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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 $v\in \Bbb C^2 ,v≠0$.
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    not getting any A s.t. A^2=0.2012-12-17
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    where to begin ....please suggest me...........2012-12-17
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    Take $A=\begin{pmatrix}a&a\\-a&-a\end{pmatrix}$ for any $a\in\mathbb{C}$, for exmaple.2012-12-17
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    2 is wrong ........ 3 is correct..........2012-12-17
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    That's right. Can you prove it?2012-12-17
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    what about 1 and 4..............2012-12-17
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    Check that $(PAP^{-1})^2=PA^2P^{-1}$. What can you conclude about 1?2012-12-17
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    I can't find 4 but1 is looking right2012-12-17
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    where to begin ....please suggest me for 4...2012-12-17

2 Answers 2

1

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?

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    4 is wrong still not getting 1.....2012-12-17
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    please guide me for 1..........2012-12-17
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    If $D$ is diagonal and $D^2=0$, then what is $D$?2012-12-17
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    D is not nilpotent.. am i right??2012-12-17
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    $D$ is nilpotent since $D^2=0$, but $D$ is *diagonal*. Can you find $D$?2012-12-17
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    D is nilpotent then D can't be diagonal2012-12-17
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    If $D=\begin{pmatrix}a&0\\0&b\end{pmatrix}$ and $D^2=0$, then what are $a,b$?2012-12-17
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    a=0 b=0,A can be zero matrix there is no limtation..2012-12-17
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    I got it.. option a is wrong.. thanks sir....2012-12-17
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    @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.2012-12-17
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    @p.haz: I got it.. option a is wrong..2012-12-17
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    @p.haz: thanks sir.....2012-12-17
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$A$ is a nilpotent matrix, so it has only $0$ as eigenvector with multiplicity $2$.