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For real non-zero $2\times 2$ matrices, can we say:

For any $A,B$, there are at most two matrices $X$ such that $XX + AX + B =0$

Is there a way to see this without going in the direction of writing the open form result of $XX + AX + B$ and solve each element is equal to zero, etc.?

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No. E.g. over $\mathbb R$, we have $\pmatrix{1&t\\ 0&-1}^2=I$ for any $t$. It also follows that $$ \pmatrix{1&t\\ 0&-1}^2+\pmatrix{0&0\\ 0&1}\pmatrix{1&t\\ 0&-1}+\pmatrix{-1&0\\ 0&0}=0 $$ for any $t$.

Edit. Another counterexample (inspired by Dietrich Burde's answer). Let $u,v$ be any two vectors such that $v^Tu=0$. Then $X=uv^T$ is a solution to the equation $X^2=0$. In particular, if the underlying field is $GF(2)$ (the field consisting of only two elements $0$ and $1$, with $1+1=0$), the equation $Y^2=0$ for a $2\times2$ matrix $Y$ has exactly four solutions $$ Y\in\left\{\pmatrix{0&0\\ 0&0},\ \pmatrix{0&1\\ 0&0},\ \pmatrix{0&0\\ 1&0},\ \pmatrix{1&1\\ 1&1}\right\}. $$ And so does the equation $X^2=I$ over $GF(2)$: just put $X=Y+I$.

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  • $\begingroup$ how did you come up with this so fast? may I ask your technique? $\endgroup$ Sep 30 '19 at 18:48
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    $\begingroup$ @independentvariable It may not be hard for someone to figure out a counterexample, but I just happen to know that $X^2=I$ has infinitely many solutions over a field of characteristic zero. $\endgroup$
    – user1551
    Sep 30 '19 at 18:52
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No, take for example $A=0$ and $$ B=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}. $$ Then $X^2=-B$ has no solution, since $X$ is nilpotent, so that $X^2=0$ but $B\neq 0$.

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  • $\begingroup$ The matrices are non zero though. $\endgroup$ Sep 30 '19 at 18:49
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    $\begingroup$ It doesn't matter really. You can adapt this example with nonzero $A$. Quadratic matrix solutions need not have any solution. $\endgroup$ Sep 30 '19 at 18:51
  • $\begingroup$ But the question is about showing it can have more than 2, isnt it? $\endgroup$ Sep 30 '19 at 19:18
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    $\begingroup$ Yes, sorry, I thought of "at least". So quadratic matrix equations can have no solution, or a number of solutions including infinity. It comes from the fact that the matrix algebra is not commutative. $\endgroup$ Sep 30 '19 at 19:22
  • $\begingroup$ Still, good for knowledge :) $\endgroup$ Sep 30 '19 at 19:22

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