51

How about Jacobi's proof?
See, e.g., Folkmar Bornemann, ``Teacher's Corner - kurze Beweise mit
langer Wirkung,'' DMV-Mitteilungen 3-2002, Seite 55 (in German, sorry). Sorry, I don't have the original reference.
The idea is simple, define $\Sigma(A)=\sum_{i=1}^n\sum_{j=i+1}^n a_{ij}^2$ for $A=(a_{ij})$ a symmetric real matrix. Then minimize the function $O(...

49

If "elementary" means not using complex numbers, consider this.
First minimize the Rayleigh ratio $R(x)=(x^TAx)/(x^Tx).$ The minimum exists and is real.
This is your first eigenvalue.
Then you repeat the usual proof by induction in dimension of the space.
Alternatively you can consider the minimax or maximin problem with the same Rayleigh ratio,
(find the ...

answered Jan 11 '13 at 14:06

Alexandre Eremenko

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36

For real $a,b,c$ and imaginary $d$ the matrix $A$ has chiral symmetry, meaning it anticommutes with a matrix $X$ that squares to the identity:
$$X=\left(
\begin{array}{cccc}
0 & 0 & 0 & -i \\
0 & 0 & i & 0 \\
0 & -i & 0 & 0 \\
i & 0 & 0 & 0 \\
\end{array}
\right),\;\;XA+AX=0,\;\;X^2=I.$$
Hence the spectrum ...

answered Aug 5 at 11:36

Carlo Beenakker

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33

How to understand the Graph Laplacian (3-steps recipe for the impatients)
read the answer here by Muni Pydi. This is essentially a concentrate of a comprehensive article, which is very nice and well-written (see here).
work through the example of Muni. In particular, forget temporarily about the adjacency matrix and use instead the incidence matrix.
Why? ...

dg.differential-geometry graph-theory gt.geometric-topology sp.spectral-theory spectral-graph-theory

31

Sometimes you can
Since there seem to be no hard results along these lines on the literature, I decided to have a look at the two shapes in the question and see if there are some relatively accessible results and happily, as it turns out, there are. In particular:
The isospectral surfaces $D_1$ and $D_2$ in the question are acoustically distinguishable: ...

25

Let me give it a try. This one only uses the existence of a maximum in a compact set, and the Cauchy-Schwarz inequality.
Let $T$ be a selfadjoint operator in a finite dimensional inner product space.
Claim: $T$ has an eigenvalue $\pm\|T\|$.
Proof: Let $v$ in the unit sphere be such that $\|Tv\|$ attains its maximum value $M=\|T\|$. Let $w$ also in the ...

24

There are examples due to Ikeda of isospectral Lens spaces which are not homotopy equivalent.
Likeliest the simplest examples are the compact connected 3-dimensional flat manifolds which are a tetracosm and didicosm. These are isospectral but not homotopy equivalent. I'm not sure if they are known to have the same spectrum for the Laplace-de Rham operator (...

23

The $k$-th eigenfunctions are actually easy to describe: In $\mathbb{C}^{n+1}$ with unitary complex coordinates $z_0,z_1,\ldots,z_n$, write $Z = |z_0|^2+\cdots+|z_n|^2$.
Now, for a given $k\ge0$, let $H_k$ be the (real) vector space of real-valued polynomials $p(z,\bar z)$ that are homogeneous of degree $k$ in the $z$-variables and degree $k$ in the $\...

22

This is a special case of a rank one perturbation or a rank one update, and there is plenty of work on such. See the nice 2010 lecture notes by Andre Ran.

answered Jul 6 '16 at 20:43

Igor Rivin

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19

Ramanujan graphs were first defined by Lubotzky, Phillips and Sarnak:
http://math1.math.huji.ac.il/~alexlub/PAPERS/ramanujan%20graphs/ramanujanGraphs.pdf
As you can see, they are $d$-regular and and all eigenvalues of the adjacency matrix, except for $\pm d$, are in $[-2\sqrt{d-1},2\sqrt{d-1}]$. This is equivalent to your 2nd definition. The name "Ramanujan"...

18

From a certain point of view the premise of the question is wrong. The study of sympletic manifolds with no additional structure is akin to differential topology rather than differential geometry. From this point of view, that there is no spectral theory of symplectic manifolds is no more surprising than that there is no spectral theory of smooth manifolds. ...

dg.differential-geometry riemannian-geometry sg.symplectic-geometry sp.spectral-theory differential-operators

18

A direct and precise way to arrive at your answer is to appeal to the theory of Fourier transforms of distributions. If I define the Fourier transform as
$$F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{ikx}\,dx,$$
then the equation $f(x+1)+f(x)=g(x)$ becomes upon Fourier transformation
$$e^{-ik}F(k)+F(k)=G(K)\Rightarrow F(k)=\frac{G(k)}{1+e^{-ik}}.$$...

answered Jun 24 '19 at 14:57

Carlo Beenakker

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18

In the open subset of $M_n(\mathbb{R})$ where the $\lambda_i$ are distinct, they are $C^{\infty}$ functions: this follows from the implicit function theorem.
On the other hand, when some eigenvalue has multiplicity $>1$ you don't get more than continuity. For example if $A=\begin{pmatrix} 0 & 1\\ 1 & t
\end{pmatrix}$ the largest $\lambda_i$ is $\...

linear-algebra ca.classical-analysis-and-odes sp.spectral-theory eigenvalues real-algebraic-geometry

18

Preliminary remark. As mentioned in the comments, I find the notion "resolvent formalism", as well as the description in the Wikipedia article, rather misleading - resolvents are not somekind of formalism, and they are certainly not a mere "technique for applying complex analysis to spectral theory" (as claimed in the Wikipedia article). ...

