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Let $N$ be a $2\times2$ complex matrix such that $N^2=0$. Prove that either $N=0$ or $N$ is similar over $\mathbb{C}$ to $$\begin{pmatrix}0&0\\1&0\end{pmatrix}$$


marked as duplicate by leo, Pedro Tamaroff, André Nicolas, user147263, Omnomnomnom Sep 27 '14 at 1:40

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  • $\begingroup$ I got to prove the first part (N=0) but the second part I have no idea how I can do this. $\endgroup$ – Henrique Apr 24 '13 at 13:47
  • 1
    $\begingroup$ Are you familiar with Jordan canonical form? If so, what can the JCF be for $N$? $\endgroup$ – vadim123 Apr 24 '13 at 13:47
  • $\begingroup$ The section which I am doesn't have Jordan Canonical Form. It only has charactristic values and polynomial and diagonalizable operator $\endgroup$ – Henrique Apr 24 '13 at 13:50
  • $\begingroup$ Seems like you are new user her. Please go through following link about how to ask question here to get maximum help math.stackexchange.com/faq $\endgroup$ – srijan Apr 24 '13 at 13:51

Hints: prove the following:
1) If $N^2=0$ then $\operatorname{Im}(N)\subseteq\ker(N)$.
2) Moreover, if $N\neq0$ and $N$ is $2\times2$, then $\operatorname{Im}(N)=\ker(N)$.
3) Take any $0\neq v\in\operatorname{Im}(N)=\ker(N)$. There exists $w\neq 0$ such that $Nw=v$ and $B=(w,v)$ form a basis to $\mathbb{C}^2$.
4) Denote $P=[I]^B_e$ - the identity matrix from basis $B$ to the standart basis $e$. What is $P^{-1}NP$?

  • $\begingroup$ Is correct to say "matrix of N relative to the basis B"? N is a matrix too. $\endgroup$ – Henrique Apr 24 '13 at 14:00
  • $\begingroup$ It is abuse of notation. I wrote it in another way now. $\endgroup$ – Dennis Gulko Apr 24 '13 at 14:04


$$N^2=0\iff \;\text{the characteristic polynomial of}\;N\;\;p_N(x)=x^2\implies\;\text{the minimal polynomial is either}\;\;x^2\;\;\text{or}\;\;x\ldots$$

Added: Trying to avoid working with the minimal polynomial: if $\,N^2=0\,$ then the only eigenvalue of $\,N\,$ is zero, so we have two possibilities: either there are two linearly independent eigevectors corresponding to this unique eigenvalue and then obviosuly $\,N=0\,$ , or else there is only one and then we can complete this eigenvector to a basis of $\,\Bbb C^2\,$ and get a matrix represention as the second option.

  • $\begingroup$ I can't use minimal polynomial. The book that I'm studing is Linear Algebra - Hoffman & Kunze and the section is 6.2 (chapter 6) $\endgroup$ – Henrique Apr 24 '13 at 13:54
  • $\begingroup$ Fair enough, though the exercise you're trying to do is in page 190 and the min. polynomial is defined in page 191...:) . Check the added part to my answer. $\endgroup$ – DonAntonio Apr 24 '13 at 14:17

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