A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

prove that an nxn nilpotent matrix always has det=0

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and det(0)=0=A^n....and A^n is connected to A somehow

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    use the fact det[A^n] = [det A]^n

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, so that would conclude the proof, because only 0^n can give zero

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    just like x^n = 0, the only solution is x must be 0

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    k thanks!

  7. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.