Sub-blocks of matrix quadratic equations

Question

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 $X$?

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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).

Answer 1

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 $Y^2=\begin{bmatrix} 0 & 0 \\ 0 & -1\\ \end{bmatrix}$.