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Let $A$ and $B$ be two matrices in $M_n$. Is the following ture:

$A$ and $B$ are similar $\iff$ $A$ and $B$ have the same jordan canonical form.

Could someone explain?

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  • $\begingroup$ for matrices over a field which is not algebraically closed there is no guarantee that there are eigenvalues in the field. For example, over $\mathbb{R}$, most rotation matrices have pure imaginary eigenvalues. Consequently, there is no Jordan form. If you're working over $\mathbb{C}$ then what has been said here is fine. Moreover, if you really mean the real Jordan form for the real case then the equivalence also holds in that context. Bottom line, similar matrices have the same eigenvalues and geometric multiplicities hence the canonical forms match-up. $\endgroup$ – James S. Cook Aug 23 '15 at 19:04
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If $A=PJP^{-1}$ is the Jordan decomposition of $A$ and $M$ is an invertible matrix such that $A=M^{-1}BM$ then we have $$ B = (MM^{-1})B(MM^{-1})=M(M^{-1}BM)M^{-1} =MAM^{-1}=M(PJP^{-1})M^{-1}=(MP)J(P^{-1}M^{-1})$$

and thus the Jordan normal form of $A$ (namely $J$) is the same as that of $B$.

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  • $\begingroup$ This is only one direction. Patrick Stevens' answer covers both directions. $\endgroup$ – amsmath Jul 22 '18 at 18:54
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It's true. Let $A$ have the same JCF $J$ as $B$ does; say $A = PJP^{-1}$ and $B = Q J Q^{-1}$. Then $J = P^{-1} A P$, so $B = Q P^{-1} A P Q^{-1}$, and so $B$ is similar to $A$.

Suppose $A$ and $B$ are similar, but for contradiction they have different JCFs. Then $A$ reduces to JCF $J$, and $B$ to JCF $K$. But $J, K$ are not similar - this is a contradiction.

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