Prove that any non-zero vector which is not a characteristic vector for $T$ is a cyclic vector for $T$. Let $T$ be a linear operator on $F^2$. Prove that any non-zero vector which is not a characteristic vector for $T$ is a cyclic vector for $T$. Hence, prove that either $T$ has a cyclic vector or $T$ is a scalar multiple of the identity operator.
I have no clue how to prove any of the statements. Can I get some help getting started?
 A: If $v \ne 0$ is not an eigenvector (or "characteristic vector") for $T$, then
$Tv \ne 0 = 0 \cdot v, \tag 1$
since $Tv = 0$ asserts that $v$ is an eigenvector corresponding to $0$; also, $v$ and $Tv$ are linearly independent over $F$ (since here $F$ is our base field); if not, we have $a, b \in F$, not both $0$, with
$av + bTv = 0; \tag 2$
now here $b \ne 0$, lest
$av = 0 \Longrightarrow a = 0, \tag 3$
since $v \ne 0$; but then $a = b = 0$, contrary to hypothesis; likewise $a \ne 0$, else
$bTv = 0 \Longrightarrow Tv = 0, \tag 4$
which violates (1); so
$a \ne 0 \ne b, \tag 5$
and thus we may write
$Tv = -\dfrac{a}{b} v; \tag 6$
that is, $v$ is an eigenvector with eigenvalue $-a/b$, again a contradiction; so $v$ and $Tv$ are linearly independent over $F$; thus, since 
$\dim_F F^2 = 2, \tag 7$
$F^2 = \text{span} \{ v, Tv \}; \tag 8$
that is, $v$ is cyclic for $T$.
Now suppose $T$ has no cyclic vector; then every $0 \ne v \in F^2$ is an eigenvector associated with some $\mu_v \in F$:
$Tv = \mu_v v, \; \mu_v \in F; \tag 9$
choosing $0 \ne w \in F^2$ linearly independent from $v$ we may write
$Tw = \mu_w w, \tag{10}$
whence
$T(v + w) = Tv + Tw = \mu_v v + \mu_w w; \tag{11}$
but also by assumption
$T(v + w) = \mu_{v + w}(v + w), \tag{12}$
and thus
$\mu_v v + \mu_w w = \mu_{v + w}(v + w) = \mu_{v + w}v + \mu_{v + w}w; \tag {13}$
it now follows from the linear independence of $v$ and $w$ that
$\mu_v = \mu_w = \mu_{v + w} = \mu \in F, \tag{14}$
and we conclude that
$Tv = \mu v = \mu I v, \; \forall v \in F; \tag{15}$
that is, $T = \mu I$ is a scalar multiple of the identity matrix $I$.
A: Hint: Consider two cases:


*

*$T$ has at least one eigenvector.

*$T$ has no eigenvector.

