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20 votes

Differentiability of Eigenvalues - Perturbation Theory

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 ...
Denis Serre's user avatar
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15 votes
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Eigenvalues of the complement of a graph

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 ...
Denis Serre's user avatar
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13 votes
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What is known about the eigenvectors of the $2^n \times 2^n$ Hadamard matrix?

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 ...
Carlo Beenakker's user avatar
10 votes

Lower eigenvectors of nonnegative matrices with zero trace

(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 &...
Noam D. Elkies's user avatar
10 votes

How to show the following matrix has eigenvalues $-d,-d+1,...,d$?

You hope to show for each $n$ that the $(n+1)\times(n+1)$ matrix $$ B_n = \begin{pmatrix} 0 &n & 0 &0 & \cdots &0 & 0 \\ 1 & 0 & n-1 &0& \cdots & 0 & ...
Kenta Suzuki's user avatar
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9 votes

Differentiability of Eigenvalues - Perturbation Theory

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 ...
Piotr Hajlasz's user avatar
9 votes
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Eigenvectors of Kronecker Product

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.
Federico Poloni's user avatar
9 votes
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The eigenvalues of the matrix $\Big(\frac{1}{\cos(k-l)\frac{\pi}{n}}\Big)_{k,l=1}^n$

Here, we verify the observation of @BrendanMcKay that the eigenvalues are $n$ with multiplicity $(n+1)/2$ and $-n$ with multiplicity $(n-1)/2$. Note that your matrix is skew-circulant, and it is known ...
Jason Gaitonde's user avatar
9 votes

How to show the following matrix has eigenvalues $-d,-d+1,...,d$?

As requested by Iosif Pinelis, here is another version of the (perfectly correct, in my opinion) answer by Kenta Suzuki. Let $V$ be the $d+1$-dimensional vector space of homogeneous polynomials of ...
მამუკა ჯიბლაძე's user avatar
8 votes

Differentiability of Eigenvalues - Perturbation Theory

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) ...
Robert Israel's user avatar
8 votes
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Eigenvalues of adjacency matrix of a k-regular graph

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 ...
M. Winter's user avatar
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7 votes

Integer eigenvectors

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 ...
Robert Israel's user avatar
7 votes

Eigenvalues of a matrix with binomial entries

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 ...
Robert Israel's user avatar
7 votes

Constructive proof of a rational version of Perron-Frobenius?

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. ...
Franka Waaldijk's user avatar
7 votes

What is known about the eigenvectors of the $2^n \times 2^n$ Hadamard matrix?

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 ...
Geoff Robinson's user avatar
7 votes
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Continuity of eigenvectors

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 ...
Alexandre Eremenko's user avatar
6 votes
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Lower bound on the entries of the Perron vector

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:
Igor Rivin's user avatar
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6 votes
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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)...
Denis Serre's user avatar
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6 votes
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Eigenvalues of a rank-one update of a symmetric matrix

Given eigenvectors and eigenvalues of the symmetric matrix $A$, you can transform to a basis where $A$ is a diagonal matrix $D$; the vector $x$ in that basis transforms to $\tilde{x}$. Then the ...
Carlo Beenakker's user avatar
6 votes
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Given an eigenvalue equation (elliptic PDE) in a ball $B_R$, prove the convergence of the first nonzero $\lambda_R$ and its eigenfunction $\phi_R$

For simplicity let me take $H$ smooth or at least $C^\alpha$ so that I don't have to worry about elliptic estimates: Limit as $R\to\infty$: Use $$ \lambda_R = \inf_{\phi \in C^\infty_c(B_R)}\frac{\...
Otis Chodosh's user avatar
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5 votes
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Constructive proof of a rational version of Perron-Frobenius?

Yes, both Theorem 1 and Theorem 2 have constructive proofs. In the following, I will work with rational numbers, but the same arguments work for any ordered field. (Note that in constructive logic, ...
darij grinberg's user avatar
5 votes
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Eigenvectors of a matrix with entries involving combinatorics

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 ...
No_way's user avatar
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5 votes

The existence of $v\in A\otimes_{\mathbb{K}}A$ such that $(a\otimes_{\mathbb{K}}1)v=(1\otimes_{\mathbb{K}}a)v$

Let $I$ be the ideal in $A\otimes A$ generated by elements of the form $a\otimes 1-1\otimes a$. (Equivalently, this is the kernel of the multiplication map $A\otimes A\to A$.) We will assume that $0\...
Neil Strickland's user avatar
5 votes
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Find a way to apply the MLE on Fisher or Covariance matrix to make cross-correlations

OK, let's turn the tables around. I'll describe in my language a physical setup, the mathematical formulation, and the answer that fit the three words I understand: "detector", "error&...
fedja's user avatar
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5 votes
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Directed graph whose adjacency matrix admits only 0 as eigenvalue

$0$ is the only eigenvalue of $A$ if and only if $A$ is nilpotent, which is equivalent to $A^n=0$. But $A^n=0$ expresses that for all $i$ and $j$, there is no path of length $n$ from $P_i$ to $P_j$. ...
Peter Mueller's user avatar
4 votes
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Eigenvector of a nonnegative matrix in closed form

If you write $A\nu=\nu$ as a system of equations it can be written as $$ \frac{\nu_i}{1-\alpha_i}=\sum_{j=1}^{n}\frac{\alpha_i\nu_j}{1-\nu_j} \qquad \forall i=1, ..., n. $$ Then the change of ...
martin cripps's user avatar
4 votes
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Solving linear system when one eigenvalue is known

Yes and no. If you modify slightly the implementation of a Krylov subspace method such as GMRES, you can construct a method with a projection subspace that contains the known eigenvector. This ...
Federico Poloni's user avatar
4 votes

"Unimodality" of the positive eigenvector of a non-negative irreducible matrix?

Let $A$ be as in the question and let $n$ denote the size of $A$, i.e. $A \in \mathbb{R}^{n \times n}$. Troughout, fix $i^* \in \{1,\dots,n\}$. We call a vector $x \in \mathbb{R}^n$... ... $i^*$-...
Jochen Glueck's user avatar
4 votes

Differentiability of eigenvalue and eigenvector on the non-simple case

I think Theorem 6.8 on page 122 in Kato: Perturbation Theory for Linear Operators may help (at least for the question concerning the eigenvalues of the symmetric $A$ and $B$ matrices). Theorem: ...
Zoltan Zimboras's user avatar
4 votes

XOR circulant matrices?

The object in question is also known as the group matrix or the Dedekind matrix, and it is closely related to the Frobenius determinant theorem which was at the origin of the representation theory.
Oleksandr  Kulkov's user avatar

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