetic Progression.

The FT and LT are widely used in Partial Differential Equations. Fourier transform is efficient for linear, constant coefficients, Cauchy problems. By this, I mean that the physical domain is ${\mathbb R}^d$, and there is a time variable $t$. Think of the Heat, Wave or Schroedinger equations. Then you apply Fourier to the space variables and receive a linear ODE in $$\frac{d\hat u}{dt}=M(\xi)\hat u(t,\xi),$$ which you analyse easily.

Once the domain has a boundary, you need the Laplace transform, because you cannot get such a simple object as an ODE. At best, you reduce your problem to a linear PDE in $(t,x_d)$, where $x_d$ is the coordinate normal to the boundary. This PDE is parametrized by the Fourier variable $\eta$ associated with the coordinates that are tangent to the boundary. In addition, you have a boundary condition at $x_d=0$, and an initial data at $t=0$. The well-posedness for $t>0$ must then be attacked through a Laplace transform in time (one could do that in the case of the Cauchy problem as well, but this is not so essential). The key words in the theory are incoming modes and Lopatinskii condition.

For the interested readers, see my book in collaboration with S. Benzoni-Gavage, The Hyperbolic Initial-Boundary Value Problem, Oxford Univ. Press (2007).

Answer by Denis Serre for Spectrum theorem for p-adic matrix analysis

2013-03-18T14:01:05Z

I think that this matter is treated in K. Kedlaya's book P-adic Differential equations, Cambridge Univ. Press (2010).

The probability for a symmetric matrix to be positive definite

2013-01-09T21:34:16Z

Let me give a reasonable model for the question in the title. In ${\rm Sym}_n({\mathbb R})$, the positive definite matrices form a convex cone $S_n^+$. The probability I have in mind is the ratio $p_n=\theta_n/\omega_n$, where $\theta_n$ is the solid angle of $\Lambda_n$, and $\omega_n$ is the solid angle of the whole space ${\rm Sym}_n$ (the area of the unit sphere of dimension $N-1$ where $N=\frac{n(n+1)}2$). These definitions are relative to the Euclidian norm $\|M\|=\sqrt{{\rm Tr}(M^2)}$ ; this is the most natural among Euclidian norms, because it is invariant under unitary conjugation.

Because $S_2^+$ is a circular cone, I could compute $p_2=\frac{2-\sqrt2}4\sim0.146$ . Is there a known close formula for $p_n$? If not, is there a known asymptotics?

More generally, we may define open convex cones $$\Lambda_n^0\subset\Lambda_n^1\subset\cdots\subset\Lambda_n^{n-1}$$ in the following way: $M\mapsto\det M$ is a homogeneous polynomial, hyperbolic in the direction of the identity matrix $I_n$. Thus its successive derivatives in this direction are hyperbolic too. The $k$th derivative defines a "future cone" $\Lambda_n^k$, these cones being nested. For instance, $\Lambda_n^0=S_n^+$. It turns out that this derivative is, up to a constant, $\sigma_{n-k}(\vec\lambda)$, where $\sigma_j$ is the $j$th elementary symmetric polynomial and $\vec\lambda$ the spectrum of $M$. Therefore $\Lambda_n^k$ is defined by the inequalities $$\sigma_1(\vec\lambda)\ge0,\ldots,\sigma_{n-k}(\vec\lambda)\ge0.$$ For instance, $\Lambda_n^{n-1}$ is the half-space defined by ${\rm Tr}M\ge0$.

Let us define again $p_{n,k}$ the probability for $M\in{\rm Sym}_n$ to belong to $\Lambda_n^k$. Thus $p_{n,0}=p_n$ and $p_{n,n-1}=\frac12$.

What is the distribution of $(p_{n,0},\ldots,p_{n,n-1})$, asymptotically as $n\rightarrow+\infty$?

Distribution of the spectrum of large non-negative matrices

2011-01-12T14:52:53Z

This question is related to that of Thurston. However, I am not interested in algebraic integers, and I wish to focus on random matrices instead of random polynomials.

