## Determinant of sum of a matrix and a nilpotent matrix

Posted: August 3, 2019 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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 $\det(P(A+N)P^{-1})=\det(PAP^{-1})$

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 $\det(A+N)=\det(P(A+N)P^{-1})=\det(PAP^{-1})=\det(A). \ \Box$