Prove diagonalizability of operator $T$ I got homework to prove some question and after almost 5 hours I gave up.
The questions are:
1) operator $T : \Bbb R^n\to \Bbb R^n$, prove that $\operatorname{Im}(T)∩ \operatorname{Ker}(T)={0}.$
2) prove or disprove : Linear operator $T : \Bbb R^n \to\Bbb R^n $, $T$ is diagonalizable if and only if $\operatorname{Im}(T)∩ \operatorname{Ker}(T)=\{0\}$.
3) $T : \Bbb R^n \to \Bbb R^n$ linear operator with all eigenvalues $=0 $, prove that $T$ is diagonalizable only if $T$ is the zero operator.
Thanks 
 A: *

*If $T$ has a basis of eigenvalues, then either $0$ is an eigenvalue or not. If it isn't, then $\operatorname{ker} T = \{ 0 \}$, and the intersection result is trivial. Otherwise, we may express the basis of eigenvectors in the form
$$B = (u_1, \ldots, u_m, v_1, \ldots, v_n)$$
where $u_1, \ldots, u_m$ are eigenvectors corresponding to $0$, and $v_i$ is an eigenvector corresponding to $\lambda_i \neq 0$ for all $i = 1, \ldots, n$. I claim that
$$\operatorname{Im} T = \operatorname{span} (v_1, \ldots, v_n).$$
To see containment in the $\subseteq$ direction, consider the image of an arbitrary element of the span of $B$. To see containment in the $\supseteq$ direction, show that $v_i = T(\lambda_i^{-1} v_i) \in \operatorname{Im} T$ for all $i$. Then,
$$\operatorname{Ker} T \cap \operatorname{Im} T = \operatorname{span} (u_1, \ldots, u_m) \cap \operatorname{span} (v_1, \ldots, v_n) = \{ 0 \},$$
by linear independence of $B$.

*Just take an invertible but not diagonalisable operator (e.g. a nilpotent operator with the identity added), and $\operatorname{Ker} T = \{ 0 \}$.

*If $0$ is the only eigenvalue of $T$, and there is a full basis of eigenvectors corresponding to $0$, then what does $T$ do to an arbitrary linear combination of these eigenvectors?
A: *

*This isn't true. Consider $T= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. We have $\ker  T = \operatorname{Im} T = \operatorname{span}\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}\right\}$.

*The direction "$T$ diagonalizable $\implies \ker T \cap \operatorname{Im} T = \{0\}$" is true, the other direction isn't.
Indeed, let $\{b_1, \ldots, b_n\}$ be a basis of eigenvectors for $T$, and take $y \in \ker T \cap \operatorname{Im} T$. There exists $x = \sum_{i=1}^n \alpha_i b_i$ such that $y = Tx = \sum_{i=1}^n \alpha_i \lambda_ib_i$.
We have
$$0 = Ty = T^2x = \sum_{i=1}^n \alpha_i \lambda^2 b_i$$
so $\alpha_i\lambda_i^2 =0$ for $i=1, \ldots, n$. Hence if $\alpha_i\ne 0$, we get $\lambda_i = 0$ so $y = \sum_{i=1}^n \alpha_i \lambda_ib_i = 0$.
For the other direction consider $$T = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$$
$T$ is not diagonalizable but $\ker T = \operatorname{span}\left\{\begin{bmatrix} 1 \\ 0  \\ 0\end{bmatrix}\right\}$ and $\operatorname{Im} T = \operatorname{span}\left\{\begin{bmatrix} 0 \\ 1  \\ 0\end{bmatrix}, \begin{bmatrix} 0 \\ 0  \\ 1\end{bmatrix}\right\}$ have trivial intersection.

*If $T = 0$ then $T$ is clearly diagonalizable. Conversely, if $T$ is diagonalizable with all zero eigenvalues, then $T$ diagonalizes to the matrix $0$, so $T = 0$.
