35
votes
Accepted
Consequences of eigenvector-eigenvalue formula found by studying neutrinos
The OP asks about generalisations and applications of the formula in arXiv:1908.03795.
$\bullet$ Concerning generalisations: I have found an older paper, from 1993, where it seems that the same ...
28
votes
Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?
For $n>2$, the span of the matrices in $\mathrm{SO}(n)$ is the full space $M_n(\mathbb{R})$ of $n$-by-$n$ matrices with real entries.
One proof is using representation theory: If we let $S\subset ...
22
votes
Expected value of determinant of simple infinite random matrix
Very nice problem!
Let me recall you that the determinant of $n \times n$ matrices with entries in $\{0,1\}$ is related to the one of $n+1 \times n+1$ matrices with entries in $\{-1,+1\}$: replace ...
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 ...
19
votes
When the sum of positive definite matrices converges, does the sum of the norm of the associate matrices converges?
Yes. The norm of a positive definite matrix does not exceed its trace, and the sum of traces is finite, since the sum of diagonal elements is finite for each of $n$ places.
19
votes
Is the determinant the only multiplicative matrix function?
It depends on what is the target space. Linear representations of ${\bf Gl}_n(k)$ do satisfy $\rho(AB)=\rho(A)\rho(B)$ by definition, and they often extend in a natural way to ${\bf M}_n(k)$.
On ...
18
votes
Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?
Let $S$ be the span of $SO(N)$ .
Then it's obvious that if $A \in S$ and $D_1, D_2 \in SO(N)$ then $D_1^{-1} A D_2 \in S$ .
Therefore it's enough to show that $B := diag(1,0,...0) \in S$ .
If $N$ ...
17
votes
Are unitarily equivalent permutation matrices permutation similar?
The answer is yes by an old theorem of Brauer. In fact, permutation matrices are conjugate over any field if and only if they are conjugate by permutation matrices. See this beautiful proof by ...
17
votes
Are unitarily equivalent permutation matrices permutation similar?
Here's a direct proof that if permutation matrices $A,B$ (of size $n\ge 0$) have the same characteristic polynomial (or equivalently are linearly equivalent, since these are diagonalizable) then they ...
17
votes
Accepted
Closed form solution for $XAX^{T}=B$
$B^{-1/2}XAX^TB^{-1/2}=I$, so $B^{-1/2}XA^{1/2}=Q$ must be orthogonal. On the other hand, for any orthogonal $Q$, it is simple to verify that $X = B^{1/2}QA^{-1/2}$ solves the equation, so this is a ...
15
votes
Accepted
is it possible to have two non-isomorphic non-regular graphs with the same adjacent spectrum and the same laplacian spectrum?
Yes, Brendan McKay showed that almost all trees have mates that are simultaneously cospectral in both adjacency and Laplacian spectra. And more.
http://users.cecs.anu.edu.au/~bdm/papers/SpectralTrees....
15
votes
Accepted
When does the determinant distribute over addition?
let me assume $A$ is invertible, then you ask when
$$\det(1+X)=1+\det X,\;\;X=A^{-1}B $$
so if $X$ has eigenvalues $x_i$, $i=1,2,\ldots n$, you would need
$$\prod_{i}(1+x_i)=1+\prod_i x_i$$
basically ...
15
votes
Accepted
Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?
Elementary proof. The linear space $E$ spanned by $SO_n$ is the orthogonal of those matrices $M$ such that $\langle M,Q\rangle:={\rm Tr}(MQ)=0$ for every $Q\in SO_n$. Let $M=SR$ be a polar ...
14
votes
Accepted
Is every real matrix conjugate to a semi antisymmetric matrix?
Yes. Every matrix can be written as the sum of a symmetric plus an antisymmetric one: $A = \frac{A+A^T}{2}+\frac{A-A^T}{2}$. Now change basis such that the symmetric part is diagonal.
14
votes
Accepted
Rank of $A\otimes B - B\otimes A$
For generic $A,B$ the matrix $A$ is invertible and $B=CA$, where $C$ is also generic. We have $(A\otimes B-B\otimes A)(u\otimes v)=Au\otimes CAv-CAu\otimes Av$. The vectors $Au$ run over the whole $\...