When considering (entrywise) non-negative matrices $M$, a natural probability measure seems to be $$\prod_{i,j=1}^ne^{-m_{ij}}dm_{ij}.$$ By Perron-Frobenius theorem, the spectral radius $\rho(M)$ is an eigenvalue, associated to a non-negative eigenvector. Almost surely, $M$ is positive and therefore this eigenvalue is simple and its eigenvector is positive.

What is the distribution of the eigenvalues of $M$ as the size $n$ goes to infinity ? What is the relevant normalization ? Should we consider $\lambda/\rho(M)$ or $\lambda/\sqrt{\rho(M)}$ or something else ? Is it the same asymptotics as in the case of the conjugates of the algebraic Perron numbers considered by Thurston ?

Note that because of the constraint $m_{ij}\ge0$, an exponential law looked more natural to me than a Gaussian. Has anyone an other suggestion of probability over non-negative matrices ?

What is this subgroup of $\mathfrak S_{12}$ ?

2010-11-12T16:30:56Z

On some occasion I was gifted a calendar. It displays a math quizz every day of the year. Not really exciting in general, but at least one of them let me raise a group-theoretic question.

The quizz: consider an hexagon where the vertices and the middle points of the edges are marked, as in the figure alt text

One is asked to place the numbers $1,2,3,4,5,6,8,9,10,11,12,13$ (mind that $7$ is omitted) on points $a,\ldots,\ell$, in such a way that the sum on each edge equals $21$. If you like, you may search a solution, but this is not my question.

Of course the solution is non unique. You may apply any element of the isometry group of the hexagon. A little subtler is the fact that the permutation $(bc)(ef)(hi)(kl)(dj)$ preserves the set of solutions (check this).

Question. What is the invariance group of the solutions set ? Presumably, it is generated by the elements described above. What is its order ? Because it is not too big, it must be isomorphic to a known group. Which one ?

Answer by Denis Serre for Modern developments in finite-dimensional linear algebra

2013-03-01T17:28:16Z

Since you lowered the level to Weyl's Inequalities (1912), it is worth mentionning the improvements of these inequalities made by Ky Fan, Lidskii and others. They culminated in a much involved conjecture by A. Horn (1961), eventually proved by Knutson & Tao on the turn of the century.

Answer by Denis Serre for When is spectral norm of AB equal to that of BA?

2013-02-22T07:25:19Z

I suppose that your norm is either the Schur-Frobenius $\|F\|=({\rm Tr}FF^T)^{1/2}$ or that subordinated to the Euclidan norm $\|F\|=\rho(FF^T)$. In either case, the equality $\|AB\|=\|BA\|$ follows from the equality of the spectra of $ABBA$ and $BAAB$ (mind that $A$ and $B$ are symmetric). This is true because both matrices are similar to $A^2B^2$ (mind that $A$ and $B$ are non-singular).

If you drop your assumption that $A,B$ are PSD, then the equality does not hold in general. A simple counter-example is given by a pair $(A,B)$ such that $AB=0_n$ but $BA\neq 0_n$.

Answer by Denis Serre for system of homogeneous matrix equations

2013-02-01T09:32:15Z

A partial answer: Let $\Sigma$ denote the manifold $x^n+y^n=0$. Away from $\Sigma$, the equation and the fact that the roots of the polynomial $X^n-x^n-y^n$ are simple, tell you that $xA+yB$ is diagonalisable, with eigenvalues that differ from each other from a multiplicative $n$th root of unity. In addition, the multiplicity of an eigenvalue is locally constant.

The complement of $\Sigma$ in ${\mathbb C}^2$ is path-connected (true for every algebraic curve), but it is not simply connected. There are many non-trivial loops around $\Sigma$. When you follow an eigenvalue $\lambda(xA+yB)$ along a loop, you end with possibly different eigenvalue. Actually, there are so many loops that you find the following: if $\lambda$ is an eigenvalue of $xA+yB$, then $\omega\lambda$ is another one for every $n$th rooth of unity $\omega$. Eeven more: because the multiplicity remains constant when you follow the loop, $\lambda$ and $\omega\lambda$ have the same multiplicity.

