18

The complete reference is Kato's book Perturbation theory .... But perhaps you need only the most basic results. Then see my book Matrices (Springer GTM #216), 2nd edition. This is Section 5.2.
Mind however that this is a much more elaborate topic than what you could think at first glance. There are at least four completely distinct aspects:
On the one ...

answered May 11 '18 at 19:49

Denis Serre

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16

If $B$ depends on a single parameter $t$ then derivating with respect to $t$ the equality
$$ B n_i =\lambda_i n_i $$
we deduce
$$\dot{B} n_i +B\dot{n}_i=\dot{\lambda}_i n_i +\lambda_i\dot{n}_i. $$
Here we assume that $\Vert n_i\Vert =1$. Hence $\dot{n}_i\perp n_i$, $\forall i$. Taking the inner product of the above equality with $n_i$ and observing ...

answered Jan 27 '16 at 23:12

Liviu Nicolaescu

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15

Edit (bis). There are two answers, depending on whether loops about vertices are allowed or not. In addition, the case of regular graphs is completely described.
If loops are allowed
The relation between matrices is
$$A+{\widetilde A}=J$$
where $J={\bf1}{\bf1}^T$ is the all-ones matrix. The first consequence is that the sum of the eigenvalues of $A$ and ${\...

13

First, one should conjugate all matrices by
$$
\begin{pmatrix}
\operatorname{diag}(\omega_1,\dots,\omega_n) & 0 \\ 0 & 1
\end{pmatrix}
$$
as this converts $S(t)$ to a rotation matrix while leaving the reflection $N$ unchanged.
The matrix $P^{-1} \operatorname{diag}(1,\dots,1,-1) P$ has a line as its -1 eigenspace and a hyperplane as its +1 ...

12

The $2^n\times 2^n$ dimensional Hadamard matrices $H_{2^n}$ are also called Sylvester matrices or Walsh matrices. There are only two distinct eigenvalues $\pm 2^{n/2}$, so the eigenvectors are not in general orthogonal. An orthogonal basis of eigenvectors is constructed recursively in A note on the eigenvectors of Hadamard matrices of order $2^n$ (1982) and ...

answered Oct 3 '19 at 19:44

Carlo Beenakker

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10

The claim is false.
In particular, we have $A=kee^T-kI$, so that $\lambda(A)=((n-1)k,-k,\ldots,-k)$, so that $\|A\|_* = 2(n-1)k$.
Now generate a random matrix $B$ such that $B_{ii}=0$, $B_{ij}=B_{ji}$ and $B_{ij} \le A_{ij}$ for every entry. It does not matter that the entries of $B$ are integers or not (to have a "clean" example, I round the entries below ...

9

First, the definition of $positive$ should be clarified. It could mean all entries strictly positive, or merely nonnegative, or that the matrix is primitive (all entries nonnegative and some power strictly positive).
I will use $\rho(M)$ to denote the spectral radius of a matrix $M$, which in case $M$ is nonnegative, is the Perron eigenvalue.
Note a ...

9

Eigenvalues are roots of the characteristic polynomial. In the case the polynomial has simple roots, they smoothly depend on the coefficients. Below is the proof in the real case. The argument below can be easily modified to more general situations.
Theorem. For $a=(a_0,a_1,\ldots,a_n)\in\mathbb{R}^{n+1}$, $a_n\neq 0$ let $P_a(x)=a_nx^n+\ldots+a_1x+a_0$. ...

answered May 11 '18 at 19:35

Piotr Hajlasz

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9

(1) No. Counterexample: the symmetric $3 \times 3$ matrix
$$
M(a,b) = \left[
\begin{array}{ccc} 0 & a & b \cr a & 0 & b \cr b & b & 0 \end{array}
\right]
$$
with $0 < b < a$ has $\lambda_3 = -a$ with 1-dimensional eigenspace
generated by $(1,-1,0)$.

answered Mar 23 '21 at 21:06

Noam D. Elkies

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8

Tl;dr: The answer to your question is, Yes, but you picked the wrong weights on the graph, which is why the eigenvalues were off.
Before I get into some generalities about how to think about this kind of approximation problem, in addition to the paper Steve Huntsman posted above, you may find this paper by Burago-Ivanov-Kurylev interesting. The main theorem ...

