# Sub-blocks of matrix quadratic equations

Suppose the following matrix quadratic equation has at least one real solution for $$X$$: $$\begin{bmatrix} A & a \\ 0 & 1\\ \end{bmatrix} X^2 + \begin{bmatrix} B & b \\ 0 & 0\\ \end{bmatrix} X + \begin{bmatrix} C & c \\ 0 & 0\\ \end{bmatrix} = 0,$$ where $$A$$, $$B$$ and $$C$$ are real square matrices, $$a$$, $$b$$ and $$c$$ are real vectors, and $$X$$ is a square matrix the same size as $$\begin{bmatrix} A & a \\ 0 & 1\\ \end{bmatrix}$$.

Now consider the following matrix quadratic equation for $$Y$$: $$A Y^2 + B Y + C = 0,$$ where $$Y$$ is a square matrix the same size as $$A$$.

Must this have a real solution for $$Y$$? If not, must this at least hold generically over all $$A,B,C,a,b,c$$ for which the first equation has a real solution for $$X$$?

Note: In the case in which $$A,B,C,a,b,c$$ are all scalars, the solutions to the original equation are as follows:

Either:

1. $$X_{1,1}$$ is a solution to $$A X_{1,1}^2+B X_{1,1}+C=0$$, $$X_{2,1}=X_{2,2}=0$$, $$X_{1,2}=-\frac{c}{AX_{1,1}+B}$$. OR
2. $$X_{1,1}=-\frac{C c}{B c-b C}$$, $$X_{1,2}=-\frac{c^2}{B c-b C}$$, $$X_{2,1}=-\frac{C^2}{B c-b C}$$, $$X_{2,2}=\frac{C c}{B c-b C}$$.

In the former case, my claim holds. In the latter case, there is no guarantee that it does (it seems).

• Unless something is off in what you wrote, $Y=X$ is a solution. Jun 22 at 22:25
• $X$ and $Y$ are matrices of different sizes, so this is impossible. I have now clarified the question to make this clear.
– cfp
Jun 23 at 10:26
• Though, perhaps you meant that $X_{1,1}=Y$. This can hold, but it does not seem to be guaranteed. See my edit to the question to cover the scalar case.
– cfp
Jun 23 at 10:47

When $$A=B=0$$ but $$C\neq 0$$,

$$A Y^2 + B Y + C = 0$$ fails to have a solution.

However, letting $$A,B,C$$ be $$1$$-by-$$1$$ square matrices,

you can set $$A=B=0\;$$; $$C=1\;$$; $$a=1\;$$; $$b=-1\;$$; $$c=0\;$$; $$X=\begin{bmatrix} 0 & 0 \\ 1 & 0\\ \end{bmatrix}$$

satisfying $$\begin{bmatrix} A & a \\ 0 & 1\\ \end{bmatrix} X^2 + \begin{bmatrix} B & b \\ 0 & 0\\ \end{bmatrix} X + \begin{bmatrix} C & c \\ 0 & 0\\ \end{bmatrix} = 0$$.

$$\\$$

@cfp If you want a more generic-looking example, $$A=\begin{bmatrix} -2 & 0 \\ 0 & 1\\ \end{bmatrix}$$, $$B=0$$, $$C=\begin{bmatrix} 0 & 0 \\ 0 & 1\\ \end{bmatrix}$$, the value of $$a$$ doesn't matter, $$b=\begin{bmatrix} 0 \\ -1\\ \end{bmatrix}$$, $$c=\begin{bmatrix} 0 \\ 0\\ \end{bmatrix}$$, $$X=\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0\\ \end{bmatrix}$$.

You can check that you cannot solve for $$Y$$ where $$AY^2+C=0$$, which is equivalent to $$Y^2=\begin{bmatrix} 0 & 0 \\ 0 & -1\\ \end{bmatrix}$$.

• This answers the first question negatively. But it looks non-generic. Do you have any idea about the second question?
– cfp
Jun 23 at 10:28
• I have edited the question to cover the scalar case. This suggests that my second claim is also false.
– cfp
Jun 23 at 10:48
• @cfp The specific case that I gave satisfies $A=B=0$ but $C\neq0$. In this case, you can solve for $X$ but not for $Y$. Jun 23 at 12:08
• Yup I got that. But $A=B=0$ is non-generic.
– cfp
Jun 23 at 12:18
• @cfp added a more generic-looking example. 2 days ago