In conclusion: for $(x,y)\not\in\Sigma$, $xA+yB$ is diagonalisable with eigenvalues at the vertices of a regular $n$-agon, and the eigenspaces have equal dimensions.

Since in your question you seem to take $n$ equal to the size of matrices, this means that the eigenvalues are simple away from $\Sigma$. Anyway, the existence of such a pair $(A,B)$ implies that $n$ (the power) divides $N$ (the matrix size).

What is $A+A^T$ when $A$ is row-stochastic ?

2013-01-04T09:18:54Z

This is motivated by this MO question.

If $A\in{\bf M}_n({\mathbb R})$ is row-stochastic (entrywise non-negative, and $\sum_j a_{ij}=1$ for all $i$), then $M:=A+A^T$ is

symmetric,
entrywise non-negative.

One finds easily the

additional property that $$\sum_{i\in I}\sum_{j\in J}m_{ij}\le|I|+|J|$$ for every index subsets $I$ and $J$,

with

equality in the extremal case: $$\sum_{i,j=1}^nm_{ij}=2n.$$

My question is whether all these four properties imply in turns that $M$ has the form $A+A^T$ for some row-stochastic $A$.

Edit. The answer is Yes when $n=2$ (obvious) or $n=3$ (more interesting).

Answer by Denis Serre for What is $A+A^T$ when $A$ is row-stochastic ?

2013-01-05T13:20:27Z

My solution for $n=3$ (upon Suvrit's request):

To begin with, we solve $A+A^T=M$ together with $A{\bf1}={\bf1}$ (no inequality for the moment). This is a linear system in $A$, which consists in $9$ equations in $9$ unknowns. However, it is not Cramer, because the set of skew-symmetric matrices $B$ such that $B{\bf1}=0$ is one-dimensional, spanned by $$\begin{pmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{pmatrix}.$$ In particular, there is a condition for solvability in $A$, but this condition is met by the assumption that $\sum_{i,j}m_{ij}=6$. Notice that $a_{ii}=\frac12m_{ii}$.

Thus there is a solution $A$, and every solution is of the form $A+aB$. There remains to find $a$ so as to satisfy the inequality $A+aB\ge0_n$. For this, let us denote $\mu$ the lower bound of $(a_{12},a_{23},a_{31})$, and $\nu$ that of $(a_{21},a_{13},a_{32})$.

Claim: we have $\nu+\mu\ge0$. This inequality allow us to find an $a$ such that $a_{12}+a,a_{23}+a,a_{31}+a,a_{21}-a,a_{13}-a,a_{32}-a\ge0$, which solves the problem.

Proof of the claim: we have $a_{12}+a_{21}=m_{12}\ge0$, $a_{12}+a_{13}=1-\frac12m_{11}\ge0$ because of the assumption that $m_{ij}\le2$, and finally $$a_{12}+a_{32}=a_{12}+a_{21}+a_{32}+a_{23}-a_{21}-a_{23}=m_{12}+m_{23}+\frac12m_{22}-1.$$ Form the assumption, this is equal to $$2-m_{13}-\frac12(m_{11}+m_{33})\ge0.$$ Finally, every sum $a_{ij}+a_{ji}$, $a_{ij}+a_{ik}$ and $a_{ij}+a_{kj}$ of elements of both sets is non-negative, hence $\mu+\nu\ge0$. Q.E.D.

Adapting this proof to higher $n$ seems difficult, but not impossible. Let us define $$s_A(I,J)=\sum_{i\in I,j\in J}a_{ij}.$$ If $A$ is any solution of $A+A^T=M$ and $A{\bf1}={\bf1}$, where $M$ meets the assumptions above, then for every $I,J$, we have $$s_A(I,J^c)+s_A(J,I^c)=|I|+|J|-s_M(I,J)\ge0.$$ Likewise, $a_{ij}+a_{ji}=m_{ij}\ge0$ for every $i,j$.

We have therefore reduced our question to the following one

Suppose that a matrix $A\in M_n({\mathbb R})$ satisfies $a_{ij}+a_{ji}\ge0$ for every $i,j$, and $s_A(I,J^c)+s_A(J,I^c)\ge0$ for every index sets $I,J$. Is it true that there exists a skew-symmetric matrix $B$, satisfying $B{\bf1}={\bf0}$, such that $A+B$ is entrywise non-negative?

