Timeline for Does the truth of any statement of real matrix algebra stabilize in sufficiently high dimensions?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2010 at 13:09 | history | edited | Ryan Reich | CC BY-SA 2.5 |
specify different eigenvalue
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| Aug 2, 2010 at 6:19 | history | edited | Ryan Reich | CC BY-SA 2.5 |
Fix fix.
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| Aug 2, 2010 at 5:37 | comment | added | Ryan Reich | Fixed!$ $ | |
| Aug 2, 2010 at 5:37 | history | edited | Ryan Reich | CC BY-SA 2.5 |
Fix construction
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| Aug 2, 2010 at 4:35 | comment | added | Joel David Hamkins | But what I don't see is that your basis statement is expressible in the language of matrix algebra independently of the dimension. For any fixed n it seems fine, but what we need is one statement in that language, whose truth varies as n increases. | |
| Aug 2, 2010 at 4:23 | comment | added | Ryan Reich | I only need one basis, so I don't have to quantify over sets of vectors. That is, $e_1, \dots, e_n$ is a real basis if for all real scalars $a_1, \dots, a_n$ we don't have $\sum a_j e_j = 0$, and if for all vectors $v$ we have real scalars $a_j, b_j$ such that $\sum a_j e_j + \sum i b_j e_j = v$. Then a matrix $A$ is "real" if each $A e_j$ is in the real span of the $e_j$ and $n$ is even iff there exists a real $A$ with $A^2 e_j = -e_j$ for each vector in the real basis. | |
| Aug 2, 2010 at 3:59 | comment | added | Joel David Hamkins | I don't follow your remarks about the basis. You can't quantify over sets of vectors, but just over vectors (and matrices and scalars). | |
| Aug 2, 2010 at 3:03 | history | edited | Ryan Reich | CC BY-SA 2.5 |
Cover complex case.
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| Aug 2, 2010 at 1:29 | history | answered | Ryan Reich | CC BY-SA 2.5 |