every eigenvalue of $T$ has only one corresponding eigenvector up to a scalar multiplication For a linear transformation $T$ on a vector space $V$ of dimension $n .$ Suppose it is given that for some vector $\mathbf{v},$ the vectors $\mathbf{v}, T(\mathbf{v}), T^{2}(\mathbf{v}), \ldots, T^{n-1}(\mathbf{v})$ are linearly independent,
then is it true that every eigenvalue of $T$ has only one corresponding eigenvector up to a scalar multiplication.
 A: The transformation $T$ satisfies some polynomial of degree at most $n$, so $$T^nv=a_0v+a_1Tv\cdots+a_{n-1}T^{n-1}v$$
Now suppose $w$ is an eigenvector with eigenvalue $\lambda$. Since $v,\ldots, T^{n-1}v$ is a basis, $$w=b_0v+\cdots+b_{n-1}T^{n-1}v$$
Substituting this in $Tw=\lambda w$,
\begin{align}
T(b_0v+\cdots+b_{n-1}T^{n-1}v)&=\lambda(b_0v+\cdots+b_{n-1}T^{n-1}v\\
b_0Tv+\cdots+b_{n-1}T^nv&=\lambda b_0v+\cdots+\lambda b_{n-1}T^{n-1}v
\end{align}
So comparing the coefficients of $T^iv$, $$b_{n-1}a_0=\lambda b_0,\quad b_0+b_{n-1}a_1=\lambda b_1,\quad \ldots, b_{n-2}+b_{n-1}a_{n-1}=\lambda b_{n-1}$$
Solving, \begin{align}
b_{n-2}&=(\lambda-a_{n-1})b_{n-1}\\
b_{n-3}&=\lambda b_{n-2}-a_{n-2}b_{n-1}=(\lambda^2-a_{n-1}\lambda-a_{n-2})b_{n-1}\\
\vdots\\
b_0&=\lambda b_1-a_1b_{n-1}=f(\lambda,a_i)b_{n-1}
\end{align}
Hence all coefficients $b_i$ are unique multiples of $b_{n-1}$ and thus unique up to a multiplicative constant.
A: Look at the basis $b_1 = v, b_2 = T(v), \ldots, b_n = T^{n-1}(v)$. In this basis the matrix of the transformation $T$ is
$$A = \pmatrix{
0 & 0 & \ldots & p_1\\
1 & 0 & \ldots & p_2\\
\vdots\\
\ldots & & 0 & p_{n-1}\\
0 & \ldots & 1 & p_n}
$$
Now note that the characteristic polynomial has degree $n$,
but there is no smaller-degree polynomial that $A$ satisfies, for if there were constants, not all zero, with
$$
\sum_{k=0}^{n-1} c_k A^k  = 0
$$
then we have
$$
\sum^{n-1} c_k A^k{e_1} = 0,
$$
and $A^k(e_1) = e_{k}$, so we have
$$
\sum^{n-1} c_k e_k = 0
$$
which is impossible because the $e_1$ are independent (here the $e_i$ are the standard basis vectors for $\Bbb R^n$ or $\Bbb C^n$).
Now let's look at the Jordan normal form of $T$ (or $A$). Suppose $T$ had two eigenvectors $v_1, v_2$ for the same eigenvalue, $s$. Then we could (using these two as the start of a Jordan basis) write the Jordan normal form of $T$ as
$$
\pmatrix{
s & 0 & \ldots & 0\\
0 & s & \ldots & 0 \\
\vdots}
$$
From this it's clear that the characteristic polynomial has the form
$$
p(x) = (x-s)^2 q(x)
$$
where $q$ is $\det(H - xI)$, where $H$ is the lower-right $(n-2) \times (n-2)$ matrix. But then it's also evident that $T$ satisfies the polynomial
$$
r(x) = (x-s) q(x)
$$
which has degree less than $n$, and that's a contradiction.
I'm pretty certain that this is far more roundabout than it needs to be, but tit gets there in the end.
A: Write the matrix for $T$ relative to $B = (v, Tv, ..., T^{n-1} v)$ to get
$$A = \begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & u_1 \\ 1 & 0 & 0 & \cdots & 0 & u_2 \\ 0 & 1 & 0 & \cdots & 0 & u_3 \\ \vdots & & \ddots & & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 & u_{n-1} \\ 0 & 0 & 0 & \cdots & 1 & u_n \end{bmatrix}$$
where we define $(u_1,...,u_n)^T = [T^n v]_B$.
Now consider $\lambda I - A$ to get
$$\begin{bmatrix} \lambda & 0 & 0 & \cdots & 0 & -u_1 \\ -1 & \lambda & 0 & \cdots & 0 & -u_2 \\ 0 & -1 & \lambda & \cdots & 0 & -u_3 \\ \vdots & & \ddots & & \vdots & \vdots \\ 0 & 0 & \cdots & -1 & \lambda & -u_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & \lambda-u_n \end{bmatrix}$$
We claim this matrix has rank at least $n-1$. To see this, first add $\lambda$ times column $1$ to column $2$, then add $\lambda$ times column $2$ to column $3$ and so on to obtain
$$\begin{bmatrix} \lambda & \lambda^2 & \lambda^3 & \cdots & \lambda^{n-1} & -u_1 \\ -1 & 0 & 0 & \cdots & 0 & -u_2 \\ 0 & -1 & 0 & \cdots & 0 & -u_3 \\ \vdots & & \ddots & & \vdots & \vdots \\ 0 & 0 & \cdots & -1 & 0 & -u_{n-1} \\ 0 & 0 & 0 & \cdots & -1 & \lambda-u_n \end{bmatrix}$$
where we see the first $n-1$ columns are clearly linearly independent.
This means $\dim \ker (\lambda I - A) \le 1$ so we are done.
