Problem with showing properties of linear transformation Let $F:K^3\rightarrow K^3$ be a linear transformation, such that

*

*$(F-\lambda I)(F-\mu I)\neq0,$

*$(F-\mu I)^2\neq 0,$

*$(F-\lambda I)(F-\mu I)^2=0.$
With this knowledge I want to show that

*

*$\lambda$, $\mu$ are eigenvalues of $F$,

*$Ker (F-\mu I)\subseteq Im(F-\mu I)$,

*$F$ has in a certain basis matrix
\begin{equation*}
\begin{pmatrix}
\lambda & 0 & 0  \\
0 & \mu & 1  \\
0  & 0 &\mu
\end{pmatrix}.
\end{equation*}
My attempt:

*

*It seems to be a consequence of the third assumption, but I am not sure.

*We know that $Ker(F-\mu I)=\{v\in K^3:F(v)-\mu v=0\}$ and $Im(F-\mu I)=\{(F-\mu I)(v):v\in K^3\}=\{(F(v)-\mu v):v\in K^3\}$. Then the inclusion seems to be trivial, but I am also not sure.

*Unfortunately I have no idea how to begin.

 A: I will assume $\lambda\neq \mu$.
Note that the factors $F-k I$ commute for all $k$.

*

*Let $v$ be a vector $v\in K^3$ such that $(F-\lambda I)(F-\mu I)\,v=u\neq 0$. Then:
$$
(F-\mu I)\,u=(F-\lambda I)(F-\mu I)^2\,v=0
$$
so $u$ is an eigenvector for $\mu$.

Analogously, let $w$ be a vector $w\in K^3$ such that $(F-\mu I)^2\,w=z\neq 0$. Then:
$$
(F-\lambda I)\,z=(F-\lambda I)(F-\mu I)^2\,w=0
$$
so $z$ is an eigenvector for $\lambda$.


*In part 3. we can see that the dimension of $Ker(F-\mu I)$ is 1, otherwise $F$ would be diagonalizable. So $u\in Ker(F-\mu I)$ from previous point is a basis of $Ker(F-\mu I)$. But $u=(F-\mu I)v'$ where $v'=(F-\lambda I)v$ thus $u\in Im(F-\mu I)$.

NOTE: I do not like this proof because it uses part 3. but right now I cannot think of something easier.


*The polynomial $(F-\lambda I)(F-\mu I)^2$ from statement 3 is the characteristic polynomial, because it has degree three, and it does not vanish when removing any of its factors. Thus $\mu$ has multiplicity 2 and the diagonal of the Jordan form of $F$ is $(\lambda,\mu,\mu)^\top$. But if $F$ were diagonalizable, then $(F-\lambda I)(F-\mu I)=0$, which is in contradiction with statement 1.

