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Equality of determinants

Let A,N \in \mathcal{M}_n(\mathbb{C}) and suppose that N is nilpotent. Show that if A,N  commute then

    \[\det(A+N)=\det(A)\]

Solution

Since \mathbb{C} is algebraically closed and A,N commute this means that A,N are simultaneously triangularizable,  there exists an invertible element P \in \mathcal{M}_n(\mathbb{C}) such that both PNP^{-1} and PAP^{-1} are triangular. Since PNP^{-1} is both nilpotent and triangular, all its diagonal entries are zero and so the diagonal entries of P(A+N)P^{-1}=PAP^{-1}+PNP^{-1} are the same as the diagonal entries of PAP^{-1}. Thus,

    \[\det(P(A+N)P^{-1})=\det(PAP^{-1})\]

because P(A+N)P^{-1}PAP^{-1} are both triangular and the determinant of a triangular matrix is the product of its diagonal entries. So,

    \[\det(A+N)=\det(P(A+N)P^{-1})=\det(PAP^{-1})=\det(A)\]

The exercise along its solution have been migrated from here.

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