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 November 21st, 2017, 10:06 AM #1 Member   Joined: Jan 2016 From: Blackpool Posts: 98 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: 98 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 bc-a^2=0 and hence contradiction. Would this be a valid proof or is there a mistake somewhere? Thanks! Thanks from Country Boy
 November 21st, 2017, 10:31 AM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,123 Thanks: 1102 $m=\begin{pmatrix}a &b \\c &d\end{pmatrix}$ $m^2 = \begin{pmatrix}a^2 + b c &b(a+d) \\c(a+d) &b c + d^2 \end{pmatrix}$ 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 Thanks from Country Boy and Jaket1

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