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Revision History for A297477

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Showing entries 1-10 | older changes
Triangle read by rows: T(n, k) gives the coefficients of x^k of the characteristic polynomial P(n, x) of the n X n matrix M with entries M(i, j) = 1 if i = 1 or j = 1, -1 if i = j > 1, and 0 otherwise. T(0, 0):= 0.
(history; published version)
#40 by Wolfdieter Lang at Sun Feb 04 13:06:42 EST 2018
STATUS

editing

approved

#39 by Wolfdieter Lang at Sun Feb 04 13:06:35 EST 2018
COMMENTS

The row sums are P(n, 1) = (-1)^(n-1)*(n-1)*2^(n-2) = A085750(n), for n >= 1, and for n=0 the row sum is 0. The alternating row sums are P(n, -1) = 2 for n=1, = -1 for n = 2 , and zero otherwise.

STATUS

approved

editing

#38 by Wolfdieter Lang at Sun Feb 04 13:04:14 EST 2018
STATUS

editing

approved

#37 by Wolfdieter Lang at Sun Feb 04 13:04:03 EST 2018
COMMENTS

Therefore the spectral radius (absolute value of the maximal eigenvectoreigenvalue) is rho_n = sqrt(n), and the spectral norm of M_n (square root of the maximal eigenvalue of (M_n)^+ M_n is also sqrt(n), for n >= 1. See the conjecture in the first comment above.

The square of the Frobenius norm (aka Hilbert-Schmidt norm) of M_n is max_{i,j=1..n} |M_n(i,j)| ^2 = 3*n - 2 = A016777(n-1), for n >= 1.

STATUS

approved

editing

#36 by N. J. A. Sloane at Sat Feb 03 12:38:26 EST 2018
STATUS

proposed

approved

#35 by Wolfdieter Lang at Sat Feb 03 05:09:31 EST 2018
STATUS

editing

proposed

#34 by Wolfdieter Lang at Sat Feb 03 05:09:07 EST 2018
COMMENTS

The norm of the matrix M appears to be sqrt(n, ), where with norm is meant the eigenvalue of the largest magnitude, negative or positive. Row sums appear to be A085750 [see below for the proof].

The eigenvalues of M_n are +1 for n = 1 and for n >= 2 they are +sqrt(n, ), -sqrt(n, ), and n-2 times -1.

Therefore the spectral radius (absolute value of the maximal eigenvector) is rho_n = sqrt(n, ), and the spectral norm of M_n (square root of the maximal eigenvalue of (M_n)^+ M_n) is also sqrt(n, ), for n >= 1. See the conjecture in the first comment above.

#33 by Wolfdieter Lang at Sat Feb 03 04:46:10 EST 2018
COMMENTS

From Wolfdieter Lang, Feb 02 32 2018 : (Start)

#32 by Wolfdieter Lang at Sat Feb 03 04:44:40 EST 2018
COMMENTS

The norm of the matrix M appears to be sqrt(n), , where with norm is meant the eigenvalue of the largest magnitude, negative or positive. Row sums appear to be A085750 [see below for the proof].

The eigenvalues of M_n are +1 for n = 1 and for n >= 2 they are +n, -n, and n-2 times -1.

Therefore the spectral radius (absolute value of the maximal eigenvector) is rho_n = n, and the spectral norm of M_n (square root of the maximal eigenvalue of M_n^+ M_n) is also n, for n >= 1. See the conjecture in the first comment above.

The square of the Frobenius norm (aka Hilbert-Schmidt norm) of M_n is max_{i,j=1..n} |M_n(i,j)| = 3*n - 2 = A016777(n-1), for n >= 1.

CROSSREFS

Cf. A016777, A085750 (row sums), A067998, A110427 (column k=2), -A131386 (column k=1), A181983 (Det M_n), A191898.

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proposed

editing

#31 by Robert Israel at Fri Feb 02 19:16:06 EST 2018
STATUS

editing

proposed