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Quotes from wiki:

The Jordan normal form is the most convenient for computation of the matrix functions (though it may be not the best choice for computer computations). Let $f(z)$ be an analytical function of a complex argument. Applying the function on a $n×n$ Jordan block J with eigenvalue $\lambda$ results in an upper triangular matrix:

$$f(J)={\begin{bmatrix}f(\lambda )&f'(\lambda )&{\tfrac {f''(\lambda )}{2}}&...&{\tfrac {f^{(n-1)}(\lambda )}{(n-1)!}}\\0&f(\lambda )&f'(\lambda )&...&{\tfrac {f^{(n-2)}(\lambda )}{(n-2)!}}\\\vdots &\vdots &\ddots &\ddots &\vdots \\0&0&0&f(\lambda )&f'(\lambda )\\0&0&0&0&f(\lambda )\end{bmatrix}},$$ where $${\displaystyle J={\begin{pmatrix}\lambda &1&0&\cdots &0\\0&\lambda &1&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\lambda &1\\0&0&0&0&\lambda \end{pmatrix}}} $$ So that the elements of the $k$-th superdiagonal of the resulting matrix are ${\displaystyle {\tfrac {f^{(k)}(\lambda )}{k!}}}$. For a matrix of general Jordan normal form the above expression shall be applied to each Jordan block.

Let $f(z)=z^n$, where $n$ is a positive integer. I want to know if the following holds for $A$ an invertible matrix:

If $${\displaystyle J={\begin{pmatrix}A &I&0&\cdots &0\\0&A &I&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&A &I\\0&0&0&0&A \end{pmatrix}}},$$ then $$f(J)={\begin{bmatrix}f(A )&f'(A )&{\tfrac {f''(A )}{2}}&...&{\tfrac {f^{(n-1)}(A )}{(n-1)!}}\\0&f(A )&f'(A )&...&{\tfrac {f^{(n-2)}(A )}{(n-2)!}}\\\vdots &\vdots &\ddots &\ddots &\vdots \\0&0&0&f(A )&f'(A )\\0&0&0&0&f(A )\end{bmatrix}}.$$

If it doesn't hold for an invertible matrix, then what are the matrics for which it holds? What if the ground field is finite field ?

Thanks for any replies.

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Let $A$ be $p\times p$. We can write your $J$ as $A\otimes I_n +I_p\otimes L$ where $L$ is the $n\times n$ matrix with $1$'s on the superdiagonal and $0$'s elsewhere. Note that the two summands commute.

The binomial theorem now gives us that $$ J^m= \sum_{r+s=m} \binom{m}{r} (A\otimes I)^r (I\otimes L)^s= \sum_{r+s=m} \binom{m}{r} A^r\otimes L^s $$ for every natural number $m$.

The matrix $L^s$ is, by the usual calculation, the matrix with $1$'s on the $s$-th superdiagonal and $0$'s elsewhere. Hence $J^m$ has $0$'s below the diagonal, and then $\binom{m}{r} A^r$ on the $(m-r)$-th superdiagonal.

This yields both your result for $z^n$ and more generally, by linearity, the result for any polynomial function $f$.

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  • $\begingroup$ Can you give more details on how to yields the result from binomial theorem ? Moreover, as I know, binomial theorem holds for $x,y $ only if $x\times y = y\times x$. It seems that $A\times L=L \times A$. Am I right? $\endgroup$ – zongxiang yi Jan 25 '18 at 10:15
  • $\begingroup$ I've put in the detail. $\endgroup$ – ancientmathematician Jan 25 '18 at 10:21
  • $\begingroup$ I am not familiar with the $\otimes$ operator. Maybe I should read some related textbook in order to understand your answer. Thanks anyway. $\endgroup$ – zongxiang yi Jan 25 '18 at 10:26
  • $\begingroup$ All you need for this is to think of it as a convenient way to express block matrices. Why not work out the powers of $J$ when there are $4$ blocks? You'll soon see what's going on. $\endgroup$ – ancientmathematician Jan 25 '18 at 10:33
  • $\begingroup$ Although the $\otimes$ operator differs from my textbook. I have work out the $3$ blocks before asking the question. But it is difficult for me to generalize the result. Now I understand your answer. It would be accepted if more details is given to explain the how to yields the result after $J^m=\sum_{r+s=m} \binom{m}{r} A^r\otimes L^s$.is obtained. $\endgroup$ – zongxiang yi Jan 25 '18 at 11:00

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