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Thread: 'Diagonalizable' proof

  1. #1
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    'Diagonalizable' proof

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

    I proved that if $\displaystyle A^{m}=0$ for some positive integer $\displaystyle m$, then the only eigenvalue of $\displaystyle A$ is $\displaystyle 0$. 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 ($\displaystyle A \neq 0$)?
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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 ($\displaystyle A \neq 0$)?

    Thank you. I got it.
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