New answers tagged linear-algebra
2
votes
What are the points of the algebra of polynomial functions on an arbitrary vector space?
I asked Andrew Snowden this question, and he pointed me to his paper "Big Polynomial rings and Stillman's Conjecture", with Dan Erman and Steven Sam. This paper assumes the axiom of choice....
- 156k
12
votes
Invertibility of a matrix defined using inner product
Let $X \in \mathbb{R}^{m \times n}$ be the matrix with columns $x_1$, $\ldots$, $x_n$. Then your matrix $A$ can be written as
$$
A = X^TX(nI - J),
$$
where $J$ is the $n \times n$ matrix with every ...
- 6,351
8
votes
Accepted
Eigenvalues of a certain combinatorially defined matrix
Let $E_m$ be the $m\times m$ all-ones matrix. We'll show that the spectrum of
$$B_n:=I_{n(n-1)}+E_{n(n-1)}-A_n,$$
the matrix whose $((i,j),(k,\ell))$ entry is $1_{i=k}+1_{j=\ell}$, consists of $n^2-3n+...
- 602
3
votes
Inequality for commuting hermitian operators
Yes. Since $S_1$, $S_2$ commute, there exists an orthonormal basis in which they are both diagonal: $S_1={\rm diag}\,(a_1,\ldots,a_n)$ and $S_2={\rm diag}\,(b_1,\ldots,b_n)$, $a_i\geqslant c_1,b_i\...
- 109k
5
votes
Accepted
Inequality for hermitian matrices
No. For example, let
\begin{align*}
p_1 = \begin{bmatrix}
1 & 0 \\ 0 & 0
\end{bmatrix}, \ \ p_2 = \begin{bmatrix}
0 & 0 \\ 0 & 1
\end{bmatrix}, \ \ S_1 = \begin{bmatrix}
1 & 1 \\ 1 ...
- 6,351
4
votes
Accepted
Eigenfunctions of the Laplace–Beltrami operator on the coadjoint orbit of $\mathfrak{su}(n)$
For the generic coadjoint orbit as you require, i.e. the full flag manifold $\operatorname{SU}(n)/\text{(max torus)}$, the paper of Yamaguchi (1979) cited at this question has not only the eigenvalues,...
- 31.5k
1
vote
Accepted
Is smoothness preserved under an isometric isomorphism?
$\newcommand\R{\Bbb R}$Counterexample: $X=\R^2$, $\|x\|_1=|u|+|v|$ and $\|x\|_2=\max(|u|,|v|)$ for $x=(u,v)\in\R^2$.
Then $\|\cdot\|_2\le\|\cdot\|_1$ and the map $\R^2\ni(u,v)\mapsto\frac12\,(u-v,u+v)$...
- 128k
3
votes
Accepted
Why is the second order correction to energy zero for a fully degenerate eigensystem?
If $H$ is fully degenerate, all eigenvalues are identical, it means that $H$ is proportional to the unit matrix. The zeroth order eigenstates can be chosen as any orthonormal basis. Degenerate ...
- 188k
4
votes
Normalizers of the principal congruence subgroups in $\mathrm{GL}(n,\mathbf Q)$
Let me add a completely different proof (of a slightly stronger result). Unlike the previous one it is not self-contained linear algebra: it uses Zariski topology, being a lattice, Borel density etc.
...
- 63.9k
9
votes
Accepted
Normalizers of the principal congruence subgroups in $\mathrm{GL}(n,\mathbf Q)$
Proposition. For every $n\ge 0$ and $m\ge 1$ the normalizer of $\Gamma_n(m)$ in $\mathrm{GL}_n(\mathbf{Q})$ is $\mathbf{Q}^*\cdot\mathrm{GL}_n(\mathbf{Z})$.
Proof. One inclusion is clear so it is ...
- 63.9k
1
vote
Accepted
1
vote
Is every automorphism of a cone diagonalisable?
Every $u\in\mathrm{SO}(3)$ preserves a lot of such cones ($u$ fixes a nonzero vector $\xi$ and hence round cones around $\mathbf{R}_+\xi$) but, up to a few exceptions, $u$ is not diagonalizable.
If ...
