Timeline for Largest Eigenvalue of a Matrix with Special Form in terms of n
Current License: CC BY-SA 4.0
11 events
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| Sep 4, 2018 at 3:20 | history | edited | dave2d | CC BY-SA 4.0 |
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| Sep 4, 2018 at 2:37 | comment | added | Noam D. Elkies | The principal minors detect the signs of the eigenvalues, not their precise sizes. For example, $\bigl({1 \; 1 \atop 1 \; 3}\bigr)$ has eigenvalues $2 \pm \sqrt 2$. | |
| Sep 4, 2018 at 2:30 | comment | added | Venkataramana | I am missing something. Computing principal minors, I seem to get (after removing column and row of zeros) the numbers 1,2, 4,4,4,.... This seems to say the eigenvalues are 1,2,2,1,1,1,...? | |
| Sep 4, 2018 at 2:07 | comment | added | AccidentalFourierTransform | presumably of interest: Perron–Frobenius theorem. | |
| Sep 4, 2018 at 1:58 | vote | accept | dave2d | ||
| Sep 4, 2018 at 1:55 | comment | added | Noam D. Elkies | Also, removing the row and column of zeros yields the inverse of the tridiagonal matrix with $-1$ on the two off-diagonals and $2$ in all diagonal entries except for a $1$ in the bottom right corner. | |
| Sep 4, 2018 at 1:50 | comment | added | Noam D. Elkies | Numerical experimentation suggests that this matrix [gp: matrix(n,n,i,j,min(i,j)-1) ] has characteristic polynomial $\sum_{k=0}^{n-1} (-1)^k {n-1+k \choose n-1-k} x^{n-k}$, and nonzero roots $-2\sum_{k=1}^{n-1} k \cos(2\pi jk/(2n-1))$ for $0<j<n$, with $j=1$ producing the maximal root. | |
| Sep 4, 2018 at 1:30 | history | edited | dave2d | CC BY-SA 4.0 |
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| Sep 4, 2018 at 1:25 | answer | added | Suvrit | timeline score: 6 | |
| Sep 4, 2018 at 1:15 | review | First posts | |||
| Sep 4, 2018 at 1:18 | |||||
| Sep 4, 2018 at 1:14 | history | asked | dave2d | CC BY-SA 4.0 |