I am trying to find a matrix X such that
$$ e^X = \left( \begin{matrix} 0 & 2021 & 0 & ... & ... & ... & 0\\ 0 & 0 & 2020 & ... & ... & ... & 0\\ 0 & 0 & ... & ... & ... & ... & 0\\ 0 & 0 & ... & ... & ... & ... & 0\\ 0 & 0 & ... & ... & ... & 3 & 0\\ 0 & 0 & ... & ... & ... & 0 & 2\\ 1 & 0 & ... & ... & ... & ... & 0\\ \\ \end{matrix}\right) $$
Which means I will need to find the log of a non-diagonalizable matrix. I know I can typically find the Jordan form of the matrix, but this one is so large, it seems very difficult. I also know the MacLaurin series for ln(X), but the matrix is not nilpotent, so the series is infinite.
I am not sure if I'm missing something in my conclusions about Jordan form and MacLaurin series. Can anyone help me identify the best way to go about this?
I will outline the essential procedure below. Define analogously an $n\times n$ matrix $A$. Let $r=(n!)^{1/n}$ and $D=\operatorname{diag}\left(\frac{r^{n}}{n!},\frac{r^{n-1}}{(n-1)!},\ldots,\frac{r^{2}}{2!},r\right)$. Then $A$ is diagonally similar to $r$ times a circulant permutation matrix: $$ DAD^{-1}=rC:=r\pmatrix{0&1\\ &0&1\\ &&\ddots&\ddots\\ &&&0&1\\ 1&&&&0}. $$ Hence one can take $\log(A)$ as $D^{-1}\log(rC)D=\log(r)I+D^{-1}\log(C)D$ and the problem boils down to finding a matrix logarithm of the circulant permutation matrix $C$.
$C$ is the companion matrix of the characteristic polynomial $x^n-1$. Its eigenvalues are therefore $\omega^k$ for $k=0,1,\ldots,n-1$, where $\omega=\exp(2\pi i/n)$ is a primitive $n$-th root of unity. So, $C$ has not any negative eigenvalue when $n$ is odd and it has exactly one negative eigenvalue when $n$ is even. It follows that $C$ has a real matrix logarithm if and only if $n$ is odd.
The eigen structures of circulant matrices are well-known. See Wikipedia for a brief description. In particular, $C=U\operatorname{diag}(1,\omega,\omega^2,\ldots,\omega^{n-1})U^\ast$, where $U$ is the unitary DFT matrix of order $n$. Therefore one can always take $$ \log(C)=U\operatorname{diag}\left(0,\frac{2\pi i}{n},\frac{4\pi i}{n},\ldots,\frac{2(n-1)\pi i}{n}\right)U^\ast $$ to obtain a complex logarithm of $C$.
When $n$ is odd (as in your case, where $n=2021$), in order to obtain a real logarithm, we may express the eigenvalues $$ 1,\omega,\omega^2,\ldots,\omega^{(n-1)/2},\omega^{(n+1)/2},\ldots,\omega^{n-2},\omega^{n-1} $$ as $$ 1,\omega,\omega^2,\ldots,\omega^{(n-1)/2},\overline{\omega}^{(n-1)/2},\ldots,\overline{\omega}^2,\overline{\omega} $$ and take $$ \log(C)=U\operatorname{diag}\left(0,\frac{2\pi i}{n},\frac{4\pi i}{n},\ldots,\frac{(n-1)\pi i}{n},\frac{-(n-1)\pi i}{n},\ldots,\frac{-4\pi i}{n},\frac{-2\pi i}{n}\right)U^\ast. $$ Then $\log(C)$ will be a weighted sum of outer products of eigenvectors where non-real terms appear in conjugate pairs. Hence it is a real matrix.