anonymous
  • anonymous
prove that an nxn nilpotent matrix always has det=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I know that at some point A^n=0, and therefore the determinant is zero, but I must also have to prove that the matrices before it becomes zero have a determinant of zero
anonymous
  • anonymous
and det(0)=0=A^n....and A^n is connected to A somehow
anonymous
  • anonymous
use the fact det[A^n] = [det A]^n

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anonymous
  • anonymous
ok, so that would conclude the proof, because only 0^n can give zero
anonymous
  • anonymous
just like x^n = 0, the only solution is x must be 0
anonymous
  • anonymous
k thanks!

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