8

The article linked by the OP proves that every degree $k$ spherical harmonic is a $k(k+1)$-eigenfunction of the Laplacian on $S^2$. The OP asks: is every eigenfunction of the Laplacian a spherical harmonic?
The answer is yes, though it seems that neither the linked article nor the Wikipedia page on spherical harmonics address this issue.
It is not hard to ...

answered Jun 21 '16 at 18:40

Paul Siegel

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8

Counterexample: the matrix $I \times I$ has eigenvectors that are not in product form, since every vector is an eigenvector of it and not every vector can be written in product form.

answered Dec 1 '19 at 17:49

Federico Poloni

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8

If $G$ is regular, then $J$ and $A_G$ are simultaneously diagonalizable (i.e. they have a common set of eigenvectors).
That is, the eigenvalues of $xA_G$ and $J$ (to the same eigenvectors) just add up to the eigenvalues of $B=J+xA_G$.
Note that the spectrum of $J$ is $\{0^{n-1}, n^1\}$, and that the eigenvalue $n$ corresponds to the eigenvector $(1,...,1)$. ...

7

Let $f(T)$ be the monic polynomial of smallest degree such that $f(A)x=0$ (which is cheap to compute if $n$ isn't too big.) Then $A^sx=\lambda x$ if and only if $f(T)$ divides $T^s-\lambda$.
So a necessary condition is that $f(T)$ has all roots of same length.
To obtain a necessary and sufficient condition, in the case $k=\mathbb Q$, requires some number ...

7

No. Consider the following example, due to Rellich 1937.
\begin{align*}
x_+(t) &:= \begin{pmatrix} \cos\frac1t \\ \sin\frac1t \end{pmatrix}, \quad
x_-(t) := \begin{pmatrix} -\sin\frac1t \\ \cos\frac1t \end{pmatrix}, \quad
\lambda_\pm(t) = \pm e^{-\frac1{t^2}},\\
A(t) &:= (x_+(t),x_-(t))
\begin{pmatrix} \lambda_+(t) & 0 \\
0 ...

7

first order perturbation theory in $\epsilon$ on the matrix B with elements $B_{nm}+\epsilon \delta_{nk}\delta_{ml}$ tells you that
$$\frac{d\vec{n}_i}{dB_{kl}}=\sum_{j\neq i}\frac{(\vec{n}_j)_k (\vec{n}_i)_l}{\lambda_i-\lambda_j}\vec{n}_j$$
with $(\vec{n}_i)_l$ the $l$-th component of the vector $\vec{n}_i$.

answered Jan 27 '16 at 17:47

Carlo Beenakker

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7

In view of Terry's comment on Michael's answer, it's perhaps worth pointing out that this monotonicity also fails if both domains are required to be convex. We can take $M_2=[0,L]^2$ as a square of side length $L$. Then $\lambda_2(M_2)=\pi^2/L^2$ (possible eigenfunction $u=\cos \pi x/L$). If we now take $M_1\subseteq M_2$ as a thin rectangle close to the ...

answered Jun 15 '16 at 3:23

Christian Remling

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7

I don't have an answer, but it appears that the eigenvalues are always real. I don't have a proof, but have checked this using Sturm sequences for $1 \le a \le b \le 30$.
You're unlikely to get "an exact analytic result", as the characteristic polynomial seems to be irreducible over the rationals unless $a=0$ (in which case $\lambda - 1$ is a factor). ...

7

The eigenvalues of a square matrix $A$ are the roots of the characteristic polynomial, and are analytic except where their multiplicities change.
Thus if (in a certain open region of parameter space) there is one eigenvalue,
of algebraic multiplicity $m$, inside a positively oriented closed contour $\Gamma$ in $\mathbb C$, and no eigenvalues on $\Gamma$, ...

