Does there exist a complex polynomial $p$ such that $p(J_k^2(a))=J_k(a)$ if $a\neq 0$? Here $J_k(a)$ is a $k\times k$ Jordan block whose diagonal element is $a\neq 0$.
What I know is just the Jordan carnonical form of $J_k^2(a)$ is just $J_k(a^2)$.
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xldd
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It is a well-known fact that you can calculate the analytic function of a matrix by using the polynomial that interpolates the function and its derivatives at the eigenvalues of the matrix.
If $f$ and $p$ are analytic functions (on a certain domain in $\mathbb{C}$, I do some hand-waving here), $A$ is a matrix, and $\lambda_1,\ldots\lambda_m$ are the distinct eigenvalues of $A$ with algebraic multiplicity $r_1,\ldots,r_m,$ then $$ f^{(n)}(\lambda_j)=p^{(n)}(\lambda_j)\; \forall n\in\{0,\ldots,r_j-1\}\; \forall j\in\{1,\ldots,m\} \Longrightarrow f(A) = p(A) $$ Now set $A=J^2(a)$ and $f(z)=\sqrt{z}.$ Then $m=1,$ $\lambda_1=a^2,$ $f(\lambda_1)=a$ and $r_1=k.$ Then $$ f^{(n)}(a^2)=p^{(n)}(a^2)\; \forall n\in\{0,\ldots,k-1\}\; \Longrightarrow p(J^2(a)) = f(J^2(a)) = \sqrt{J^2(a)} = J(a) $$ (Note that we have to make sure that we use the branch of the square root that yields $a$ for $\sqrt{a^2}$, but not the branch that yields $-a.$ The derivatives must be selected accordingly.)
This shows that we can simply use the Taylor polynomial of degree $k-1$ for the square root, at the point $a^2.$ $$ p(x) = \sum_{n=0}^{k-1}\frac{f^{(n)}\!\left(a^2\right)}{n!} \left(x-a^2\right)^n = a+\frac{1}{2a}\left(x-a^2\right)-\frac{1}{8a^3}\left(x-a^2\right)^2+\ldots $$
Reinhard Meier
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To show that such a $p_0$ exists, start by noting that $J := J_k^2(a) - aI$ is upper-triangular and Toeplitz. From there, one can recursively show that there exist polynomials $q_j(x)$ such that $q_j(J) = J_k(0)^j$ for $j = k-1,k-2,\dots,1$.
– Ben Grossmann Dec 27 '22 at 00:18