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November 21st, 2017, 11:06 AM   #1
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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 11:09 AM.
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November 21st, 2017, 11:30 AM   #2
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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!
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November 21st, 2017, 11:31 AM   #3
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$m=\begin{pmatrix}a &b \\c &d\end{pmatrix}$

$m^2 = \begin{pmatrix}a^2 + b c &b(a+d) \\c(a+d) &b c + d^2 \end{pmatrix}$

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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