(Remark that for such $B$, one has $s_B(I,J^c)+s_B(J,I^c)\equiv0$.)

A side remark: this set of assumptions about $A$ is redundant. All of them derive from the smaller set of inequalities $$a_{ij}+a_{ji}\ge0,\quad\forall i,j,\qquad s_A(I,I^c)\ge0,\quad\forall I.$$ As a matter of fact, one has $$s_A(I,J^c)+s_A(J,I^c)=s_A(I\setminus J,I\setminus J)+s_A(J\setminus I,J\setminus I)+s_A(I\cap J,(I\cap J)^c)+s_A((I\cup J)^c,I\cup J)$$ and $s_A(K,K)\ge0$ follows from $a_{ij}+a_{ji}\ge0$.

Answer by Denis Serre for Old books still used

2012-12-28T17:22:01Z

That depends if you speak of research books or advanced text book. In the second category, I should place

Rudin's Real and complex analysis (1966),
J.-P. Serre's Cours d'Arithmétique (1970) (hope you will forgive me),
Lang's Algebra (1st Edt 1965).

In the first category, I see

Kato's Perturbation theory of linear operators (1966),
Courant & Hilbert's Methods of Mathematical Physics (1924),
Courant & Friedrich's Supersonic Flow and Shock Waves (1948),
V. I. Arnold's Mathematical methods of classical mechanics (1974).

Pseudonyms of famous mathematicians

2010-11-07T17:56:35Z

Many mathematicians know that Lewis Carroll was quite a good mathematician, who wrote about logic (paradoxes) and determinants. He found an expansion formula, which bears his real name (Charles Lutwidge) Dodgson. Needless to say, L. Carroll was his pseudonym, used in literature.

Another (alive) mathematician writes under his real name and under a pseudonym (John B. Goode). (That person, by the way, is Bruno Poizat: it's no secret, even MathSciNet knows it.)

What other mathematicians (say dead ones) had a pseudonym, either within their mathematical activity, or in a parallel career ?

Of course, don't count people who changed name at some moment of their life because of marriage, persecution, conversion, and so on.

Edit. The answers and comments suggest that there are at least four categories of pseudonyms, which don't exhaust all situations.

Professional mathematicians, who did something outside of mathematics under a pseudonym (F. Hausdorff - Paul Mongré, E. Temple Bell - John Taine),
People doing mathematics under a pseudonym, and something else under their real name (Sophie Germain - M. Le Blanc, W. S. Gosset - Student)),
Professional mathematicians writing mathematics under both their real name and a pseudonym (B. Poizat - John B. Goode),
Collaborative pseudonyms (Bourbaki, Blanche Descartes)

Answer by Denis Serre for The distribution of roots of elliptic polynomial

2012-12-28T20:46:40Z

First of all, the notion of ellipticity refers to the principal part of the differential operator. In terms of the symbol, this means a property for the so-called principal symbol. Thus we may assume that $p$ is a homogeneous polynomial, of degree $2m$.

Then the answer to your question is Yes, when $y\in{\mathbb R}^{n-1}$ is non-zero, the univariate polynomial $z\mapsto p(y,z)$ has $m$ roots of positive (resp. negative) imaginary part. Here is the proof. First of all, there are $2m$ roots by assumptions, and none of the roots are real. If $n\ge3$ there imaginary parts keep a constant sign as $y$ varies, because the complement of the origin is connected. Thus let $m_\pm$ be (constant) number of roots of positive/negative imaginary parts. We have $m_-+m_+=2m$. Because the roots $z_j(-y)$ are nothing but the $-z_j(y)$ (by homogeneity), we have $m_-=m_+$. Therefore $m_\pm=m$.

If $n=2$ instead, one build the strongly elliptic polynomial $q(x,t)=p(x)+t^{2m}$ in $3$ variables. The previous analysis applies to $q$ and gives the result for $p$.

