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Since similarity of matrices' is an equivalence relation, doesn't that imply that given any polynomial equation involving similar matrices you can substitute in any similar matrices' and the equation will still hold?

For example, given $A,B,C\in M^F_{n\times n}$

if $B \cong C$ then: $$A \cong B^2+5B+3I \iff A \cong C^2+5C+3I$$


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up vote 3 down vote accepted

What you say is correct (although it doesn't follow just from the fact that similarity is an equivalence relation, but from the fact it is an equivalence relation preserved by polynomials, that is if $A\cong B$, then for any polynomial $p$, $p(A)\cong p(B)$).

If $B=PCP^{-1}$ and for some polynomial $p$, $p(B)=DAD^{-1}$, then $p(B)=p(PCP^{-1})=Pp(C)P^{-1}$, so $p(C)=(P^{-1}D)A(D^{-1}P)=(P^{-1}D)A(P^{-1}D)^{-1}$

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Ah, this works because the result of multiplying invertible matrices is an invertible matrix. Nice. – Robert S. Barnes Jun 22 '12 at 15:30
@Robert: Actually, it works because conjugation is a ring homomorphism. That is, because for all invertible matrices $P$ and all matrices $A$ and $B$, we have $P(A+B)P^{-1} = PAP^{-1}+PBP^{-1}$, and $P(AB)P^{-1} = (PAP^{-1})(PBP^{-1})$. The fact that the product of invertible matrices is invertible does not enter into it. – Arturo Magidin Jun 22 '12 at 15:42

Having the same element in the upper lefthand corner is also an equivalence relation on $n\times n$ matrices, but you would hardly expect that it would be preserved in polynomials. E.g., $$\pmatrix{1&0\\0&0}^2=\pmatrix{1&0\\0&0}\;,$$ but $$\pmatrix{1&1\\1&1}^2=\pmatrix{2&2\\2&2}\;,$$ with a different element in the upper lefthand corner.

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I think I may have written the question in a confusing way. When I wrote a polynomial equation what I'm talking about is not equality but similarity. I.E. if B is similar to C, then A is similar to f(B) iff A is similar to f(C). – Robert S. Barnes Jun 22 '12 at 13:37
@Robert: Not quite: you were suggesting that polynomials preserve equivalence relations, in particular similarity. I’m pointing out that even if they do preserve similarity, this could not have anything to do with similarity being an equivalence relation, because they don’t preserve this very simple equivalence relation. – Brian M. Scott Jun 22 '12 at 13:41
OK, so idea is correct, but it's a property of similarity itself, not a result of similarity being an equivalence relation, right? – Robert S. Barnes Jun 22 '12 at 13:49
@Robert: Yes, that’s right. – Brian M. Scott Jun 22 '12 at 13:51

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