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Math Help - 'Diagonalizable' proof

  1. #1
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    'Diagonalizable' proof

    Let A \in M_{n}(R) such that A^{m}=0 for some positive integer m. If A is a nonzero matrix, show that A is not diagonalizable.

    I proved that if A^{m}=0 for some positive integer m, then the only eigenvalue of A is 0. I think it can be proven by contradiction. But I'm still struggle.
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  2. #2
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    You've almost got it. If, by contradiction, A is diagonalizable and all its eigenvalues are 0, then what's A's corresponding diagonal matrix? Do you see why that is a contradiction ( A \neq 0)?
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  3. #3
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    Quote Originally Posted by Defunkt View Post
    You've almost got it. If, by contradiction, A is diagonalizable and all its eigenvalues are 0, then what's A's corresponding diagonal matrix? Do you see why that is a contradiction ( A \neq 0)?

    Thank you. I got it.
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