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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 |
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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 |
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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 |
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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). |