Trace of a matrix equals sum of its determinants?

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The trace of a diagonalizable matrix equals the sum of its eigenvalues because the trace is defined as the sum of the diagonal elements, which correspond to the eigenvalues in a diagonal matrix. The discussion highlights the importance of understanding the algebraic properties of the trace, particularly its similarity invariance, meaning it remains constant under similarity transformations. This property allows for the conclusion that the trace of any diagonalizable matrix is equal to the sum of its eigenvalues. The participants express a need for a formal proof and clarification on the trace of diagonal matrices. Overall, the relationship between the trace and eigenvalues is a fundamental concept in linear algebra.
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If a matrix is diagonalizable, how does its trace equal the sum of its eigenvalues?

I can't find a proof for this anywhere.
 
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What is the trace of a diagonal matrix? What are the algebraic properties of trace?
 
Ah I see, the trace is similarity invariant. thanks
 
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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