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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.

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  • $\begingroup$ $x$ is a scalar? $\endgroup$
    – sai-kartik
    Aug 13, 2020 at 17:06
  • $\begingroup$ @sai-kartik No, $x$ is a $n\times 1$ column vector. $\endgroup$
    – K.defaoite
    Aug 13, 2020 at 17:10
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    $\begingroup$ $\mathbf{A}\underline{x}=\mathbf{B}\underline{x}$ implies $\mathbf{A}=\mathbf{B}$ since $\underline{x}\neq 0$ does it not? So it seems the statement holds trivially... $\endgroup$
    – K.defaoite
    Aug 13, 2020 at 17:16
  • $\begingroup$ @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 $A \neq B$. $\endgroup$
    – Zhanxiong
    Aug 13, 2020 at 17:27
  • $\begingroup$ @K.defaoite Have you ever heard of singular matrices ? $\endgroup$ Aug 13, 2020 at 17:28

1 Answer 1

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$(A-B)x=0$.

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

$\cdots$

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

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.

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

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