A linear operator $T$ on a complex vector space $V$ has characteristic polynomial $x^3(x-5)^2$ and minimal polynomial $x^3(x-5)$. Choose correct options.
The operator induced by $T$ on quotient space $V/\ker(T-5I)$ is nilpotent. Where $I$ is the identity operator.
The operator induced by $T$ on quotient space $V/\ker(T-5I)$ is a scalar multiple of the identity operator.
My attempt:
From given information, we get $$T = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5 \\ \end{pmatrix}$$
Now $$T-5I = \begin{pmatrix} -5 & 1 & 0 & 0 & 0 \\ 0 & -5 & 1 & 0 & 0 \\ 0 & 0 & -5 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix}$$. Here we see that $\operatorname{Rank}(T-5I) = 3$ and by Rank-nullity theorem, $dim\ker(T-5I) = 2$
I have no more knowledge how to proceed it further. Please help me.
You are correct that the Jordan normal form theorem implies that there's a basis $e_1,\dots,e_5$ of $V$ such that the matrix of $T$ in this basis is what you gave.
Then, especially after calculating the dimension of the kernel, you can read off that specfically $K:=\ker(T-5I)={\rm span}(e_4,e_5)$.
So, the cosets $e_1+K,\ e_2+K,\ e_3+K$ form a basis of the quotient space $V/K$, and $T$ acts on them as in the first $3\times 3$ block of the matrix of $T$: $$T(e_1+K)=0+K,\quad T(e_2+K)=e_1+K,\quad T(e_3+K)=e_2+K\,.$$