Now, the non-homogeneous case. We still know that $z\mapsto p(y,z)$ has $2m$ roots, with a constant number $m_\pm$ of positive/negative imaginary part. Fix $y\ne0$ real and d consider the polynomial $p_s(y,z)=s^{-2m}p(sy,sz)$. When $s\rightarrow+\infty$, $p_s$ tends to the homogeneous part $p_\infty$ of degree $2m$. The assumption implies that $P_\infty$ is strongly elliptic, to which the previous analysis applies. By continuity of the roots as functions of $\frac1s$, we find again that $m_\pm=m$.

Which functions of one variable are derivatives ?

2011-05-30T15:11:08Z

This is motivated by this recent MO question.

Is there a complete characterization of those functions $f:(a,b)\rightarrow\mathbb R$ that are pointwise derivative of some everywhere differentiable function $g:(a,b)\rightarrow\mathbb R$ ?

Of course, continuity is a sufficient condition. Integrability is not, because the integral defines an absolutely continuous function, which needs not be differentiable everywhere. A. Denjoy designed a procedure of reconstruction of $g$, where he used transfinite induction. But I don't know whether he assumed that $f$ is a derivative, or if he had the answer to the above question.

Answer by Denis Serre for Given a sequence of real numbers,do the following conditions suffice to guarantee convergence to 0?

2012-12-20T16:21:46Z

Just a very quick argument which reduces the possibilities: Let $\Omega\subset{\mathbb R}\cup\{\pm\infty\}$ be the $\omega$-limit set of the sequence, that is the set of limits of "converging" sub-sequences. It is a non-void closed set by construction. The property $x_{a+1}-x_a\rightarrow0$ tells us that $\Omega$ is a connected set. The property $x_{2a}-2x_a\rightarrow0$ tells us that $2\Omega=\Omega$. Therefore $\Omega$ can only be equal to one of the four sets $$\{0\},\quad[0,+\infty],\quad[-\infty,0],\quad{\mathbb R}.$$

Edit. It was commented that the second property gives only an inclusion, of $2\Omega$ into $\Omega$. Actually, it does give also the reverse inclusion (hence the equality), when combined with the first property: Let $\ell$ be the limit of some subsequence $x_{n_k}$. Because of the first property, we may suppose that $n_k=2m_k$ is even. Then $\ell/2$ is the limit of $x_{m_k}$, hence $\ell/2\in\Omega$.

Answer by Denis Serre for When are cones of matrices "generated" by vectors?

2012-12-19T12:16:55Z

Let $K$ be a closed convex cone in ${\bf Sym}_n({\mathbb R})$. I assume a generic cone: non void interior, strictly convex. Let $$K^0=\{ S\in{\bf Sym}_n({\mathbb R})\quad|\quad{\rm Tr}(SH)\ge0,\quad\forall H\in K\}.$$ be its dual. Then $K=(K^0)^0$. If $K=Z^0$ for some conical set $Z$ (that is, $tZ=Z$ for $t>0$), it is necessary that $Z\subset K^0$ and $Z$ contains the extremal lines of $K^0$.

Therefore the cone $K$ has the property that you request if, and only if, the extremal lines of its dual $K^0$ are spanned by matrices of the form $xx^T$.

This happens to be true if $K=K^0={\bf Sym}_n^+$, but it fails in general. Just take any finite set of lines in an open half-space, not all of them being spanned by a rank-one symmetric matrix, take $C$ their convex hull, and choose $K=C^0$. Then $K^0=C$ has an extremal line not spanned by an $xx^T$. Such a $K$ does not share the expected property.

Answer by Denis Serre for Minimum eigenvalue of a Affine Combination of two Hermitian matrices

2012-12-19T06:14:39Z

This is related to so-called hyperbolic polynomials, studied by L. Gaarding in the fifties. More generally, let $\lambda(\xi)$ be the least eigenvalue of $A(\xi)=\sum_\alpha\xi_\alpha A^\alpha$, where $A^\alpha$ are Hermitian matrices and $\xi$ is a real vector. Then $\lambda$ is a concave function. It is generically strictly concave, except in the radial directions of course because of homogeneity $\lambda(s\xi)=s\lambda(\xi)$ for $s>0$. The strict concavity is related to the lack of commutativity of pairs $(A^\alpha,A^\beta)$. For instance, the Pauli matrices yield $\lambda(\xi)=-|\xi|$, which is clearly strictly concave away from rays ${\mathbb R}^+\xi$. On the opposite, if $[A^\alpha,A^\beta]=0$ for every pair, then $\lambda$ is piecewise linear.

