Given $$A,B\in M_3(C)$$ and $$rank(A)=rank(A^2),rank(B)=rank(B^2) $$ and $A^3$ is similar to $B^3$.
and all eigenvalue of $A,B$ are real number
can I prove that $A$ is similar to $B$?
What does the rank imply?
Given $A^3$ is similar to $B^3$ and $rank(A)=rank(A^2),rank(B)=rank(B^2)$ , prove that $A$ is similar to $B$
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linear-algebra
1 Answers
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Hint: The rank implies that in both $A$ and $B$, the geometric multiplicity of the eigenvalue $0$ coincides with the algebraic multiplicity. That is: in their Jordan forms, if $A$ and $B$ have $0$ as an eigenvalue, then the associated Jordan block has size at most $1$.
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0How do you handle the case where $A,B$ don't have eigenvalue $0$? – 2017-02-10
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0@quasi one Jordan block at a time. The statement holds over $M_n$, not just $M_3$. The fact that the eigenvalues are real is important. – 2017-02-10
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0A neat approach is to consider an "analytic" cube-root function around a point. – 2017-02-10
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0I don't follow. Do you mean some kind of matrix power series? – 2017-02-10
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0@quasi yeah, but that's not necessary. All you really **need** to do is show that for any Jordan block of a non-zero eigenvalue, $J^3$ is a similar to a Jordan block of the same shape (but of a new eigenvalue). – 2017-02-10