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

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The discussion focuses on proving that if A is a normal matrix, then A and its transpose A^T share the same nullspace. The relationship A^TA = AA^T is central to this proof. The user seeks clarification on how to manipulate the equations A^TAx = 0 and AA^Tx = 0 to demonstrate this property. They also inquire about the implications of multiplying by x^T from the left and its relation to norms. The conversation emphasizes understanding these mathematical relationships to establish the proof effectively.
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Hello,

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

Thank you.
 
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A^TAx = 0 = AA^Tx for the vectors in the nullspace right?
 
don't understand your hint.
 
multiply both by x^T from the left, what is the resulting expression in terms of norms?
 
yes, I understand, thank you
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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