If $A^kX=B^kY$ for all $k$, is $X=Y$?

Question

This one is more general than the one I asked before.

Given invertible matrices $A,B$ and matrices $X,Y$ all with size $n$, such that $A^k X = B^k Y$ for $k=1,2,...,2n$. Does it follow that $X = Y$?

I have no idea where to work. Thanks for any help.

Answer 1

Let $$C= \begin{pmatrix} A & 0 \\0 & B \\ \end{pmatrix} \,.$$

Then $C$ is invertible with inverse $C^{-1}= \begin{pmatrix} A^{-1} & 0 \\0 & B^{-1} \\ \end{pmatrix} \,.$

Let $$P(x)=x^{2n}+a_{2n-1}x^{2n-1}+\dots+a_0$$ be the characteristic polynomial of $C$. Then, since $C$ is invertible we have $a_0 \neq 0$.

Thus

$$I_{2n}=-\frac{1}{a_0}\left[ C^{2n}+a_{2n-1}C^{2n-1}+\dots+a_1C\right] \,.$$

Let $$Q(x)=-\frac{1}{a_0} \left[ x^{2n}+a_{2n-1}x^{2n-1}+\dots+a_1x\right] \,.$$

Since $Q(C)=I_{2n}$ from the definition of $C$ we get $Q(A)=Q(B)=I_n$.

From $A^kX=B^kY, \forall 1 \leq k \leq 2n$ we get $Q(A)X=Q(B)Y$, thus $X=Y$:

$$X=I_nX=Q(A)X=-\frac{1}{a_0} \left[ A^{2n}X+a_{2n-1}A^{2n-1}X+\dots+a_1AX\right]=-\frac{1}{a_0}\left[ B^{2n}Y+a_{2n-1}B^{2n-1}Y+\dots+a_1BY\right]=Q(B)Y=I_nY=Y \,.$$