14
votes
Question on whether, "An entire function, nowhere zero, has an entire logarithm," holds for matrix-valued entire functions as well
Counterexample: Consider the entire function $$ A(z) = \pmatrix{e^z & 0\cr
z & 1\cr}$$
An entire logarithm of $A(z)$ must have eigenvalues $z + 2\pi i n$ and $2 \pi ...
13
votes
Accepted
Differentiability of operator norm
It need not be differentiable everywhere. Let $P$ and $Q$ be mutually orthogonal self-adjoint projections. Then the norm of $P+ t Q$ is 1 for $|t| \leq 1$ and $|t|$ for $|t| > 1$.
However, $\|A + ...
13
votes
Consequences of eigenvector-eigenvalue formula found by studying neutrinos
I want to add to the list of places where this identity has been used previously.
For a recent one, see arXiv:1710.02181 “State transfer in strongly regular graphs with an edge perturbation”.
...
13
votes
One observation of special type of square matrix exponentiation
The answer is quite simple. First observe that $A$ is triangular, hence the spectrum is on the diagonal. From your assumptions, $1$ is a simple eigenvalue and the other eigenvalues belong to $[0,1)$. ...
12
votes
Accepted
Finding the nearest matrix with real eigenvalues
I have no idea what is going on, but your conjecture is not correct. This is more transparent perhaps in the complex version. Consider
$$
A = \begin{pmatrix} i & a \\ 0&-i \end{pmatrix} , \...
12
votes
Matrix elements of exponential of tridiagonal matrices
Yes! Most methods to compute exponentials of large sparse matrices are based on computing directly $\exp(A)b$ for a given vector $b$ rather than the full matrix $\exp(A)$. Just take $b$ as a vector of ...
11
votes
Accepted
When the sum of positive definite matrices converges, does the sum of the norm of the associate matrices converges?
You can bound $\|A_k\| \leq C(n)\max_{i,j} |(A_k)_{ij}|$ for some function of the dimension only $C(n)$, because all norms are equivalent in finite dimension. If I am not mistaken $C(n)=\sqrt{n}$, but ...
11
votes
Accepted
Generalized Hölder's inequality for operator (subordinate) norms
Actually, there is a much stronger result, known as the Riesz-Thorin Theorem:
The subordinate norm $\|A\|_p$ is a log-convex function of $\frac1p$.
In other words,
$$\left(\frac1r=\frac\theta{p}+\...
11
votes
Is the linear span of special orthogonal matrices equal to the whole space of $N\times N$ matrices?
Clearly the span of $SO(n)$ contains the tangent space of $SO(n)$, the antisymmetric matrices. But it also contains the rotations by $\pi$ in any 2-plane, so contains the diagonal matrices with even ...
11
votes
Accepted
How to see that the determinant of this matrix is nonzero for all primes?
As discussed in the comments, I don't see how to extract the desired matrix from the original question (about spanning the vector space). However, the matrix being nonzero IS equivalent to the ...
11
votes
Accepted
Is it true that $\lVert A\rVert \leq \lVert A^2\rVert$ for $A\in \operatorname{SL}(2, \mathbb{R})$ when $\operatorname{trace}(A)>2$?
We can do this by a calculation. The assumptions on the determinant and trace are equivalent to having eigenvalues $\lambda,1/\lambda$, with $\lambda>1$. We can rotate the first eigenvector to the $...
11
votes
Generating $\mathbf{PGL}_2(\mathbb{Z})$ and $\mathbf{PGL}_2(\mathbb{Q})$
For $PGL_2(\mathbb{Z})$, you can use general properties of arithmetic groups due to Borel and Harish-Chandra, which will carry over to rings of $S$-integers of number fields. Alternatively, you can ...
10
votes
Accepted
Solving $\text{trace}\left[\left(I + pY\right)^{-1} \left(I - p^{2}Y\right)\right] = 0$ for scalar $p$
Write $Y=SDS^{-1}$, where $D$ is diagonal (since $X_1$ and $X_2$ are psd, $Y$ is diagonalizable). Then, observe that
\begin{equation*}
f(p) = \text{tr}(S^{-1}(I+pSDS^{-1})^{-1}SS^{-1}(I-p^2SDS^{-1})...
10
votes
Expected value of determinant of simple infinite random matrix
I agree with user39115!
I will give a heuristic from random matrix theory because we know the global behaviour of the eigenvalue. First
$$A=p 1 +\sqrt{N(p-p^2)}\frac{B}{\sqrt{N}} $$ where $1$ is the ...
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