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If $N$ is an $n\times n$ nilpotent matrix such that $N^k=0$ for some integer $k$. Is it true that $(DN)^k=0$ for any diagonal matrix $D$?

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2 Answers 2

up vote 6 down vote accepted

This is false. A matrix is nilpotent iff all its eigenvalues are zero. However, $DN$ need not have all eigenvalues zero.

For example, look at: $$N=\begin{bmatrix} 6 & -9 \\ 4 & -6 \end{bmatrix}$$ $$D=\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$$

$N$ is nilpotent, but their product has eigenvalues 0 and -6 and thus isn't nilpotent.

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No. An important source of leverage in a situation like this is that nilpotence is a coordinate-independent condition (that is, it's invariant under change of basis) but being diagonal isn't. So up to change of basis you can replace "diagonal" with diagonalizable, and then it's easy to write down a counterexample, since e.g. any matrix with distinct eigenvalues is diagonalizable over an algebraically closed field. Thus if $D = \left[ \begin{array}{cc} a & b \\\ c & d \end{array} \right]$ is our mystery diagonalizable matrix, we can take $N = \left[ \begin{array}{cc} 0 & 1 \\\ 0 & 0 \end{array} \right]$ and compute that $$DN = \left[ \begin{array}{cc} 0 & a \\\ 0 & c \end{array} \right]$$

which fails to be nilpotent if $c \neq 0$. And it's easy to choose appropriate values of $a, b, d$ so that $D$ has distinct eigenvalues.

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