- 63.9k
8
votes
Accepted
Is every automorphism of a cone diagonalisable?
No, see Example 6.1 in Tam's and Schneider's paper On the Core of a Cone-Preserving Map. For completeness, I will write the counterexample here, but the details can be found in the paper.
Let $V = \...
- 716
7
votes
Accepted
Matrix-tree theorem for inverse matrices
There is a formula as a sum of forests, i.e., collections of trees, for arbitrary minors of any size and row/column selection, for arbitrary matrices. So it can be applied for the numerator and the ...
1
vote
Eigenvalues of positive matrices in $\mathrm{SL}(d,\mathbb{Z})$
You are looking for Pisot matrices.
You may take a look at Avila and Delecroix - Some monoids of Pisot matrices.
5
votes
Eigenvalues of positive matrices in $\mathrm{SL}(d,\mathbb{Z})$
There is also in size 3 (found by checking all $3\times 3$ matrices with entries in $\{1,2,3\}$): the matrix $\begin{pmatrix}1&1&1\\2&1&2\\2&3&1\end{pmatrix}$, with ...
- 63.9k
7
votes
Eigenvalues of positive matrices in $\mathrm{SL}(d,\mathbb{Z})$
Did you look at any examples before asking the question?
I sat down to generate a positive integer matrix with determinant 1, and found this 4-by-4 example quickly:
$$\begin{pmatrix}1 & 2 & 3 &...
- 781
3
votes
Accepted
Lower spectral radius of matrices with an invariant subspace
This is false; the result is not perfectly analogous because the definition of the spectral radius contains a maximum. Here is a counterexample.
Take
$$
A_1 = \begin{bmatrix}2 & 0\\ 0 & 1\end{...
- 20.2k
6
votes
Accepted
Prove that $ n \leq d+1 $ under ordering constraints in $\mathbb{R}^d$
For $d=2$ the maximal $n$ is $3=d+1$. Indeed, between 4 points $x_1,x_2,x_3,x_4$ either one of them, $x_k$, belongs to the convex hull of three others, then $\langle \theta,x_k\rangle$ can not be the ...
- 109k
7
votes
Accepted
Maximizing trace subject to two equality constraints
If ${\rm rank} \, X>1$, there exists a two-dimensional space $\mathcal{L}$ such that the restriction of $X$ to $\mathcal{L}$ is positive definite. The space of Hermitian operators, which vanish on ...
- 109k
4
votes
Prove that $ n \leq d+1 $ under ordering constraints in $\mathbb{R}^d$
This is not a complete answer, but too large to fit reasonably as a comment. As shown by Fedor Petrov, it is not true that $n \leq d+1$, since for $d = 3$ we can construct $5$ such points and ...
- 716
2
votes
Invertibility of one matrix constructed by order n subgroup of symmetric group
The answer is no. Consider the subgroup generated by (1243). The matrix you have will be cyclic. See Determinant of cyclic matrix, proof without eigenvectors for a discussion of the determinant.
It is ...
- 287
4
votes
Invertibility of one matrix constructed by order n subgroup of symmetric group
I presume you mean invertible in $M_{n\times n}(\mathbb{Q})$. Here's an example.
$$\left(\begin{matrix}
1&2&3&4&5&6\\
2&1&3&4&5&6\\
1&3&2&4&5&...
- 16.2k
2
votes
Accepted
Can the derivative of eigenvectors with respect to its components be taken as zero if all eigenvalues are equal?
I’ll try to explain it in physicists words:
A matrix with all eigenvalues equal is proportional to the identity matrix.
The eigenvectors are maximally degenerate, as every arbitrarily oriented ...
- 3,671
0
votes
Accepted
How to efficiently replace the Paterson-Stockmeyer algorithm and reduce non-scalar multiplications
There are faster methods if you focus on a specific prescribed polynomial and you allow precomputation to find an optimal algorithm. The idea is based on selecting linear combinations that contain ...
- 20.2k
2
votes
Accepted
Generalization of a result of Kostant related to Gauss decomposition and Toda lattices
I believe that the answer to your question is yes and it follows from Proposition 8.1 in our paper The Mirković–Vilonen basis and Duistermaat–Heckman measures (with Baumann and Knutson). The notation ...
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