Determinant of sum of a matrix and a nilpotent matrix

Posted: August 3, 2019 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: , ,

Let M_n(\mathbb{C}) be the ring of n\times n matrices with entries from \mathbb{C}, the field of complex numbers.
Let A,N \in M_n(\mathbb{C}) and suppose that N is nilpotent, i.e. N^k=0 for some integer k \ge 1. Then \det(N)=0. But what can we say about \det(A+N) ? Is it equal to \det(A) ? Not necessarily, of course! For example, consider

A =\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}, \ \ \  N=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}.

Then N^2=0 but \det(A+N)=-1 \ne 0=\det(A). See that in this example, AN \ne NA.

Problem. Let A,N \in M_n(\mathbb{C}) and suppose that N is nilpotent. Show that if A,N commute, i.e. AN=NA, then \det(A+N)=\det(A).

Solution. since \mathbb{C} is algebraically closed and A,N commute, A,N are simultaneously triangularizable, i.e. there exists an invertible element P \in 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


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). \ \Box


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