quadratic equations on 2 by 2 matrices

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.?

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 $$X=Y+I$$.

• how did you come up with this so fast? may I ask your technique? Sep 30 '19 at 18:48
• @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. Sep 30 '19 at 18:52

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 $$B\neq 0$$.

• The matrices are non zero though. Sep 30 '19 at 18:49
• It doesn't matter really. You can adapt this example with nonzero $A$. Quadratic matrix solutions need not have any solution. Sep 30 '19 at 18:51
• But the question is about showing it can have more than 2, isnt it? Sep 30 '19 at 19:18
• 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. Sep 30 '19 at 19:22
• Still, good for knowledge :) Sep 30 '19 at 19:22