The determinant of an 𝑛×𝑛 matrix is equal to the 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 of its eigenvalues The trace is equal to the 𝑠𝑢𝑚 of its eigenvaluespic.twitter.com/sWdUWUeP0y
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Provided that the matrix M is diagonazible with diagonal matrix D M=S^(-1)D S detM=det(S^(-1)D S)= det(D S S^(-1))=detD="product of eigenvalues" (using detAB=detBA) trM=tr(S^(-1)D S)= tr(DS S^(-1))=trD= "sum of eigenvalues" (using trAB=trBA, cyclic property of the trace).
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Also for matrices that cannot be diagonalized but can be brought in e.g. a "Jordan-normal form". ("Near dIagonal form").
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You are correct!
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