I have seen some answers for the question:
Q. A linear operator $T$ on a complex vector space $V$ has characteristic polynomial $x^3(x-5)^2$ and minimal polynomial $x^2(x-5)$. State whether the following propositions true or false.
(i) The operator induced by $T$ on the quotient space $V / \operatorname{Ker}\left(T-5 I\right)$ is nilpotent, where $I$ is the identity operator,
(ii) The operator induced by $T$ on the quotient space $V / \operatorname{Ker}(T)$ is a scalar multiple of the identity operator,
but none of them made me clear.
However, I got the points:
- $V=\operatorname{Ker}(T^2) \oplus \operatorname{Ker}(T-5I)$ by primary decomposition theorem,
- The quotient space $V / \operatorname{Ker}\left(T-5 I\right)$ is a 3-dimensional space as it is naturally isomorphic to $\operatorname{Ker}\left(T^2\right)$,
So, let me see it as a linear span of $\{e_1+\operatorname{Ker}\left(T-5 I\right), e_2+\operatorname{Ker}\left(T-5 I\right),e_3+\operatorname{Ker}\left(T-5 I\right)\},$ where I feel fair to choose $e_1,e_2,e_3$ from the ordered basis $\mathscr B=\{e_1,e_2,e_3,e_4,e_5\}$ associated with the Jordan canonical matrix (representation) $$J_T=\begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5 \\ \end{pmatrix},$$
Here, I couldn't capture the claim ''the operator $\bar T$ induced by $T$ on the quotient space $V / \operatorname{Ker}\left(T-5 I\right)$ is identical to the block $[\bar T]=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}''$, even though, I feel $\bar T(e_i+ \operatorname{Ker}\left(T-5 I\right))=Te_i+ \operatorname{Ker}\left(T-5 I\right),~i=1,2,3$, seemingly addressing the first three columns of $J_T$.
I would be grateful to you if you have a clear cut explanation for both of the propositions.
Thanks in advance.
