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Math Help - [SOLVED] Nilpotence

  1. #1
    Super Member redsoxfan325's Avatar
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    [SOLVED] Nilpotence

    A nilpotent matrix is any n\times n matrix A where \exists~k such that A^k=0. I need to prove that if A is nilpotent, then A+I is invertible (where I is the identity matrix). I am not allowed to use any theorems regarding determinants, trace, linear (in)dependence, eigenvalues, etc. I am only supposed to use basic matrix operations and properties of invertible matrices.

    I wrote this proof, but it doesn't feel right, though I can't place my finger on the error.

    Proof: Assume for a contradiction that A+I is not invertible; that is, \forall~B, B(A+I)\neq I.

    Thus, BA+B\neq I\implies BA\neq I-B. Multiplying through (on the right) by A^{k-1} gives us BA\cdot A^{k-1}\neq (I-B)A^{k-1}\implies (I-B)A^{k-1}\neq0.

    However, choosing B=I implies that (I-B)A^{k-1}=0, giving us a contradiction. Therefore A+I is invertible. \square

    Is that a valid proof, or did I screw up somewhere?
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  2. #2
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    Quote Originally Posted by redsoxfan325 View Post
    A nilpotent matrix is any n\times n matrix A where \exists~k such that A^k=0. I need to prove that if A is nilpotent, then A+I is invertible (where I is the identity matrix). I am not allowed to use any theorems regarding determinants, trace, linear (in)dependence, eigenvalues, etc. I am only supposed to use basic matrix operations and properties of invertible matrices.

    I wrote this proof, but it doesn't feel right, though I can't place my finger on the error.

    Proof: Assume for a contradiction that A+I is not invertible; that is, \forall~B, B(A+I)\neq I.

    Thus, BA+B\neq I\implies BA\neq I-B. Multiplying through (on the right) by A^{k-1} gives us BA\cdot A^{k-1}\neq (I-B)A^{k-1}\implies (I-B)A^{k-1}\neq0.

    However, choosing B=I implies that (I-B)A^{k-1}=0, giving us a contradiction. Therefore A+I is invertible. \square

    Is that a valid proof, or did I screw up somewhere?
    no, it's not a valid proof because BA\neq I-B does not necessarily imply that BA\cdot A^{k-1}\neq (I-B)A^{k-1}.

    the proof is quite easy: A^k=0 implies that A^m=0, for all m \geq k. so we may assume that k is odd and thus: I=A^k+I=(A+I)(A^{k-1}-A^{k-2} + \cdots -A + I).
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  3. #3
    Super Member redsoxfan325's Avatar
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    Thank you. It always seems so obvious once I see the solution...
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