7

Yes. Let the size of your matrix be $n$.
Your condition implies that there is an $n-1\times n-1$ submatrix whose determinant is not identically equal to $0$.
Assume without loss of generality that this is the submatrix formed by the first $n-1$ rows and columns. Then we can set $u_n=1$ and find a vector $u(z)$ such that
$M(z)u(z)=0$ by solving the system of $...

answered Jul 22 '20 at 0:13

Alexandre Eremenko

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6

There are some quantifiers unclear in your question, but regardless of how
to read it, your assertion is false. -- The smallest counterexample with
blocks of pairwise distinct size all of whose eigenvalues are integers has
blocks of size $5$, $8$ and $12$, and set of eigenvalues $\{-10,-6,0,16\}$.
This can be found with GAP as follows:
First we write a ...

answered Dec 4 '15 at 19:00

6

There won't be a nontrivial closed form in general. A trivial "closed-form" is mentioned below.
The Hadamard product $A \circ B$ is a principal submatrix of the Kronecker product $A\otimes B$. Thus, let $A=U\Lambda U^{-1}$ and $B=VDV^{-1}$ (I assumed $B$ is diagonalizable and not fully arbitrary as required in the question). Then, we have
\begin{equation*}
...

6

For an integer matrix $A$ and an integer eigenvalue $\lambda$, $A - \lambda I$ is an integer matrix, and Gaussian elimination will produce rational eigenvectors; multiply by a common denominator and you'll get an integer eigenvector.
But the question in your second paragraph is different: there is an eigenvector with entries $-1, 0, 1$ iff there are two ...

6

A course in constructive algebra by Mines, Richman & Ruitenburg, covers the constructive development of discrete fields and their algebraic completion, and even their valuation-metric completion.
This is a very direct and clear development, showing that for discrete fields (such as $\mathbb{Q}$), the algebraic closure (in this case $\mathbb{A}$) is ...

6

### Can a vector-function $v:\mathbb{R}^n\to \mathbb{R}^n$ be an eigenvector of its own Jacobian matrix?

I am surprised that Ramesh's answer was voted up. Its first paragraph does not convince me. Here is my analysis:
Consider an integral curve of $v$, that is a curve $t\mapsto X(t)$ so that $\dot X=v(X)$. Then
$$\frac{d}{dt}v=(\dot X\cdot \nabla)v=\lambda v.$$
This shows that $t\mapsto v(X(t))$ keeps a constant direction. Therefore the curve is a straight ...

6

It seems to me that $H_{N}$ is the character table of an elementary Abelian $2$-group of order $2^{n}$ (with respect to a suitable ordering of elements). As such, its rows are orthogonal by the orthogonality relations for group characters. Also, it is clear, by induction that $H_{N}$ is symmetric.Hence we have $H_{N}H_{N}^{t} = 2^{n}I$ (since it is a ...

answered Oct 5 '19 at 19:54

Geoff Robinson

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5

Search for the term "G-circulant matrix".

5

Take $M_1$ to be a region consisting of two "blobs" connected by a narrow channel of length 1 and width $\epsilon$. Now choose $u_1$ and $u_2$ such that each is 1 in one of the blobs, 0 in the other, and linearly interpolated in the narrow channel. It follows that both $\int |\nabla u_1|^2$ and $\int |\nabla u_2|^2$ are order $\epsilon$. Consequently $N_{M_1}...

5

In fact for a fixed $n$, the matrices $M(l, n)$ for $l>0$ commute with each other and thus are simultaneously diagonalisable. For your second question, if $\{p_j(y)\}$ is a sequence of polynomials satisfying
\begin{eqnarray}
\left(\frac{t}{\sinh t}\right)^y=\sum_{j=0}^\infty p_j(y)t^{2j}.
\end{eqnarray}
then the $i$-th entry of an eigenvector ...

5

This seems to be answered in the accepted answer to this question: The height of the Perron-Frobenius eigenvector
For convenience, here is the estimate:

answered May 10 '18 at 15:45

Igor Rivin

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