Timeline for What is known about the eigenvectors of the $2^n \times 2^n$ Hadamard matrix?
Current License: CC BY-SA 4.0
9 events
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| Oct 4, 2019 at 17:46 | vote | accept | MCH | ||
| Oct 4, 2019 at 17:45 | comment | added | MCH | Thanks. Is there a more meaningful interpretation of the eigenvectors? See the edit in the question above. | |
| Oct 4, 2019 at 6:27 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Oct 3, 2019 at 21:58 | comment | added | Federico Poloni | Yes, if I am not missing anything. If $H_2 = QDQ^*$, with $Q$ orthogonal (which exists since $H_2$ is symmetric), then $H_{2^n} = (Q\otimes Q \otimes \dots \otimes Q)(D\otimes D \otimes \dots \otimes D) (Q\otimes Q \otimes \dots \otimes Q)^*$. I tested this quickly with Octave and it seems to work. | |
| Oct 3, 2019 at 21:53 | comment | added | Carlo Beenakker | would that give you an orthogonal basis ? (as I understood the cited papers, that was the aim, to provide an efficient orthogonalization) | |
| Oct 3, 2019 at 21:52 | comment | added | Federico Poloni | Aren't the eigenvectors easy to compute from the fact that $H_{2^n} = \underbrace{H_2 \otimes H_2 \otimes \dots \otimes H_2}_{\text{$n$ times}}$? You can diagonalize a Kronecker product factor by factor. | |
| Oct 3, 2019 at 21:34 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Oct 3, 2019 at 19:51 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Oct 3, 2019 at 19:44 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |