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Let $A$ be complex $2×2$ matrices s.t. $A^2=0$. Which of the following statements are true?

  1. $PAP^{-1}$ is diagonal for some $2×2$ real matrix $P$.
  2. $A$ has $2$ distinct eigenvalues in $\Bbb C$.
  3. $A$ has $1$ eigenvalue in $\Bbb C$ with multiplicity $2$.
  4. $Av=v$ for $v\in \Bbb C^2 ,v≠0$.

marked as duplicate by vadim123, Martin, Stahl, Henry T. Horton, achille hui May 11 '13 at 5:11

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  • $\begingroup$ not getting any A s.t. A^2=0. $\endgroup$ – pankaj Dec 17 '12 at 13:22
  • $\begingroup$ where to begin ....please suggest me........... $\endgroup$ – pankaj Dec 17 '12 at 13:35
  • $\begingroup$ Take $A=\begin{pmatrix}a&a\\-a&-a\end{pmatrix}$ for any $a\in\mathbb{C}$, for exmaple. $\endgroup$ – Dennis Gulko Dec 17 '12 at 13:35
  • $\begingroup$ 2 is wrong ........ 3 is correct.......... $\endgroup$ – pankaj Dec 17 '12 at 13:40
  • $\begingroup$ That's right. Can you prove it? $\endgroup$ – Dennis Gulko Dec 17 '12 at 13:40

I will sum the above discussion here:
1. Show that $(PAP^{-1})^2=PA^2P^{-1}$. Since $A^2=0$, what can you conclude about $A$, if indeed there exists such $P$?
2. You said that it is wrong. Can you show why? (Hint: what is the def of an eigenvalue?)
3. Is coorect, as you said. It should follow from the proof you used in 2.
4. If there esists $v\in\mathbb{C}^2$ such that $Av=v$, what is $A^2v$? can that happen?

  • $\begingroup$ 4 is wrong still not getting 1..... $\endgroup$ – pankaj Dec 17 '12 at 14:18
  • $\begingroup$ @Dennis,I am new in mathematics field , so as far as I know about $2$,since $A^2=0$ , so $A$ will satisfy the equation $x^2=0$. So minimal polynomial of $A$ will divide this polynomial $x^2$.If A has two distinct eigenvalues then the minimal polynomial will not divide $x^2$.so $2$ false. $3$ will be true against this logic.Am I right? One thing I want to clarify I am not the person who asked the question,I made it because our name is same. $\endgroup$ – p.haz Dec 17 '12 at 14:36
  • $\begingroup$ please guide me for 1.......... $\endgroup$ – pankaj Dec 17 '12 at 14:37
  • $\begingroup$ If $D$ is diagonal and $D^2=0$, then what is $D$? $\endgroup$ – Dennis Gulko Dec 17 '12 at 15:18
  • $\begingroup$ D is not nilpotent.. am i right?? $\endgroup$ – pankaj Dec 17 '12 at 16:27

$A$ is a nilpotent matrix, so it has only $0$ as eigenvector with multiplicity $2$.


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