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November 21st, 2017, 10:06 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 96 Thanks: 2  Prove matrices question
Prove that there is no matrix: \[b\in m_{2}\mathbb{(c)}\] such that \[b^{2}=\]\[\left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right]\] any help appreciated, I've created a generic matrices b with coefficients a b c d and have 4 equations but don't know where to go from here. Last edited by Jaket1; November 21st, 2017 at 10:09 AM. 
November 21st, 2017, 10:30 AM  #2 
Member Joined: Jan 2016 From: Blackpool Posts: 96 Thanks: 2 
Hi after looking at this question I think I have a solution: My 4 equations after multiplying bxb= $a^{2}+bc=0$ $ab+bd=1$ $ca+dc=0$ $cb+d^{2}=0$ By rearranging we have a^{2}=d^{2} hence a=d or a=d, if a=b then ca+cd=0=ca+ca=2ca=0 so a=0 or c=0, if a=0 then d=0 so ab+bd=0$\neq$1 id c=0 then a^2=0 which implies a=0 contradiction again If a=b then we have bc+a^2=0 and bca^2=0 and hence contradiction. Would this be a valid proof or is there a mistake somewhere? Thanks! 
November 21st, 2017, 10:31 AM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,039 Thanks: 1063 
$m=\begin{pmatrix}a &b \\c &d\end{pmatrix}$ $m^2 = \begin{pmatrix}a^2 + b c &b(a+d) \\c(a+d) &b c + d^2 \end{pmatrix}$ so $a^2 + b c = 0,~(1)$ $c(a+d)=0,~(2)$ $b c + d^2 = 0,~(3)$ $b(a+d) = 1,~(4)$ $a+d \neq 0~,(4)$ so $c = 0,~(2)$ $a^2  d^2 = 0,~(1)(3)$ $a^2 = d^2$ $a+d \neq 0,~(4)$ $a = d \neq 0$ $a^2 + b c = 0,~(1)$ $a^2 = 0$ $a = 0$ contradiction 

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