Prove that $col(A^n) + nul(A^n) = \mathbb{R}^n$ Consider a linear transformation such that $A: \mathbb{R}^n \rightarrow \mathbb{R}^n$, is it possible to prove the following identity?
\begin{align}
col(A^n) + nul(A^n) = \mathbb{R}^n
\end{align}
I have been struggling on this problem for a week and I appreciate if anyone can point a direction or provide a proof for this.
Thank you very much!
 A: I'm going to prove something slightly more general. You might apply the following theorem to $K=\mathbb{R}$, $E=\mathbb{R}^n$ and $u$ the endomorphism of $\mathbb{R}^n$ canonically associated to $A$ to get $\ker(A^n)\oplus\mathrm{im}(A^n)=\mathbb{R}^n$.
Theorem. Let $K$ be a field, let $E$ be a $K$-vector space of dimension $n\geq 1$, and let $u\in \mathscr{L}(E)$ be an endomorphism of $E$. Then $$\ker(u^n)\oplus \mathrm{im}(u^n)=E.$$
Proof.
Let $P\in K[X]$ be the minimal polynomial of $u$, and write $P=X^m Q,$ with $m\geq 0$ and $Q(0)\neq 0$.
Set $N=\ker(u^m)$ and $C=\ker(Q(u))$. Then $N,C$ are stable by $u$, and $$E=N\oplus C,$$ since $X^m$ and $Q$ are coprime polynomials.
Since $N$ and $C$ are stable by $u$, we get by double restriction an endomorphism $u_N$ of $N$, and an endomorphism $u_C$ of $C.$
Fact. $u_N$ is nilpotent and $u_C$ is an automorphism of $C$.
Indeed, $X^m$ annihilates $u_N$ by definition, so $u_N$ is nilpotent
Since $Q(0)\neq 0$, we may write $Q=XR+a,$ with $a\neq 0$. Since $Q$ annihilates $u_C$, we get $$0=u_C\circ R(u_C)+aId_C,$$ that is $$ u_C\circ (-a^{-1}R(u_C))=Id_C.$$ Hence $u_C$ is invertible.
By definition of $m$, we have $u_N^m=0$. Since $m\leq n$, we get $u_N^n=0$.
But for all $x\in N$ and $y\in C$ , we have
$$u^n(x+y)=u^n(x)+u^n(y)=u_N^n(x)+u_C^n(y)=u_C^n(y).$$
Since $u_C$ is invertible, so is $u_C^n$. Hence we get $$u^n(x+y)=0\iff u_C^n(y)=0\iff y=0,$$ meaning that $\ker(u^n)=N.$
Moreover, $\mathrm{im}(u^n)=\mathrm{im}(u_C^n)=C,$ since $u_c^n$ is invertible. Consequently, $$E=N\oplus C=\ker(u^n)\oplus\mathrm{im}(u^n).$$
PS. Modifying slightly the proof, one can in fact show that if $X^m$ is the greastest power of $X$ dividing the minimal polynomial of $u$, then for all $k\geq m$, $E=\ker(u^k)\oplus\mathrm{im}(u^k)$.