Strict concavity occurs for instance when the least eigenvalue is simple for every $\xi\ne0$, or if it has a constant multiplicity. Then $\lambda$ is analytic away of the origin, with a Hessian matrix of rank $n-1$. It turns out that this property implies a so-called Strichartz inequality for the solutions of the system of Partial Differential Equations $$\frac{\partial u}{\partial t}+\sum_\alpha A^\alpha\frac{\partial u}{\partial x_\alpha}=0.$$ Obviously, the symbol of the system is $\det(\tau I_n+A(\xi))$.Thus the characteristic manifold is related to the eigenvalues of $A(\xi)$, in particular to $\lambda(\xi)$.

The other eigenvalues satisfy more involved inequalities such as Weyl, Lidskii, Ky Fan -type inequalities. For instance, the sum of the $k$ least eigenvalues is concave too. This is a part of Alfred Horn's conjecture, now a theorem thanks to the work of many people, including Knutson & Tao.

The last part of the question, that about Toeplitz-Hausdorff theorem, is unclear. ${\mathbb S}$ is not a singleton, so what means \begin{align} \lambda(t)=(1-t)x_1+tx_2, ~~[x_1,x_2]\in \mathbb{S} \qquad ? \end{align} The equality certainly holds true for teh particular point $(x_1,x_2)$ obtained by taking $x$ a unit eigenvector associated with $\lambda(t)$, but what else ? To see some deep relations between Toeplitz-Hausdorff and hyperbolic polynomials, have a look to our paper with Th. Gallay, Numerical measure of a complex matrix, in Comm. Pure Appl. Math. 65 (2012), no. 3, 287–336.

Answer by Denis Serre for how to find/define eigenvectors as a continuous function of matrix?

2012-12-12T07:30:25Z

The example given by Anthony Quas reveals a phenomenon discussed in Kato's book Perturbation Theory for Linear Differential Operators. The point is the following:

If the symmetric matrix depends analytically upon one parameter, then you can follow analytically its eigenvalues and its eigenvectors. Notice that this requires sometimes that the eigenvalues cross. When this happens, the largest eigenvalues, as the maximum of smooth functions, is only Lipschitz.
On the contrary, if the matrix depends upon two or more parameters, the eigenvalues are at most Lipschitz when crossing happens, and the eigenvectors cannot be chosen continuously. A typical example is $$(s,t)\mapsto\begin{pmatrix} s & t \\ t & -s \end{pmatrix},$$ whose eigenvalues are $\pm\sqrt{s^2+t^2}$. Up to the shift by $I_2$, Quas' example is just a piecewise $C^1$ section of this two-parameters example, and it inherits its lack of continuous selection of eigenvectors.
Likewise, if analyticity is dropped, a $C^\infty$-example by Rellich shows that eigenvectors need not be continuous functions of a single parameter. Of course, Quas' example can be recast as a $C^\infty$ one, by flatening the parametrisation at $t=0$, say by replacing $t$ by $s$ such that $t={\rm sgn}(s)\cdot e^{-1/s^2}$.

Side remark: Kato's result is only local. If the domain is not simply connected, it could happen that a global continuous selection of eigenvectors is not possible. This is classical in the exemple above if you restrict to the unit circle $s^2+t^2=1$; then the eigenvalues $\pm1$ are global continuous functions, but when following an eigenvector, it experiences a flip $v\mapsto -v$ as one makes one turn.

Being a subgroup: proof by character theory

2011-11-05T11:10:22Z

Let me first cite a theorem due to Frobenius:

Let $G$ be a finite group, with $H$ a proper subgroup ($H\ne (1)$ and $G$). Suppose that for every $g\not\in H$, we have $H\cap gHg^{-1}=(1)$. Then $$N:=(1)\cup(G\setminus\bigcup_{g\in G}gHg^{-1})$$ is a normal subgroup of $G$.

