Timeline for Under what conditions a linear automorphism is an isometry of some norm?
Current License: CC BY-SA 3.0
18 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Jun 19, 2015 at 3:01 | review | Close votes | |||
| Jun 19, 2015 at 6:29 | |||||
| Jun 15, 2015 at 15:13 | vote | accept | Asaf Shachar | ||
| Jun 13, 2015 at 22:36 | answer | added | Asaf Shachar | timeline score: 6 | |
| Jun 13, 2015 at 19:08 | comment | added | Pietro Majer | 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, 2015 at 17:00 | comment | added | Asaf Shachar | The direction: $A$ is an isometry w.r.to some norm $\Rightarrow A$ is diagonalizable. (I know how to do it only in the case where it is given $A$ is an isometry w.r.t some inner product). | |
| Jun 13, 2015 at 16:55 | comment | added | Pietro Majer | What do you still need to prove? | |
| Jun 13, 2015 at 16:44 | comment | added | Pietro Majer | Sure, over $\mathbb{C}$. And, yes, I refer to any norm. | |
| Jun 13, 2015 at 16:36 | comment | added | Asaf Shachar | @Pietro: Maybe you mean diagonalizable over $\mathbb{C}$. A rotation by $90^0$ is an isometry w.r.t to the Euclidean norm but not diagonalizable over $\mathbb{R}$. As I noted in the question (see remark) this condition is known to be equivalent in the case of being an isometry w.r.t to an inner product. There is still something which need to be demonstrated if we only assume our operator is an isometry w.r.t some norm which is not induced by inner product. | |
| Jun 13, 2015 at 16:14 | comment | added | Pietro Majer | 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, 2015 at 15:49 | comment | added | Pietro Majer | 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 13, 2015 at 15:12 | comment | added | Asaf Shachar | Two questions: 1) Is it trivial that $\Lambda$ , $\Delta$ commute? (you seem to be assuming this). 2) why is $ n\Lambda^{n-1}\Delta$ the dominating factor in the sum? In particular, at least all the real eigenvalues of $\Lambda$ are of size 1, so powers of $\Lambda$ do not increase the "size" of its matrix elements. So unless $\Delta ^2 = 0$ it seems to me that the other factors become more dominant. | |
| Jun 13, 2015 at 13:31 | review | Close votes | |||
| Jun 14, 2015 at 0:24 | |||||
| Jun 13, 2015 at 13:14 | comment | added | YCor | So to conclude $T$ preserves a norm iff it preserves a scalar product. | |
| Jun 13, 2015 at 11:58 | comment | added | Alex Ravsky | So if the matrix $\Lambda$ is non-degenerated and $\Delta\ne 0$ then the matrix $n\Lambda^{n-1}\Delta$ (and so the matrix $J^n$ too) gets unbounded when $n$ goes to infinity. | |
| Jun 13, 2015 at 11:57 | comment | added | Alex Ravsky | It seems the following. I shall follow similarly to example which you propose and I extend the comment by Loïc Teyssier. There exists a non-degenerated matrix complex matrix $T$ such that $T^{-1}AT=J$, where $J$ is a Jordan form of the matrix $A$. Then $J=\Lambda+\Delta$, where $\Lambda$ is a diagonal matrix and $D=\|d_{ij}\|$ is a $0-1$-matrix with $d_{ij}=0$ when $j\ne i+1$. Let $n$ be an arbitrary natural number. Then $$J^n=(\Lambda+\Delta)^n=\Lambda^n+ n\Lambda^{n-1}\Delta+\frac {n(n-1)}2 n\Lambda^{n-2}\Delta^2+\dots.$$ | |
| Jun 13, 2015 at 9:21 | comment | added | Loïc Teyssier | It seems to me that the phenomenon you describe with the matrix $A$ will repeat itself whenever you have a non-diagonalisable endomorphism (over $\mathbb C$). | |
| Jun 13, 2015 at 8:47 | history | asked | Asaf Shachar | CC BY-SA 3.0 |