every eigenvalue of $T$ has only one corresponding eigenvector up to a scalar multiplication

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.