The proof is fascinating. One never proves directly that $N$ is stable under the product and the inversion. Instead, one constructs a complex character $\chi$ over $G$, with the property that $\chi(g)=\chi(1)$ if and only if $g\in N$. This ensures (using the equality case in the triangle inequality) that the corresponding representation $\rho$ satisfies $\rho(g)=1$ if and only if $g\in N$. Hence $N=\ker \rho$ is a subgroup, a normal one!

Does anyone know an other example where a subset $S$ of a finite group $G$ is proven to be a subgroup (perhaps a normal one) by using character theory? Is there any analogous situation when $G$ is infinite, say locally compact or compact?

Edit: If the last argument, in the proof that $S$ is a subgroup, is that $S$ is the kernel of some character, then $S$ has to be normal. Therefore, an even more interesting question is whether there is some (family of) pairs $(G,T)$ where $T$ is a non-normal subgroup of $G$, and the fact that $T$ is a subgroup is proved by character theory. I should be happy to have an example, even if there is another, character-free, proof

Comment by Denis Serre

2013-05-22T11:44:48Z

Just terminology. The matrix $P^{-T}\circ P$ is the gain array matrix associated with $P$. It was studied by C. R. Johnson & H. Shapiro.

Comment by Denis Serre

2013-05-18T14:14:55Z

Is the pointwise convergence $x_n(t)\rightarrow x(t)$ in $X$ a claim or an assumption? If it is a claim, it is false.

Comment by Denis Serre

2013-05-17T11:08:08Z

The French version of te game is "Le cochon qui rit". English translation "The laughing pig".

Comment by Denis Serre

2013-05-14T16:29:02Z

I eventually delete my answer. It seems that I described the set of involutory matrices! Fortunately, this was not a doctoral dissertation; see the MO question about urban legends...

Comment by Denis Serre

2013-05-14T15:41:52Z

Sebastian, I changed deeply my answer, because there was a mistake in calculations. It is still nteresting, I hope, but in a different way.

Comment by Denis Serre

2013-05-06T05:25:02Z

@Joel. Yes, I can!

Comment by Denis Serre

2013-05-05T20:07:03Z

Actually, I should like to accept your answer. Unfortunately, I accepted already JHI's.

Comment by Denis Serre

2013-05-05T20:06:07Z

Beautiful! I'm especially impressed that you found a way to explain it in a convincing way by using 2-D figures.

Comment by Denis Serre

2013-04-29T21:03:17Z

(A2) doesn't work if three of the four subsquares are aligned.

Comment by Denis Serre

2013-04-25T14:15:48Z

Related to this question is the observation that even if you succeed to draw a "general triangle" as described above, you can still "prove" that it has to equal sides (hence two equal angles). Of course, you cheat somewhere, but it is very subtle. This was shown to me by my math teacher when I was 12, and I never forget the argument. This teacher claimed that "Geometry is the art of making correct reasoning from wrong pictures"; in French "la Géométrie est l'art de raisonner juste sur des figures fausses".

Comment by Denis Serre

2013-04-20T13:07:29Z

@S. Sra. If you multiply modulo $2$, you cannot distinguish between $+1$ and $-1$. Therefore the minors are defined only modulo $2$, which means that they are either $0$ or $1$. Since every matrix should be TU modulo $2$, this notion in not interesting in ${\mathbb Z}_2$. It is only interesting in $\mathbb Z$, in which the product of TU matrices is not even unimodular in general. So the question about multiplication is just not a good one.

Comment by Denis Serre

2013-04-20T03:39:47Z

@qianchi. Of course you're right. I'll edit.

Comment by Denis Serre

2013-04-12T06:27:36Z

MO is not designed for posing exercises.

Comment by Denis Serre

2013-03-29T15:07:27Z

However, this answer is somehow duplicate of that by Yemon Choi.

Comment by Denis Serre

2013-03-29T15:04:30Z

This comment finds a wide extension in the notion of numerical measure of a matrix, which is supported by the numerical range. See Th. Gallay & D. S. Comm. Pure Appl. Math. 65 (2012), pp 287-336.