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Say I have a sequence of linear operators $A_1,...,A_n$ on a (real) vector space $V_1$. I suspect that there's a second vector space $V_2$, and an operator $A$ on $V_1\oplus V_2$, such that $A_i=(\pi_1 A^i)|_{V_1}$ for $i=1,...,n$, where $\pi_1:V\rightarrow V_1$ is projection onto $V_1$.

Is there a way to confirm my suspicion, with or without finding such $A$ and $V_2$?

Say we confirmed my suspicion, and an oracle gave us $V_2$ along with an operator $\Delta$, and they told us that $A=C\Delta$ for some operator $C$ that acts diagonally on $V_1\oplus V_2$. They even gave us bases for $V_1,V_2$, on which $C$ also acts diagonally. How would you find $C$?

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  • $\begingroup$ Do you mean $A_i=A^i$ for each $i$ or are you fixing $i$ in some way? $\endgroup$
    – Simon Wadsley
    Commented Aug 23, 2023 at 8:50
  • $\begingroup$ Thank you for your question. I mean for all $i$, I have edited the question to reflect this $\endgroup$
    – Dror Speiser
    Commented Aug 23, 2023 at 8:54
  • $\begingroup$ If $A_1=A\mid_{V_1}$, that means that $V_1$ is invariant under $A$, so the action of $A$ on $V_2$ is irrelevant. Did you mean, perhaps, that you need to take a composition of $A^i\mid_{V_1}$ with the projection onto $V_1$? $\endgroup$
    – Ilya Bogdanov
    Commented Aug 23, 2023 at 9:10
  • $\begingroup$ Ah! Yes! You're correct. Thanks! $\endgroup$
    – Dror Speiser
    Commented Aug 23, 2023 at 9:14

1 Answer 1

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An affirmative answer to the first question is easy. Set $V_2=\bigoplus_{i=1}^{n-1} V_1$. Define $A(v,0,0,\dots,0)=(A_1v,v,0,0,\dots)$. Then define the action of $A$ on $0\oplus V_1\oplus 0\oplus\dots$ to get $A^2(v,0,\dots)=A(A_1v,v,0,\dots)=(A_2v, 0, v,0,0,\dots)$. Proceed on in he same manner.

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