Let $A,\ B,\ C$ as square matrices. Is there a general formula for finding $B,C\ $ such that: $A=B+C$ where $BC=CB$
I am trying to extend to a particular case of matrix exponentials.
Example:
Find the matrix exponential $e^{At}$ where $A=\left(\begin{matrix}1&a\\0&1\\\end{matrix}\right)$
Since $A=I+\left(\begin{matrix}0&a\\0&0\\\end{matrix}\right)$
And there is a formula where $e^Be^C=e^{B+C}$ if $BC=CB$
Then $$e^{At}=e^{Bt+Ct}={(e}^{Bt})(e^{Ct})$$ such that $B=I$ and $C=\left(\begin{matrix}0&a\\0&0\\\end{matrix}\right)$
Now $$e^{Bt}=\left(\begin{matrix}e^t&0\\0&e^t\\\end{matrix}\right)$$
$$e^{Ct}=e^{0t}(I+Ct+\frac{C^2t^2}{2}\ldots)$$
Since $C$ has only one (multiple) eigenvalues of 0, a special application of the Cayley-Hamilton theorem implies that ${(C-0I)}^2=0$ (nilpotent)
Then $$e^{Ct}=e^{0t}(I+Ct)$$
Finally $$e^{At}=\left(\begin{matrix}e^t&0\\0&e^t\\\end{matrix}\right)(I+\left(\begin{matrix}0&at\\0&0\\\end{matrix}\right))$$
This method only works for the case $BC=CB=IC=CI$ That’s why I am asking if for a general method of extracting $B,C$ which commutes.
One idea: first upper-triangularize $A$ (namely $A'$). Then, as $A$ is invertible, all entries of the diagonal of $A'$ are non-zero so we can choose some $k\neq 0$ such that $A'=(A'-kI)+kI$ and $B':=A'-kI$ is still invertible. Therefore $B'$ and $C':=kI$ commute and $A'=B'+C'$.
Finally, as $A=D^{-1}A'D$ for some invertible $D$ then setting $B:=D^{-1}B'D$ and $C:=D^{-1}C'D$ you finally find that
$$ A=D^{-1}A'D=D^{-1}(B'+C')D=B+C\\ BC=D^{-1}B'C'D=D^{-1}C'B'D=D^{-1}C'D D^{-1}B'D=CB $$ ∎
P.S.: to upper-triangularize $A$ maybe you will need to use complex-valued matrices.