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Let $A$ and $B$, be two $n\times n$ matrices such that $A=P^{-1}BP$ for some invertible matrix $P$. Show that both $A$ and $B$ have the same minimal polynomial.

I'm not sure how to approach this. The only method I know to find minimal polynomial of a matrix is to estimate from its characteristic polynomial using the fact that characteristic polynomial and minimal polynomial must have same roots. Any hints/suggestions?

  • @DietrichBurde I've seen that answer you've linked. It seems to use way more machinery than I know. I've just taken a first course in LA –  Mar 19 '18 at 19:24
  • Well, since you have to prove this for any invertible matrix, you won't be able to calculate the specific polynomial. You need to show that that $A$ and $P^{-1}AP$ satisfy the same polynomials. – saulspatz Mar 19 '18 at 19:24
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    @user296602 That answer talks about "characteristic polynomial", not minimal polynomial. I don't understand why people are closing this as duplicate. –  Mar 19 '18 at 19:24
  • Hint: $(P^{-1}AP)^2 = P^{-1}APP^{-1}AP = P^{-1}A^2P$ – saulspatz Mar 19 '18 at 19:28
  • @saulspatz That makes sense. So I could just plug in $P^{-1}$ and $P$ at the front and end of the minimal polynomial, to express it in that form and show that if $A$ satisfies the minimal polynomial and so does $B$. BUT that doesn't prove that it is indeed the "minimal polynomial for $B$". B could satisfy equations other than the minimal polynomial, too. –  Mar 19 '18 at 19:31
  • For any polynomial $f$ we have $$f(A) = f(P^{-1}BP) = P^{-1}f(B)P$$ so $f(A) = 0$ if and only if $f(B) = 0$. Now it should be clear that $A$ and $B$ have the same minimal polynomial. – mechanodroid Mar 19 '18 at 19:31
  • @mechanodroid Ah, we need to show it both ways I guess. $A$ satisfies the minimal polynomial of $B$ and $B$ satisfies the minimal polynomial of $A$, to make the proof complete. I get the idea though, thanks –  Mar 19 '18 at 19:33
  • @Blue The above shows that $A$ and $B$ satisfy the same set of polynomials. The monic polynomial of minimal degree in that set is the minimal polynomial for both $A$ and $B$. – mechanodroid Mar 19 '18 at 19:35
  • Right, but now you prove it in the other direction. In fact, you've already proved it, because $A = P(P^{-1}AP)P^{-1}.$ – saulspatz Mar 19 '18 at 19:36
  • @saulspatz Thanks, that helped! –  Mar 19 '18 at 19:37

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