Liouville Formula Proof: Det(exp(A))=exp(tr(A))

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The discussion focuses on the proof of the equation det(exp(A)) = exp(tr(A)), where A is a diagonalizable matrix. The initial step involves diagonalizing matrix A as A = CDC^(-1). Participants confirm that the property (CDC^(-1))^n = CD^(n)C^(-1) simplifies the proof, allowing the factors C and C^(-1) to cancel out, leading to a straightforward conclusion. This establishes a clear relationship between the determinant of the matrix exponential and the trace of the matrix.

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  • Understanding of matrix diagonalization
  • Familiarity with matrix exponentiation
  • Knowledge of determinants and traces of matrices
  • Basic linear algebra concepts
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Diophantus
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Would anyone be so kind as to offer a small hint for constructing a proof of:
det(exp(A)) = exp(tr(A))
where A is a square (diagonalizable) matrix and exp(A) = I + A + (1/2)A^2 + ...
 
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As a first step, diagonalize A. That is, write it as A=CDC^(-1)
 
Thank you

Very nice. Of course (CDC^(-1))^n = CD^(n)C^(-1). The C and C^(-1) factor out and disappear then the rest just follows naturally. Thank you agin!:smile:
 

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