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Dec
13
comment Finding $n$ scalars such that $\det{(cI-A)}=0$ without eigenvalues
Perhaps you can use the multilinearity of the determinant function to prove that $\det(xI-A)$ is a degree-$n$ polynomial in $x$.
Dec
10
comment How to prove that $A^2$ is a symmetric matrix
For $2\times2$ matrices, however, that "PS" is correct if $A$ is noninvertible. Let $A=uv^T$ for some vectors $u$ and $v$. The condition $A^{2n+1}=(AA^T)^nA$ gives $(v^Tu)^{2n} uv^T = (v^Tv)^n (u^Tu)^n uv^T$. By Cauchy-Schwarz inequality, one can assert that $u,v$ are linearly dependent. Hence $A$ is symmetric and so is $A^2$. This result generalises to the case of a general rank-1 square matrix $A$ as well.
Dec
8
comment A good source for linear algebra on matrices
QMUL is Queen Mary University of London, not Queen Margaret University (QMU), and the latter is located in Edinburgh, not London.
Dec
8
awarded  Caucus
Dec
8
comment Gram matrix of Gaussian kernel is not positive definite
It is well known that the Gaussian kernel is strictly positive definite. As long as your sample points $x_1,\ldots,x_N$ are drawn from a continuous distribution, the probability that $K$ is positive definite is $1$. So, what you encountered is clearly a numerical artefact (if not a programming error). When the interval $[a,b]$ is too narrow, or when $\gamma$ is not big enough, it's very easy for $K$ to become ill-conditioned, although it should be positive definite in theory.
Dec
8
comment $\| AB\|_\square \leq 1$ implies $\| BA\|_\triangle \leq 1$
@loupblanc Thanks a lot. Minimal norms (for fixed matrices!) ... of course ... that makes much sense. I have looked for useful results about minimal norms before answering the question, but quickly abandoned the search because such norms (the usual ones, not Householder's) are minimal for all matrices.
Dec
8
comment Applying modulus to determinant
Hmm, how? What is your answer anyway?
Dec
8
comment Applying modulus to determinant
The same approach applies. Find an $n$ such that the remainder of $21n$ divided by 29 is 1. That way, $21n$ is equal to 1 in modular arithmetic and hence $n$ is the inverse/reciprocal of 21, modulo 29.
Dec
8
comment Applying modulus to determinant
mod 26 is a bit weird. Anyway, 6x16-15x5=21. Can you find an integer $n$ such that when $21n$ is divided by 26, the remainder is 1? If so, that $n$ is the answer.
Dec
7
comment $\| AB\|_\square \leq 1$ implies $\| BA\|_\triangle \leq 1$
@loupblanc Thanks. Initially I thought this is a known result, but I couldn't find it anywhere in the books offhand. Google showed nothing either. The closest thing I can find is the characterisation of radial matrices. Do you know if this is a known result?
Dec
7
comment Matrix norm induced by a vector norm.
In its present form, the answer to your question is yes, by a circular reasoning. Call the induced norm $N(\cdot)$. It is easy to see that $N(I)=1$. Let $Q=I$ and $F(M)=N(M)I$. Then $\|F(M)Q\|_1=\|N(M)I\|_1=N(M)$. You need to be more specific about $F$ to make your question meaningful.
Dec
7
comment Dividing an obtuse triangle into acute triangles
@EdwardJiang How to prove your claim that seven is minimum?
Dec
7
answered $\| AB\|_\square \leq 1$ implies $\| BA\|_\triangle \leq 1$
Dec
7
revised Spectral radius inequality
added 39 characters in body
Dec
7
answered Quick question: matrix with norm equal to spectral radius
Dec
6
comment Prove that matrices commute
This is very nice. When I read the question, my first thought was to invoke Cayley-Hamilton theorem, which, given your answer, is surely a big overkill.
Dec
6
answered Two two-by-two positive matrices both have positive entries
Dec
6
answered How can I show that these matrices don't commute
Dec
6
revised Showing that the exponential map $\mathrm{exp}:\mathfrak{sl}(2,\mathbb{R})\to\mathrm{SL}(2,\mathbb{R})$ is not surjective
added 25 characters in body
Dec
5
answered How to prove that $A^2$ is a symmetric matrix