17

Sorry for the necromancy, but the most basic application of the spectral theorem has to be the second derivative test in multivariable calculus, no?
If $f:\mathbb{R}^n \to \mathbb{R}$ is a smooth function, then its Hessian $D^2f$ is a symmetric bilinear form at each point of $\mathbb{R}^n$. By the spectral theorem, it has an orthogonal basis of ...

17

This is just the details of the first step of Alexander Eremenko's answer (so upvote his answer if you like mine), which I think is by far the most elementary. You only need two facts: A continuous function on a compact set in $R^n$ achieves its maximum (or minimum), and the derivative of a smooth function vanishes at a local maximum. And there's no need for ...

17

These are usually known as the Laplacian, the normalized Laplacian and the unsigned Laplaian. All three are positive semidefinite. If the graph is regular, they all provide the same information.
If the graph is not regular they are, in general, independent. The normalized Laplacian is the right tool for the analysis of random walks. The spectral information ...

17

Johnson graphs do not cause difficulty to existing programs. Actually they are rather easy; nauty can handle them up to tens of millions of vertices, and so can other programs such as Traces and Bliss.
The difficulty that Babai refers to is more theoretical. The Johnson graph $J(v,t)$, which has $n=\binom vt$ vertices, has the property that $\Theta(n^{1/t})...

17

The characteristic variety (i.e. vanishing locus of the symbol) of a symplectomorphism invariant scalar differential equation is a real projective hypersurface invariant under the group of projectivized linear symplectic transformations. This group acts transitively on the real points of projective space, so preserves no hypersurface.

dg.differential-geometry riemannian-geometry sg.symplectic-geometry sp.spectral-theory differential-operators

16

Of course, Igor's answer points the way to working out the answer the OP wanted, but it may not be clear, even after you have got the eigenvalues, what the corresponding eigenfunctions are, or that they have a simple geometric interpretation analogous to the one for the sphere, as the OP asks.
The nice way to describe the first eigenfunctions on the $n$-...

answered Nov 22 '12 at 14:38

Robert Bryant

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16

In fact, the original motivation behind Lubotzky--Phillips--Sarnak's construction of Ramanujan graphs was in analogy with modular curves $Y(N)=\mathbb H^2/\Gamma(N)$ for the principal congruence subgroups $\Gamma(N)\subseteq\operatorname{PSL}(2,\mathbb Z)$. So the answer is yes, there is a continuous analogue, but in fact it came first!
Let me give a few ...

16

An equivalent trick : Let $J:= \operatorname{diag}(1,i,-1,-i)$. Then $J^*AJ=iB$ where $B$ is real and skew-symmetric. Hence the spectrum of $iB$ (thus that of $A$) comes by pairs $\pm\lambda$.

15

One simple example with a special matrix, which has somehow "a continuum" as eigenvalue...
Consider some function $ f(x) = K + ax + bx^2 + cx^3 + ... $ having a nonzero radius of convergence. Then think of the infinite matrix of the form
$$ \small \begin{bmatrix}
K & . & . & . & \cdots \\\
a & K & . & . & \cdots \\\
...

15

I was informed by Sugata Mondal at the MPI that Scott Wolpert proved the following result in his 1994 Annals paper Disappearance of cusp forms in special families:
Theorem 5.14. The eigenvalues of the Laplacian above $\tfrac14$ on a closed hyperbolic surface vary nontrivially under analytic deformations.
That is, if $g_t$ is a real-analytic path of ...

15

Suppose $f(x,y,z)$ is symmetric (in the following, symmetric tout court always means "symmetric w.r.to the three variables $(x,y,z)$") . Then $\mathcal{L}f(x,y,z):=(x+y)(x+z)f(x-1,y,z)-x^2f(x,y,z)$ is already symmetric w.r.to $(y,z)$, so it is symmetric if and only if it is symmetric w.r.to $(x,y)$, that is, after simplifications, if and only if $f$ ...

15

The keyword is the Cartan decomposition in the theory of symmetric spaces.
In short, when an eigenvalue is simple (its multiplicity is $1$) it is locally an analytic function. But when the eigenspace is degenerate (the multiplicity is greater than $1$), the eigenvalue function is not differentiable. The problem is essentially one of choosing branches: if ...

linear-algebra ca.classical-analysis-and-odes sp.spectral-theory eigenvalues real-algebraic-geometry

14

As far as I know, this is a very delicate question. That is, already in two dimensions, there can be only finite discrete spectrum. (See Phillips-Sarnak and Wolpert.)
Even on the modular curve $SL(2,\mathbb Z)\backslash \mathfrak H$, it was highly non-trivial to prove existence of infinitely-many $L^2$ eigenvalues, apparently requiring Selberg's invention ...

14

Using symmetry, you can extend any Dirichlet eigenfunction on the upper half-sphere to the entire sphere $\mathbb{S}^{N-1}$. Therefore, the spectrum of the upper hemisphere is a subset of the spectrum of the full sphere. You are searching for the spherical harmonics which vanish on the great circle $x_N \equiv 0$. The reference that I've seen that explicitly ...

14

Perhaps I can contribute to the history part the question, since I was quite close to the Institut Fourier at that time and was very interested in their work (I am a physicist). Grenoble now has several different research groups doing graph theory (like G-SCOP, Institut Fourier, GIPSA-lab, LIG) but I think L'Institut Fourier was the early one for graph ...

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