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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31

Here are the results that you are probably looking for.
The first one is for positive definite matrices only (the theorem cited below fixes a typo in the original, in that the correct version uses $\prec_w$ instead of $\prec$).
Theorem (Prob.III.6.14; Matrix Analysis, Bhatia 1997). Let $A$ and $B$ be Hermitian positive definite. Let $\lambda^\downarrow(X)...

23

The answer is 'no'. The generic pair $A$ and $B$ of $4$-by-$4$ Hermitian symmetric matrices will not have any nonzero real linear combination that has a double eigenvalue.
For a specific example, take
$$
A = \begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&-2&0\\0&0&0&2\end{pmatrix}
\quad \text{and}\quad
B = \begin{...

answered Jan 1 '18 at 17:29

Robert Bryant

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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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21

The type of matrix you have written down is called Jacobi matrix and people are still discovering new things about them basically their properties fill entire bookcases at a mathematics library. One of the reasons is the connection to orthogonal polynomials. Basically, if $\{p_n(x)\}_{n\geq 0}$ is a family of orthogonal polynomials, then they obey a ...

21

This is a well-known fact. A simple proof : setting $y=B^{1/2}x$, we have $\|y\|^2\le y^TB^{-1/2}AB^{-1/2}y$, that is $I_n\le B^{-1/2}AB^{-1/2}$. The eigenvalues of the latter symmetric matrix are thus $\ge1$. Its inverse $B^{1/2}A^{-1}B^{1/2}$ has eigenvalues $\le1$, that is $B^{1/2}A^{-1}B^{1/2}\le I_n$. This gives $z^TB^{1/2}A^{-1}B^{1/2}z\le\|z\|^2$. ...

19

This is true. Indeed, you can estimate the sum of all $n^2$ elements of $A$ rather than individual elements. (Thanks to thomashennecke for observing this, my original answer dealt with the row sums of $A$ instead of the sum of all $n^2$ elements of $A$.) Moreover, you do not need the elements to be integer; it suffices to have them uniformly bounded away ...

19

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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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

17

Yes, it is true that the largest eigenvalue is bounded by the largest absolute row sum or column sum. You can check Gershgorin circle theorem.
Actually, all the eigenvalues lie in the union of all Gershorin circles.

17

Counterexample: let $k=7$, and let $B$ be the circulant matrix with $B_{ij}=1$ iff $i-j \in \{1,2,4\} \bmod 7$. Then $X_B$ is $I + \frac12 J$, with characteristic polynomial $(x-1)^6 (x-\frac92)$. Or use $B+I$ instead to get $I + \frac13 J$, with characteristic polynomial $(x-1)^6 (x-\frac{10}{3})$.

answered Sep 28 '16 at 23:39

Noam D. Elkies

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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

The smallest eigenvalue can go up or down when an edge is removed.
For "down": $G=K_n$ for $n\ge 3$.
For "up": Take $K_n$ for $n\ge 1$ and append a new vertex attached to a single vertex of the original $n$ vertices. Now removing the new edge makes the smallest eigenvalue go up.
Both of these follow from the fact that the smallest eigenvalue of a ...

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

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

Amongst the polynomials that can arise as characteristic polynomials of tridiagonal matrices with zero diagonal, one finds the Hermite polynomials. Schur showed that Hermite polynomials of even degree are irreducible and that their Galois groups are not solvable. Hence there can be no closed form expression for the zeros in terms of the $b_i$'s in general.

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 ...

13

There exists a simple graph $G=(V,E)$ together with a vertex $v\in V$ such that $\operatorname{det}(A(G))$ and $\operatorname{det}\left(A(G\backslash\{v\})\right)$ are equal and non-zero.
Example. The determinant of the adjacency matrix of the complete graph $K_n$ is known to be $(-1)^{n-1}(n-1)$. Let $G'=K_5$ be the complete graph on the vertex set $[1,5]$...

12

A cute fact that is trivial to prove is this: define the characteristic polynomial of a matrix $M$ by $\phi_M(x) = |xI-M|$. Then for any $A$ and any $s$, $$\phi_{A+sJ}(x) = (1-s)\phi_A(x)+s\phi_{A+J}(x).$$
The proof only needs the fact that a determinant doesn't change when a multiple of one row is added to a different row, so I don't see why it wouldn't be ...

12

I have no idea what is going on, but your conjecture is not correct. This is more transparent perhaps in the complex version. Consider
$$
A = \begin{pmatrix} i & a \\ 0&-i \end{pmatrix} , \quad a>0.
$$
This matrix is already in (complex) Schur form, and the obvious procedure to make the eigenvalues real and similar in spirit to what you propose ...

answered Jul 4 '17 at 19:28

Christian Remling

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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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11

As mentionned by other answers, simple eigenvalues are $C^\infty$, while non-simple ones are not. Let me add however two important properties which you can find in Kato's book Perturbation theory of linear operator.
The first one is that each $\lambda_j$ is a Lipschitz function. This statement is still valid if you replace ${\bf Sym}_n({\mathbb R})$ by a ...

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

10

I think you mean dense in $[0,\infty)$, since the spectral radius of a nonnegative integer matrix must be at least 1 (the product of all nonzero eigenvalues must be a nonzero integer). You are effectively asking whether Perron numbers are dense in $[1,\infty)$, and this is easy to see. For example, let $A_n$ be the companion matrix of $x^n-x-1$ and $\...

10

If $A$ has eigenvalues $\lambda_1 \geq \dots \geq \lambda_n$ and $B$ has eigenvalues $\mu_1 \geq \dots \geq \mu_n$, then the largest eigenvalue of $A+B$ can take any value between $\lambda_1+\mu_1$ (the maximal value) and $\max_{1 \leq i \leq n} \lambda_i+\mu_{n+1-i}$ (the minimal value). That these inequalities are necessary comes from the Weyl ...

10

This problem is essentially the same as this one. In particular, let $J$ be the anti-diagonal identity matrix, and $P^{-1}$ be the matrix mentioned in the link above. Then, the matrix in the current post is nothing but
\begin{equation*}
B_n = JP^{-T}J^T.
\end{equation*}
Since $J^TJ=I$, we can recover eigenvectors and values of $B_n$ using the derivation ...

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 ...

10

I take this from my comments above and add something.
The question is about rational points on the surface $S$ given by
$$ \Delta(a,b,c) := (ab+bc+ca)^3 - 27(abc)^2 = z^2 $$
in the weighted projective space ${\mathbb P}_{1,1,1,3}$. Along the line
$b+c = 4a$ the left hand side factors as a square times $b^2 + 18bc + c^2$,
and the equation effectively reduces ...

10

In the case $n=4$ you could have $D = \pmatrix{0 & 0 & 0 & 1\cr 0 & 0 & 1 & 0\cr
0 & 1 & 0 & 0\cr 1 & 0 & 0 & 0}$, in which case $A_1 A_2 A_3 A_4 A_1 A_2 A_3 A_4 = \pmatrix{0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 1 & 1 & 0\cr 1 & 0 & 0 & 1\cr}$ has eigenvalues $0$ ...

10

One can prove this statement along the following lines.
Prove that ${\rm Trace}(M(l,n)) = 1+l +\cdots + l^{n-1}$.
Prove that $M(l,n)^p = M(l^p,n)$.
Clearly, these statements 1 and 2 together imply that the eigenvalues
are $1,l,\ldots, l^{n-1}$.
We will assume throughout that $l\geq 2$, otherwise the statement is obvious.
The proof of the first statement ...

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