Which of the following is/are true?
$(1)$ It is possible for a generalised eigen vector of a linear operator $T$ to correspond to a scalar that is not an eigen value of $T$.
$(2)$ Any linear operator on a finite dimensional vector space has a Jordan Canonical Form.
$(3)$ A cycle of generalised eigen vector is linearly Independent.
$(4)$ There is exactly one cycle of generalised eigen vectors corresponding to each eigen value of a linear operator on a finite dimensional vector space.
My attempt :-
For $(1)$
I think it is false
It is enough to prove that if $x$ is a generalised eigen vector of $T$ corresponding to scalar $c$, then there is an eigen vector corresponding to $c$ which would imply $c$ must be an eigen value of $T$ (Right ? please correct me if I am wrong)
Proof :-
Let $x$ be a generalised eigen vector of rank $m$ corresponding to scalar $c$. Then
$(T-cI)^mx=0$ and $(T-cI)^{m-1}x\neq 0$
Let $u=(T-cI)^{m-1}x $ .
Then $u\neq 0$ and $(T-cI)u=0$.
Thus the result follows.
To my surprise, $(1)$ is given True in the answer. Don't know what's happening !
$(2)$ False since the characteristics polynomial has to split into linear factors.
$(3)$ True.
Proof :- Let a cycle of generalised eigen vector be
$\{x , (T-cI)x, (T-cI)^2x,..., (T-cI)^{m-1}x\}$
Let $U=T-cI$ and
$c_0x+ c_1Ux + c_2U^2x+...+c_{m-1}U^{m-1}x=0$ for some scalars $c_i$
Operating both sides by $U^{m-1}$, we get
$c_0U^{m-1} x=0$ since all other vectors are zero.
This gives $c_0=0$ since $U^{m-1}x\neq 0$
Proceeding similarly, $c_i=0 , \forall i$
So the set is LI
$(4)$ I don't really understand this option but as far as the wikipedia (https://en.m.wikipedia.org/wiki/Generalized_eigenvector) article suggest there could be infinitely ways of choosing a generalised eigen vector of certain rank corresponding to an eigen value thus leading to different cycle of generalised eigen vector.
But again the answer is True.
Conclusion :- The answer given (true options) are $1,3,4$.
My main problem is in option $1$ and $4$.
Please help. Thanks for your time.