show $V = C_x$ iff there are no multiple eigenvalues $L: V\to V$ a diagonalizable linear operator on finite-dim vector space.

show that $V = C_x$ iff there are no multiple eigenvalues



here $C_x = \operatorname{span} \{x, L(x), L^2(x), \cdots\}$

basically it is a cyclic subspace generated by x that belongs to V.



my thought:

since L is diagonalizable Lx = ux for x is eigenvector and u is evalue

=> if V = C_x

V = C_x = span {x, L(x), L^2(x),.....}

= span {x, ux, (ux)^2,.....}

how does this show no multiple eigenvalues

&lt;= ???
You first need to be clear about what the mystery $x$ is; if it is the zero vector for instance, you'll have a hard time getting $C_x$ to be $V$, no matter how simple the eigenvalues are. I'll assume you omitted a quantifier: there exists a vector $x$ with $C_x=V$ if and only if the characteristic polynomial has no multiple roots (which is a better thing to say than that there are no multiple eigenvalues, since different notion of multiplicity exist for them).

Now since $L$ is diagonalisable, you may suppose $x$ is expressed in a basis of eigenvectors for $L$. Then $L$, and consequently any polynomial in $L$, acts independently on each of the coordinates of$~x$. One thing that is useful, is that the dimension of $C_x$ equals the degree$~d$ of the minimal degree monic polynomial$~P$ such that $P[L](x)=0$: that equation allows $L^d(x)$ to be expressed as linear combination of $x,L(x),\ldots,L^{d-1}(x)$, and once this happens you can see by induction that the same is true for $L^i(x)$ whenever $i\geq d$. So you must show that that $d=n$ is obtained for some vector$~x$ if and only if the eigenvalues associated to each of the coordinates are all distinct.

For a given nonzero coordinate of $x$, whose position has an associated eigenvalue$~\lambda$, check that $P[L]$ makes that coordinate zero if and only if $\lambda$ is a root of$~P$. So in order to have $P[L](x)=0$, it is necessary and sufficient that $P[\lambda]=0$ whenever $\lambda$ is the eigenvalue associated to (the position of) a nonzero coordinate of$~x$.

Now I trust that you can show that the minimum degree a nonzero$~P$ must have in order to satisfy this is equal to the number of different such eigenvalues$~\lambda$. Then you can complete the proof.
Let $x=\sum_{i=1}^n α_i e_i$, where $e_i$ are the eigenvectors of $L$ and $α_i$ non-zero constants. Then

$$L^kx=\sum_{i=1}^nλ_i^kα_ie_i$$

and now you can argue about the representability of any other vector $y=\sum_{i=1}^n \beta_ie_i$ by considering the Vandermonde matrix of the eigenvalues in the system
$$
y=\sum_{i=0}^{n-1} c_iL^i(x)\iff 
\pmatrix{β_1\\\vdots\\β_n}=
\pmatrix{α_1&amp;&amp;0\\&amp;\ddots\\0&amp;&amp;α_n}
\pmatrix{1&amp;λ_1&amp;\dots&amp;λ_1^{n-1}\\\vdots&amp;&amp;\vdots\\1&amp;λ_n&amp;\dots&amp;λ_n^{n-1}}
\pmatrix{c_0\\\vdots\\c_{n-1}}
$$