# Tag Info

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### 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 ...
Accepted

### The sum of squared logarithms conjecture

Is there anything wrong with the following argument? First of all, by scaling all $x_i$ and all $y_i$ by a positive constant, we may safely assume that $\prod_i x_i = \prod_i y_i =1$. The result ...

### 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”. ...
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} , \...
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.

### 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 ...

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, $\|A + ... 11 votes Accepted ### No arbitrary product of matrices has eigenvalue 1? In the case$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 ...
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 ...