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Conference rusure::math

Title:	Mathematics at DEC

Moderator:	RUSURE::EDP

Created:	Mon Feb 03 1986
Last Modified:	Fri Jun 06 1997
Last Successful Update:	Fri Jun 06 1997
Number of topics:	2083
Total number of notes:	14613

2070.0. "Crux Mathematicorum 2174" by RUSURE::EDP (Always mount a scratch monkey.) Mon Oct 14 1996 10:40

T.R Title User Personal
Name Date Lines

2070.1 Have I missed something? CHEFS::STRANGEWAYS Andy Strangeways@REO DTN 830-3216 Mon Oct 14 1996 11:32 29

2070.2 another proof 63509::YODER MFY Wed Feb 05 1997 14:54 18

T.R	Title	User	Personal Name	Date	Lines
2070.1	Have I missed something?	CHEFS::STRANGEWAYS	Andy Strangeways@REO DTN 830-3216	`Mon Oct 14 1996 11:32`	29
2070.2	another proof	63509::YODER	MFY	`Wed Feb 05 1997 14:54`	18
	The matrix A satisfies its characteristic polynomial, so we can write a[n]A^n + ... + a[0]A^0 = 0 where a[n] /= 0. Choose k, 0<=k<=n, such that a[k]/=0 and a[r]=0 for r<k. Then we can drop terms a[r]A^r for r < k, and write a[n]A^n + ... + a[k]A^k = 0 Multiplying by A^(n-k), a[n]A^(2n-k) + ... + a[k]A^n = 0 Using A^(n+1) = 0, all terms but the last are seen to be zero, so a[k]A^n = 0 Hence A^n = 0.

The matrix A satisfies its characteristic polynomial, so we can write

  a[n]A^n + ... + a[0]A^0 = 0

where a[n] /= 0.  Choose k, 0<=k<=n, such that a[k]/=0 and a[r]=0 for r<k.
Then we can drop terms a[r]A^r for r < k, and write

  a[n]A^n + ... + a[k]A^k = 0

Multiplying by A^(n-k),

  a[n]A^(2n-k) + ... + a[k]A^n = 0

Using A^(n+1) = 0, all terms but the last are seen to be zero, so

  a[k]A^n = 0

Hence A^n = 0.