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For matrix A, A*A = 0 implies A = 0

  1. Jan 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Given that [itex]A[/itex] is a square matrix and [itex]A^\dagger[/itex] is its Hermitian conjugate, and that [itex]A^\dagger A = 0[/itex], show that [itex]A = 0[/itex].

    3. The attempt at a solution

    Let [itex]\{|i\rangle\}[/itex] be some orthonormal basis. Find the matrix elements of [itex]A[/itex] by taking [itex]0 = \langle j| A^\dagger A |j \rangle = \displaystyle\sum_n \langle j|A^\dagger |n \rangle \langle n|A|j \rangle = \displaystyle\sum_n | \langle n|A|j \rangle |^2 \ge | \langle i|A|j \rangle |^2 = |A_{ij}|^2[/itex], so [itex]A_{ij} = 0[/itex] and [itex]A = 0[/itex].

    3. Question

    Is there a way to prove the result that does not rely on finding matrix elements?
     
  2. jcsd
  3. Jan 15, 2014 #2

    Office_Shredder

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    Yes, the proof is very similar but does not require picking a specific vector to multiply by. Suppose that Av is not zero. Then by observing
    [tex] \left<v| A^{\dagger} A| v \right> = \left<Av| Av \right> [/tex]
    You should be able to prove that this number is both zero and non-zero, giving a contradiction.
     
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