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I am a professor of Mathematics at the University of Pisa, Italy.


Aug
11
awarded  Enlightened
Aug
11
awarded  Nice Answer
Jul
24
comment Writing the quotient of solutions of linear ODEs as the solution of a nonlinear ODE
no, $X$ and $Y$ are square matrices, and $Y^{-1}$ is the inverse of $Y$.
Jul
23
comment Writing the quotient of solutions of linear ODEs as the solution of a nonlinear ODE
The given example in some form also works for linear systems: if $X'=A(t)X$, and $Y'=B(t)Y$, and we define $U:=XY^{-1} $, then $U'=AU-UB$.
Jun
14
comment Under what conditions a linear automorphism is an isometry of some norm?
Also: if all orbits of $A$ are bounded, then we can renorm $\mathbb{R}^n$ by $\|x\|_A:=\sup_{k\in \mathbb{Z}} \|A^k x\|$, which is an $A$-invariant equivalent norm. So the following are equivalent: (i) $A$ is an isometry w.r.to some norm (ii) all $A$-orbits are bounded (iii) $A$ is diagonalizable over $\mathbb{C}$ with all eigenvalues of modulus $1$.
Jun
13
comment Under what conditions a linear automorphism is an isometry of some norm?
Use e.g. the spectral radius formula (for both $A$ and $A^{-1}$) to prove that the spectrum of $A$ is in the unit circle. If $A$ is not diagonalizable (over $\mathbb{C}$ ) there is a vector $v\in \mathbb{C}^n$ such that $A^kv$ diverges (use the Jordan form), but since $v=x+iy$ with $x$ and $y$ in $\mathbb{R}^n$, and $A^kv=A^kx+iA^ky$, either $A^kx$ or $A^ky$ diverge. This is impossible if $A$ is an isometry w.r.to some norm on $\mathbb{R}^n$.
Jun
13
comment Under what conditions a linear automorphism is an isometry of some norm?
What do you still need to prove?
Jun
13
comment Under what conditions a linear automorphism is an isometry of some norm?
Sure, over $\mathbb{C}$. And, yes, I refer to any norm.
Jun
13
comment Under what conditions a linear automorphism is an isometry of some norm?
Also: iff all orbits of $A$ are bounded (that is, $\sup_{k\in\mathbb{Z}}\|A^k x\|<+\infty$, for any $x\in\mathbb{R}^n$) .
Jun
13
comment Under what conditions a linear automorphism is an isometry of some norm?
If I'm not wrong $A$ is an isometry w.r.to some norm iff it is diagonalizable, with all eigenvalues of modulus $1$
Jun
8
awarded  Nice Answer
Jun
4
comment Power series with matrix coefficients
that's what I'm saying: in the given assumptions one can't exclude these simple cases, for which the answer is clearly negative.
Jun
3
comment Power series with matrix coefficients
I think you need some more assumptions to have an affirmative the answer. What if $A=0$, or what if e.g. $B= C^{-1}$ are constant matrices and commute with $A$?
Jun
3
revised Sums of arctangents
edited body
May
29
awarded  Nice Answer
May
26
awarded  Popular Question
May
18
awarded  Nice Question
May
14
awarded  Yearling
May
6
comment Orthogonal complements of intersections of closed subspaces
Yes, it's easy that $(K_1+K_2)^\perp =K_1^\perp \cap K_2^\perp$ holds for linear subspaces, and you can generalize by induction to n subspaces. Then if you take another orthogonal, you get closures (and you can also rename $H_j:=K_j^\perp$)
May
5
comment Orthogonal complements of intersections of closed subspaces
In general it is not true even for $n=2$, because $(H_1\cap H_2)^{\perp}$ is always closed, while the sum of two closed subspaces may fail to be closed. (For instance, in $H:=X\times X$ take the graph of the zero operator as $H_1^{\perp}$ and the graph of a dense, non-surjective bounded operator $T$ on $X$ as $H_2^{\perp}$: then $H_1^{\perp} + H_2^{\perp}= X\times T(X)$ is a dense not closed subspace).