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In this question $A$ and $B$ are non-zero $2\times 2$ real matrices. Prove, or find a counter example to the statement: For any $A$ and $B$ there is at most two matrices $X$ such that $X^2+AX+B=0$.

I have the impression that it should be false and so I've been searching for counter examples by letting $A$ and $B$ be particularly simple matrices (like all entries zero except one) but every thing I've tried so far has made $X$ have only one solution. If it were true I'm not sure you'd go about proving it in a relatively simple way. Any hints would be great!

Jay
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    Take the matrix with $a_{11}=a_{21}=a_{22}=0$ and $a_{12}=b$, where $b$ is any real number and square it. – Shoutre Nov 23 '15 at 17:24
  • @Shoutre so you'd get the zero matrix which would just leave $AX+B=0$ but then surely if we fix $A,B$ then you can find $X$ or $b$? – Jay Nov 23 '15 at 17:35
  • Well, you said "for any $A$ and $B$", so in particular take $A$ and $B$ the null matrix and then you get the equation $X^{2}=0$. Using the matrix above, you get infinite solutions. Even simpler counter-example than the answer below. – Shoutre Nov 23 '15 at 17:47
  • @Shoutre Yeah I should have but it in the question title that it requires $A,B$ be non zero matrices. But I just tried to apply this to one of the answers: with $(X+I_n)^2=0$ you can let $X$ be in the form $a_{11}=a_{21}=a_{22}=-1$ and $a_{12}=b-1$ so that $(X+I_n)^2=0$ for all real $b$. – Jay Nov 23 '15 at 17:52
  • Oh, okay then. If you know some abstract algebra, it is a nice exercise to think if this statement is true if we swap the ring of $2x2$ matrices with an integral domain $A$ or a field $F$. – Shoutre Nov 23 '15 at 17:54
  • $X^2=I$ has many solutions, a lot of them (like reflections) being obvious ones. – Marc van Leeuwen Nov 23 '15 at 21:02
  • @MarcvanLeeuwen I keep forgetting that matrices can be thought of as linear maps. It makes things more intuitively obvious. I'm trying to think of what the solutions to $X^2=I$ would look like. So - like you say - reflections in any line or maybe a rotation by $\pi$ radians (I guess centred anywhere in the plane). Though I can't think of any others. – Jay Nov 23 '15 at 22:31
  • @Jay: Well, the solutions to $X^2=I$ in dimension$~2$ are, apart from the scalar solution $\pm I$, precisely the diagonalisable matrices with (simple) eigenvalues $+1$ and $-1$, which are in a sense all reflections, but not necessarily orthogonal ones (reflecting parallel to a line that is not necessarily orthogonal to the axis; it can have any direction except parallel to the axis). – Marc van Leeuwen Nov 24 '15 at 09:27

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This is false. Consider $A=2I_n$ and $B=I_n$. We then have $$X^2+2X+I_n =0 \implies (X+I_n)^2 = 0$$ which clearly has infinite solutions.

Adhvaitha
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Let $A \colon= I_2$, the $2 \times 2$ identity matrix, and let $B \colon= -I_2$.

Then you'll see that the equation $X^2 + AX + B = 0$ will have more than two solutions $X$.

Check it out for yourself.

Saaqib Mahmood
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  • It seems I completely missed using identity matrices, thanks for the answer! – Jay Nov 23 '15 at 17:53
  • This is very nice, but a little subtle. It illustrates the issue with all diagonal matrices. – copper.hat Nov 23 '15 at 17:53
  • @copper.hat what issue do you mean? Sorry if it's obvious, I've only just started learning it and so I'm not entirely sure I know what you're referring to :) – Jay Nov 23 '15 at 19:17
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    If you use diagonal matrices then you have two independent quadratic equations, hence you can take any diagonal $X$ with any pair of solutions on the diagonal. Hence with just diagonal $A,B,X$ you have 4 solutions (generically). The point is that with the diagonal matrices, you get two independent quadratics. – copper.hat Nov 23 '15 at 19:19
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If you stick to polynomials with scalar (i.e., multiples of the identity) coefficients, and $K$ is an infinite field then every polynomial equation over $K$ of degree $n>1$ has infinitely many solutions in $M_n(K)$. The reason is essentially the Cayley-Hamilton theorem, which forces matrices to satisfy an equation with a lot less degrees of freedom than the matrices have $(n$ rather than $n^2$). But a more formal argument is as follows: assuming without loss of generality our polynomial $P$ to be monic, the companion matrix $C_P$ of $P$ is a solution of the matrix equation $P[X]=0$, and so are all its conjugates, of which there are a lot because $C_P$ is clearly not central. More precisely, the matrices commuting with $C_P$ are only the polynomials in $C_P$, because its minimal and characteristic polynomials are both$~P$; then the conjugation orbit of $C_P$, being in bijection with $GL(n,K)/Z(C_P)$ where $Z(C_P)$ is the centraliser of $C_P$ in $GL(n,K)$, has dimension $n^2-n$.

Community
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Marc van Leeuwen
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The non-symmetric algebraic Riccati equation is (1) $XAX+B_1X+XB_2+C=0_2$ where $A,B_1,B_2,C$ are given in $M_2(\mathbb{C})$ and the unknown is $X\in M_2(\mathbb{C})$. If $A,B_1,B_2,C$ are generic (to get an idea, consider random choices) matrices, then this equation can be rewritten (2) $X^2+AX+B=0$. Now, if $A,B$ are generic, then (2) admits exactly $6$ solutions in $M_2(\mathbb{C})$. That is to say that we find an infinity of solutions only in exceptional cases.

In the particular case when moreover $A,B$ commute (note that $B$ is a polynomial in $A$), the solutions necessarily commute with $A,B$ and (2) admits exactly $4$ solutions (cf. the copper.hat's comment).