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NILPOTENT AND
INVERTIBLE
Hanpen Robot
Today’s topic !
1 − 𝑥 −1
= 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1
.
𝑖𝑓 𝑥 𝑛
= 0 𝑡ℎ𝑒𝑛
Let’s proof !
𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1
𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1
+ 𝑥 𝑛
𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1
𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1
+ 𝑥 𝑛
𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1
𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1
+ 𝑥 𝑛
Remember 𝑥 𝑛
= 0
𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1
𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1
+ 0
Remember 𝑥 𝑛
= 0
𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1
𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1
∴ 1 − 𝑥 𝑆 = 1
𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1
𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1
∴ 𝑆 = 1 − 𝑥 −1
1 − 𝑥 −1
= 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1
Nilpotent and invertible

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Nilpotent and invertible

  • 2. Today’s topic ! 1 − 𝑥 −1 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1 . 𝑖𝑓 𝑥 𝑛 = 0 𝑡ℎ𝑒𝑛
  • 4. 𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1 𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1 + 𝑥 𝑛
  • 5. 𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1 𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1 + 𝑥 𝑛
  • 6. 𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1 𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1 + 𝑥 𝑛 Remember 𝑥 𝑛 = 0
  • 7. 𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1 𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1 + 0 Remember 𝑥 𝑛 = 0
  • 8. 𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1 𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1 ∴ 1 − 𝑥 𝑆 = 1
  • 9. 𝑆 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1 𝑥𝑆 = 𝑥 +∙∙∙ +𝑥 𝑛−1 ∴ 𝑆 = 1 − 𝑥 −1
  • 10. 1 − 𝑥 −1 = 1 + 𝑥 +∙∙∙ +𝑥 𝑛−1