Proof That A & A^T Have the Same Nullspace (Kernel)

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    Kernel Nullspace Proof
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Discussion Overview

The discussion revolves around proving that a matrix \( A \) and its transpose \( A^T \) have the same nullspace (kernel) under the condition that \( A \) is a normal matrix, defined by the property \( A^TA = AA^T \). The conversation includes mathematical reasoning and exploration of the implications of this property.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant asks for help with the proof that if \( A \) is normal, then \( A \) and \( A^T \) have the same nullspace.
  • Another participant suggests that for vectors in the nullspace, the equations \( A^TAx = 0 \) and \( AA^Tx = 0 \) hold true.
  • A participant expresses confusion regarding a hint provided in the discussion.
  • Another participant proposes multiplying both sides by \( x^T \) from the left to derive an expression in terms of norms.
  • One participant indicates understanding after the clarification provided.

Areas of Agreement / Disagreement

The discussion includes some agreement on the mathematical properties of normal matrices and their implications for nullspaces, but it also contains points of confusion and uncertainty regarding the proof process.

Contextual Notes

Participants have not fully resolved the proof steps or the implications of the normality condition on the nullspace. There are missing assumptions and dependencies on definitions that have not been explicitly stated.

Who May Find This Useful

Readers interested in linear algebra, particularly those studying properties of matrices and their transposes, may find this discussion relevant.

stanley.st
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Hello,

can you help me with the proof? If A is normal
[tex]A^TA=AA^T[/tex]
then A and A^T have the same nullspace (kernel). And
[tex]||Ax||=||A^Tx||[/tex]

Thank you.
 
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[itex]A^TAx = 0 = AA^Tx[/itex] for the vectors in the nullspace right?
 
don't understand your hint.
 
multiply both by [itex]x^T[/itex] from the left, what is the resulting expression in terms of norms?
 
yes, I understand, thank you
 

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