Similar Matrices and Equivalence Relations

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

Correct?

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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.