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What can deduce from the given fact that the following matrices commute

$\left( \begin{array}{ccc} 0 & 0 & ... & 0 \\ 1 & 0 & ... & 0 \\ 0 & 1 & ... & 0 \\ so \,\ & on&... \end{array} \right)$ : an $n\times n$ Jordan block with diagonal $=0$

and the matrix $\left( \begin{array}{ccc} a & 0 & ... & 0 \\ b & a & ... & 0 \\ c & b & ... & 0 \\ so \,\ & on&... \end{array} \right)$: an $n\times n$ matrix where the column entries shift one place down as we move one column across

And that they are in this for wrt the same basis (which I know is not surprising since commuting matrices are simultaneously triangularizable).

What implications does this have if we interpret them as linear operators?

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  • $\begingroup$ The second matrix is a polynomial in the first one with $a$, $b$, $c$... as coefficients. $\endgroup$ Nov 21, 2011 at 0:57

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I just gave an answer on this an hour ago, if one of the matrices, call it $M,$ has only full Jordan blocks, as in your first matrix $ M \; \; = \; \; \left( \begin{array}{ccc} 0 & 0 & ... & 0 \\ 1 & 0 & ... & 0 \\ 0 & 1 & ... & 0 \\ so \,\ & on&... \end{array} \right)$ : an $n\times n$ Jordan block with diagonal $=0,$ then all matrices that commute with it can be written as a polynomial in $M.$ By the Cayley-Hamilton theorem this polynomial need have degree no larger than $n-1.$

Indeed, your second matrix is $$ A = a_0 I + a_1 M + a_2 M^2 + a_3 M^3 + \cdots + a_{n-1} M^{n-1},$$ with your letters $ a = a_0, \; b = a_1, \; c = a_2.$

Note that, for $M,$ the characteristic polynomial and the minimal polynomial are the same, that is, $M^n = 0$ but no lower power.

See: Commuting linear maps

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