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I am not sure about this question. I have two claims and I need to know if anyone of them is correct.

Claims:

a) There exists a real matrix for which the characteristic polynomial is: $t(t-1)(t^2+t+1)$ and the minimal polynomial is $t(t-1).$

b) Let $A$ and $B$ be $n\times n$ matrices. If there exists a polynomial $q$ such that $q(A)=0$ but $q(B) \neq 0$, then $A$ and $B$ are not similar.

What I think:

a) Basically, the minimal polynomial is the monic polynomial of smallest degree that "resets" a matrix, so the minimal polynomial divides a polynomial say $q$(some matrix)$=0$, and it also divides the characteristic polynomial. Because of the last sentence, I think that the claim is true.

b) Similar matrices are matrices that follow $A=x^{-1}Bx$, and because of that if $q(A)=0$ and $q(B) \neq 0$ then they are not similar. so I think that this claim is correct a well.

Are both claims true?

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  • $\begingroup$ for a) think about that $t$ divides the characteristic polynomial, but if $t$ is minimal, the matrix is $0$! The characteristic and minimal polynomial have the same roots. for b) the key is "$A$ and $B$ are similar $\iff$ the characteristic and minimal polynomials are equal" $\endgroup$ – Martín Vacas Vignolo Jan 10 '18 at 12:20
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The first claim would violate several important facts about minimal and characteristic polynomials, so it cannot be true. For instance, both polynomials have the same set of roots over any field$~K$, which is the set of eigenvalues that the $K$-linear operator defined by the matrix has; here for $K=\Bbb C$ the purported characteristic polynomial has roots that the minimal polynomial does not have. Also, with minimal polynomial $X(X-1)$, a matrix must be diagonalisable over$~\Bbb R$ (even over $\Bbb Q$) with eigenvalues $0,1$, and this is not compatible with the characteristic polynomial that corresponds to a matrix not diagonalisable over$~\Bbb R$ (because it has an irreducible factor of degree$~2$).

The second claim is true: similar matrices can be considered to describe the same linear operator$~\phi$ with respect different bases, which makes it clear that they have the same set of annihilating polynomials (namely those that annihilate $~\phi$).

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  • $\begingroup$ thank you very much for your elaborated answer. it helped me a lot! $\endgroup$ – BeginningMath Jan 10 '18 at 14:52
  • $\begingroup$ i have a small questions about combining the two questions if you may: is it true that if two matrices has the same characteristic polynom then they are similar? $\endgroup$ – BeginningMath Jan 10 '18 at 14:56
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    $\begingroup$ @BeginningMath No. For instance, all nilpotent matrices (of the same dimension) have the same characteristic polynomial. $\endgroup$ – José Carlos Santos Jan 10 '18 at 15:17
  • $\begingroup$ thank you very much again! $\endgroup$ – BeginningMath Jan 10 '18 at 19:00
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    $\begingroup$ @BeginningMath: Although the characteristic polynomial in general does not characterise similarity (as José Carlos replied), nor will adding any simple information like the minimal polynomial, let me add some positive facts. (1) Having the same characteristic polynomial implies similarity if (and only if) its factorisation has no repeated irreducible factors. (2) Having a given characteristic polynomial leaves only a finite number of possible classes of similarity. For full details, see the structure theorem for finitely generated modules over a principal ideal domain (here $K[X]$). $\endgroup$ – Marc van Leeuwen Jan 11 '18 at 6:42
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a) The claim is false. If you express the characteristic polynomial $p(x)$ as the product $q_1(x)^{n_1}\cdots q_k(x)^{n_k}$ of irreducible polynomials, then the minimal polynomial can be expressed as $q_1(x)^{m_1}\cdots q_k(x)^{m_k}$, where each $m_j$ belongs to $\{1,2,\ldots,n_j\}$.

2) The claim is true. If $A=SBS^{-1}$, then $q(A)=Sq(B)S^{-1}$.

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  • $\begingroup$ thank you very much for explaining and correcting my mistake $\endgroup$ – BeginningMath Jan 10 '18 at 14:52

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