Let $V=\mathbb{R}^3$ be a vector space with standard basis. Given a linear map $A\in End(V)$. Fitting's theorem says that there exists a unique reduction $V=N\oplus R $ for which $A|_N\in End(N)$ is nilpotent and $A|_R\in End(R)$ is invertible. I'm following this far, but if we where to apply this to an example, let $A$ be represented by the matrix $\left( \begin{array}{ccc} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array}\right)$. How would one determine $N,R,A|_R,A|_N$? I guess that $N,R$ is of lower dimension than $dim(V)=3$, and hence $A|_R,A|_N$ would not be a square matrix, or am I all wrong? Is it then even possible to explicitly determine $A|_R,A|_N$?
Text on Fittings theorem: https://planetmath.org/fittingslemma
I suppose that $V=\mathbb R^3$ instead of $\mathbb R$, otherwise the vector space is one-dimensional and $A$ is represented not by a $3\times3$ matrix but by a $1\times1$ matrix.
In general, suppose $m$ is the smallest positive integer such that $\ker(A^m)=\ker(A^{m+1})$. Then $N=\ker(A^m)$ and $R=\operatorname{range}(A^m)$. If you don't want to calculate $m$, it is also guaranteed that $N=\ker(A^n)$ and $R=\operatorname{range}(A^n)$ because $m\le n$. In your case, $m$ is obviously equal to $1$. Therefore $N=\ker(A)=\operatorname{span}\{(1,-1,1)^T\}$ and $R=\operatorname{range}(A)=\{(x,y,0)^T:x,y\in\mathbb R\}$. It follows that $A|_N=0$ and $A|_R$ is represented by the matrix $\pmatrix{1&1\\ 0&1}$ with respect to the ordered basis $\{(1,0,0)^T,(0,1,0)^T\}$.