Jordan Cell as Jordan Form implies Commuting Matrices are in Polynomial

Question

If $X$ is a matrix such that its Jordan form is a single Jordan cell, then show that all matrices $Y$ that commute with $X$ are polynomials in $X$ (there is a polynomial $f$ such that $Y=f(X)$).

Answer 1

Outline of proof: It suffices to show that $$ J = \pmatrix{0&1\\&0&1\\&&\ddots\\&&&0} $$ commutes with a matrix $A$ if and only if $A$ has the form $$ A = \pmatrix{ a_0&a_1&a_2&\cdots\\ &a_0&a_1 & \cdots\\ &&\ddots&\ddots\\ \\ &&&&a_0} = a_{n-1}J^{n-1} + \cdots + a_1 J + a_0I $$ (this can be done by computation). From there, we note that the matrix $B$ commutes with $S(J + \lambda I)S^{-1}$ if and only if $S^{-1}BS$ commutes with $J$.

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In response to your comment: confirm that for arbitrary matrices $A,B$, $I$ the identity, and $\lambda \in \Bbb C$, we have $$ AB = BA \iff (A + \lambda I)B = B(A + \lambda I) $$