# For $n \times n$ matrices $A$ and $B$, if $(A^i - B^i)x = 0$ for $i = 1, \dots,\ n$, does $(A^{n+1} - B^{n+1}) x = 0$?

Conjecture: Let $$A,B \in \mathbb{R}^{n \times n}$$ and $$x \in \mathbb{R}^n \setminus \{0\}$$. If

$$\left( A^{i} - B^i \right) \, x = 0, \quad \forall i \in \{ 1, \dots, n \}$$

then

$$\left( A^{j} - B^{j} \right) \, x = 0, \quad \forall j \in \{ n+1, n+2, \dots \}$$

My initial goal is to prove this for $$j = n+1$$, i.e., $$\left(A^{n+1} - B^{n+1}\right)\,x = 0.$$

Since $$\left(A - B\right)\,x = 0,$$ we have $$A\,x = B\,x,$$ and similarly, $$A^{i}\,x = B^{i}\,x,\quad i = 1,\ \dots,\ n.$$

So $$\left(A^{n+1} - B^{n+1}\right)\,x = A^{n+1}\,x - B^{n+1}\,x = A^{n}\,A\,x - B^{n}\,B\,x = \left(A^{n} - B^{n}\right)\,A\,x = \left(A^{n} - B^{n}\right)\,B\,x,$$ and similarly,

$$\left(A^{n+1} - B^{n+1}\right)\,x = \left(A^{n+1-i} - B^{n+1-i}\right)\,A^{i}\,x = \left(A^{n+1-i} - B^{n+1-i}\right)\,B^{i}\,x,\quad i = 1,\ \dots,\ n.$$

I'm not entirely sure where to go from here.

• $x$ is a scalar? Aug 13, 2020 at 17:06
• @sai-kartik No, $x$ is a $n\times 1$ column vector. Aug 13, 2020 at 17:10
• $\mathbf{A}\underline{x}=\mathbf{B}\underline{x}$ implies $\mathbf{A}=\mathbf{B}$ since $\underline{x}\neq 0$ does it not? So it seems the statement holds trivially... Aug 13, 2020 at 17:16
• @K.defaoite Surely not. Here $x$ is a given $x$ but not all $x \neq 0$. Consider $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, B = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}$, and $x = (0, 1)^T$. Then $Ax = Bx$ but $A \neq B$. Aug 13, 2020 at 17:27
• @K.defaoite Have you ever heard of singular matrices ? Aug 13, 2020 at 17:28

$$(A-B)x=0$$.

$$(A-B)Ax=A^2x-BAx=A^2x-B^2x=0$$

$$\cdots$$

$$(A-B)A^{n-1}x=A^nx-{BA}^{n-1}x=A^nx-B^nx=0$$

Hence $$(A-B)$$ annihilates all of $$x,Ax,A^2x,\ldots,A^{n-1}x$$. But $$A^n$$ can be written in terms of lower terms since $$A$$ satisfies a minimal polynomial.

Thus $$(A-B)A^nx=0$$, so $$A^{n+1}x={BA}^nx=B^{n+1}x$$.

etc.

• Lovely argument! Aug 13, 2020 at 17:45
• By "minimal polynomial", I think you mean "characteristic polynomial"? Jan 18, 2023 at 1:20