Stability by $T$ of a sequence defined by $\ker T^r$ if spectrum(T)=$\{0,\lambda\}$ Suppose that the only eigenvalues of T are 0 and λ, where λ $\neq$ 0.
Let $W = T^r(V )$, r satisfies 
ker $T^r$=ker $T^{r+1}$. 
Show that $T(W) ⊆ W$, and that the restriction of T to W has $λ$ as its only eigenvalue. Let S denote the restriction of $(T − λI)$ to W. Show that 0 is the only eigenvalue of S.
I can show $T(W) ⊆ W$, but how to show the two only eigenvalues? Can anyone give a hint?
 A: In a Chinese mathematical contest, a problem is given:
If $A \in M_n(F)$, where $F$ is a field (no need to be an algebraic close field), then $A$ is similar to $\left( 
\begin{array}
{lcr} 
A & 0 \\ 
0 & B \\ 
 \end{array}
\right) $with $A$ invertible and $B$ nilpotent.
Since $F$ is a field,the finite dimensional vector space $F^{(n)}$ is both Noetherian and Artinian. The ascending chain $Ker T \subseteq kerT^2 \subseteq kerT^3...$must terminate, and so does the descending chain $ImT \supseteq ImT^2 \supseteq ImT^3...$must terminate. Since $dim(ImT)+dim(kerT)=n$, they must terminate at same integer $r$. Therefore, we have $ImT^r=ImT^{r+1}=ImT^{r+2}$...and so on.
Now we prove $F^{(n)}=ImT^{r} \oplus KerT^{r}$. Since the sum of $dim{(ImT^r)}$ and $dim{(KerT^r)}$ is $n$, we only need to prove $M=ImT^{r} \cap KerT^r = \{0\}$. Since if $\alpha \in M$, $\alpha = T^r(\beta)$, $0=T^{r}(\alpha)=T^{2r}(\beta)$, so $\beta \in ker T^{2r}=ker T^r$ (ascending chain comes stable), $\alpha = T^n(\beta)=0$, $F^{(n)}=ImT^{r} \oplus KerT^{r}$.
If we pick a base $\{a_1,a_2,...,a_r\}$ in $ImT^{r}$, $r$ equals the dimension of $ImT^{r}$, the matrix on the basis is invertible since $T(ImT^r)=ImT^{r+1}=ImT^{r}$, we denote the matrix as $A$. Then, we pick a base $\{b_1,b_2,b_3,...,b_s\}$ in $KerT^{r}$, the matrix on the basis is nilpotent,we denote it as $B$. If we gather $\{a_i\}$ and $\{b_i\}$, the base of $ImT^r$ and $KerT^r$ together to form a new base $\{c_i\}$, the matrix on the $\{c_i\}$ corresponding to the linear transformation $T$ is the quasi-diagonal matrix
$\left( 
\begin{array}
{lcr} 
A & 0 \\ 
0 & B \\ 
 \end{array}
\right) $.
If we want to use the terminology "restriction", then $A$ is your desired matrix because $A$ is the matrix on a basis of $ImT^r$. $B$ is the restriction on $KerT^r$.
Actually It's Fitting's lemma(see https://en.wikipedia.org/wiki/Fitting_lemma).
Abstract form: $F^{(n)}=ImT^{r} \oplus KerT^{r}$
Concrete matrix form: If $A \in M_n(F)$, where $F$ is a field (no need to be algebraically closed), then $A$ is similar to $\left( 
\begin{array}
{lcr} 
A & 0 \\ 
0 & B \\ 
 \end{array}
\right) $with $A$ invertible and $B$ nilpotent.
Come back to your question. Since $T$ has eigenvalue $0, \lambda \neq 0$, so you see $A$ has eigenvalue $\lambda$ and $B$ has eigenvalue $0$ (Since $A$ is invertible and $B$ nilpotent). Consider the matrix
$\left(
\begin{array}
{lcr} 
\lambda I_r-A & 0 \\ 
0 & \lambda I_{n-r}-B \\ 
 \end{array}
\right) $ is similar to the matrix $\lambda I-T$, and we restrict $\lambda I-T$ on $ImT^r=W$ we get matrix $\lambda I_r-A$, with eigenvalue $0$. Then please finish it yourself!
