"Vieta's-formulas" on matrices

Question

Let matrices $X $ and $ Y$ satisfy the equality $Z^2+AZ+B=0$ where $A , B$ are real matrices and $\det(X - Y)\ne 0$ Prove

1. $ \operatorname{tr} X + \operatorname{tr} Y = - \operatorname{tr} A$
2. $\det X \det Y = \det B $

Proof 1) $(X+Y)(X-Y)+XY-YX+A(X-Y)=0\Rightarrow X+Y+(XY-YX)(X-Y)^{-1}+A=0$

As $ \operatorname{tr} PQ= \operatorname{tr} QP$ we have $\operatorname{tr} (XY-YX)(X-Y)^{-1}=0$ entailing the result.

2. We have $(X+A)X=-B$ and $(Y+A)Y=-B$. The result follows if $\det XY=0.$

If $\det X\ne 0$ and $\det Y\ne0$ then $X-Y=B(Y^{-1}-X^{-1})$ Assuming that $X$ and $Y$ are permutable, we can easily obtain the result.

Is it possible to achieve this result without such an assumption, or is it necessary to look for a counterexample?

Answer 1

The two equalities do not necessarily hold when $X-Y$ is singular. Here is a random example: $$ X=\pmatrix{1&0\\ 1&1}, \ Y=\pmatrix{2&0\\ 1&1}, \ A=\pmatrix{-3&1\\ -1&1}, \ B=\pmatrix{1&-1\\ -2&-2}. $$ One can easily verify that $$ \begin{aligned} (X+A)X&=\pmatrix{-2&1\\ 0&2}\pmatrix{1&0\\ 1&1}=-B,\\ (Y+A)Y&=\pmatrix{-1&1\\ 0&2}\pmatrix{2&0\\ 1&1}=-B.\\ \end{aligned} $$ However, $\operatorname{tr}(X)+\operatorname{tr}(Y)=5\ne2=-\operatorname{tr}(A)$ and $\det(X)\det(Y)=2\ne-4=\det(B)$.

Answer 2

Thanks a lot! How did I not realize straightaway? How to fix the first statement

Let $X=P+Q, Y=P-Q\Leftrightarrow X+Y=2P, X-Y=2Q $.

Now $$X^2-Y^2+A(X-Y)=0\Leftrightarrow $P+QPQ^{-1}+A=0\Rightarrow \operatorname{tr} X + \operatorname{tr} Y = - \operatorname{tr} A$$

Proof of the second was completed by user 1551 in comment.