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Given that $N$ is nilpotent, then $N$ is similar to $N^2$ if and only if $N=0$ ($N$ is the zero matrix). The "$\Longleftarrow$" direction is easy enough, because $N = 0 = N^2$ so they're equal and trivially similar.

But showing $N \text{ nilpotent and } N \sim N^2\Longrightarrow N=0$ is turning out to be more difficult. I'm comfortable using well-known facts about $n \times n$ nilpotent matrices such as:

\begin{align} N \text{ nilpotent } &\Longleftrightarrow \text {all eigenvalues }=0\\ &\Longleftrightarrow p_N = \lambda^n\\ &\Longleftrightarrow m_N =\lambda^k\,,\, \text{ for some k, }1\leq k \leq n\\ &\Longleftrightarrow {\rm tr}N^q = 0\,,\, \text{ where }q\in \mathbb{N}\text{, minimal, s.t. } N^q=0 \end{align}

Note that $p_N$ above is the characteristic and $m_N$ is the minimal polynomial. My hunch is to use that trace equals 0 for some power $q$ and the fact that each entry $c_{ij}$ of $N^2$ is \begin{align} c_{ij} = \sum_{k=1}^n n_{ik} n_{kj} \end{align} to get some cancellations but I can't quite see it yet. Any thoughts?

svenprime
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4 Answers4

14

Denote matrix similarity by the symbol $\sim$. This is an equivalence relation. In addition, if $A\sim B$, then $A^2\sim B^2$. So, if $N\sim N^2$, then $N\sim N^2\sim N^4\sim N^8\sim\cdots\sim N^{2^m}$ for every positive integer $m$. Since $N$ is nilpotent, $N^{2^m}$ is eventually zero. Hence $N\sim0$, meaning that $N=0$.

user1551
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7

Without even getting into minimal polynomials, this can be seen directly from the definitions of matrix similarity and nilpotency. Let $k > 0$ be the smallest positive integer such that $N^k = 0$, and suppose that $N\sim N^2$. Then there exists an invertible matrix $M$ such that $$MN^2M^{-1} = N$$ If $k$ is even, raising both sides to the power of $k/2$ establishes $N^{k/2} = 0$ which contradicts the minimality of $k$ so $N$ must be zero. Try figuring out the details for $k$ odd.

E G
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  • The even case makes sense. I'm trying to do the odd case similarly but I seem to be going in circles. If $k$ is odd, then for some $j$, $k=2j+1$ so:

    \begin{align} MN^{2(2j+1)}M^{-1}&=N^{2j+1}\ MN^{4j}N^2M^{-1}&=N^{2j}N\ MN^{4j}M^{-1}NMM^{-1}&=N^{2j}N\ MN^{4j}M^{-1}N&=N^{2j}MN^2M^{-1}\ \cdots \end{align}

    Could you please clarify more? I'm sure it's staring me in the face.

     – svenprime Apr 19 '21 at 02:07
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    @svenprime Instead of raising both sides to the power of $2j+1$ in the first line, try the power $j+1$ and derive a similar contradiction to the minimality of $k = 2j+1$. – E G Apr 19 '21 at 02:46
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    Ah, I got it now. \begin{align} MN^{2(j+1)}M^{-1}&=N^{j+1}\ MN^{2j+1}NM^{-1}&=N^{j+1} \end{align} Where $N^{2j+1}$ on the LHS is zero so that means that N^{j+1} is zero, again contradicting the minimality of $k:=2j+1$. – svenprime Apr 19 '21 at 13:16
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(I’ll use $x$ for the variable for the minimal/characteristic polynomial rather than $\lambda$, because I usually reserve $\lambda$ to denote an eigenvalue)

Similar matrices have the same minimal polynomial. If $N$ is nilpotent, then as you note the mininal polynomial is $x^k$ for some $k\gt 0$. If $k$ is even, $k=2m$, then since $(N^2)^m=N^{2k}=0$, the minimal polynomial of $N^2$ would divide $x^m$, which means it is different from the minimal polynomnial of $N$. So it is impossible for $k$ to be even. Thus, $k$ is odd, say $k=2m+1$.

Then $(N^2)^{m+1} = N^{2m+2} = N^{2m+1}N=0$, so the minimal polynomial for $N^2$ divides $x^{m+1}$. That means that $x^{2m+1}$ divides $x^{m+1}$, whichn in turn means that $2m+1\leq m+1$. But this can only happen if $m=0$, so the minimal polynomial for $N$ is $x$, hence $N=0$.

Arturo Magidin
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2

Increasing powers of nilpotent matrices are of decreasing ranks (until $N^k=0$ )

It can be easily seen from Jordan forms of nilpotent matrices. See more detailed description here.
Two non-zero matrices of different ranks can't be similar.

Conclusion: the given in the question nilpotent matrix is zero matrix.

Widawensen
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