On a finite-dimensional vector space $V$, when does $V = \operatorname{Ker}(A) \bigoplus \operatorname{R}(A)$? This was a question given on a take-home exam that I have turned in. The deadline has passed, and I want to see if I gave a sufficient argument.
The question is posed as follows:

Is it true that $V = \operatorname{Ker}(A) \bigoplus \operatorname{R}(A)$? Give a condition in terms of the eigenvalues of $A$ under which the claim holds.

Here, $V$ is assumed to be finite-dimensional, and $A$ is a square matrix that is the representation of some linear operator.
My solution is summarized as follows: If $0$ were not an eigenvalue of $A$, the claim holds, as the kernel of $A$ is zero-dimensional. The columns of $A$ then span $V$ by the Invertible Matrix Theorem, and any $v \in V$ can be expressed as a unique linear combination of the columns of $A$.
Otherwise, suppose the kernel of $A$ is $p$-dimensional. Choose a basis for this kernel, and note that the eigenvectors corresponding to the eigenvalue $0$ are precisely those vectors in the basis of $\operatorname{Ker}(A)$. Now construct a basis for each eigenspace $E_{\lambda_i}$. These build a collection of invariant subspaces under $A$ and form the range. So, consider specifically the basis of the range
$$ \operatorname{R}(A) = \operatorname{span} \{ v_{r_1, 1}, v_{r_1, 2}, \ldots , v_{r_{m_i}, m_i} \} $$
where each eigenspace has dimension $m_i$, and the sum of these dimensions is $n - p$ (by Rank-Nullity). Now consider the linear combination
$$ 0 = \sum_i c_{r_i, i} v_{r_i, i} + \sum_j d_{k_j, j} v_{k_j, j} $$
where $r_i$ indexes those vectors in the range and $k_j$ those in the kernel. Now apply $A$ on both sides. Since the second sum is in the kernel, its image is $0$. The first sum contains vectors that form a basis for the range, so each $c_{r_i, i} = 0$. Finally, the second sum forms a basis of the kernel, so each $c_{k_j, j} = 0$. We conclude $0$ has a unique representation, and $V = \operatorname{Ker}(A) \bigoplus \operatorname{R}(A)$.
While I did rush to get this in, I think the condition can be summarized as a TL;DR: Each eigenspace needs enough eigenvectors.
Are there flaws in my solution? Is there a condition nicer than "having enough eigenvectors?"
 A: Consider a Nilpotent matrix
$$A = 
\begin{pmatrix}
0 & 1 &0 \\
0 & 0 &1  \\
0 & 0 & 0
\end{pmatrix}$$
Then $ker(A)=\langle e_{1}\rangle$ ,while $R(A)=\langle e_{1};e_{2}\rangle$ So the decomposition mentionned is not true in general.
In fact  $V=Ker(A)\oplus R(A)$ if and only if $R(A)=R(A^{2})$ which is equivalent to $Ker(A)=Ker(A^{2})$ if and only if $Ker(A)\cap R(A)=\{0\}$ .suppose that we have the decomposition,then $A$ can be written in bloc matrix form as follows:
$$A = 
\begin{pmatrix}
B & 0\\
0 & C
\end{pmatrix}$$
Where $B$ is the zero matrix $B:Ker(A)\to \ker(A)$ and $C:R(A)\to R(A)$ is invertible .The characterestic polynomial of $A$,denoted by $P_{A}$ equals $P_{B}\times P_{C}=x^{\text{dim} Ker(A)}\times P_{C}$ , where of course $P_{C}$ has non zero constant term ($P_{C}(0)\neq 0$)from this decomposition, u can see that a necessary condition for the existence of the decomposition is that  the algebraic and geometric multplicities of the eigenvalue 0 are equal.Conversely u can show that this condition is also sufficient.
