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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 14:40

T.RTitleUserPersonal
Name
DateLines
2070.1Have I missed something?CHEFS::STRANGEWAYSAndy Strangeways@REO DTN 830-3216Mon Oct 14 1996 15:3229
2070.2another proof63509::YODERMFYWed Feb 05 1997 17:5418
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.