Minimal Polynomial and Jordan Basis Claim: Assume $A:V\rightarrow V$ is an endomorphism with $\dim V=d$. The minimal polynomial of this linear transformation is $m(t)=(t-\lambda)^d$. Choose $v$ such that $(A-\lambda)^{d-1} v\neq 0$. Then, $V$ has basis $B=\{v, (A-\lambda)v, \cdots, (A-\lambda)^{d-1} v\}$ such that the representation of $A$ on this basis is in Jordan Forms, e.g.:
\begin{pmatrix}
\lambda & 1 &0\\
0 & \lambda &1\\
0 & 0 & \lambda
\end{pmatrix}
I tried to prove this statement by induction. When $d=1$, we get $v$ as the only element in basis $B$, and the corresponding Jordan Form will be a single block consisting of $\lambda$. However, I am not sure how to proceed later. Also, I do not know why the representation of $A$ on this basis is in the Jordan Form when $d\geq 2$.
Any help will be appreciated.

Let $v$ with $(A-\lambda )^{d-1}v\ne 0$. Denote
$$
v_i := (A-\lambda)^i v, \quad i=0\dots d-1.
$$
Then it holds for $i=0\dots d-2$
$$
A v_i = (A-\lambda) v_i + \lambda v_i = \lambda v_i + v_{i+1}.
$$
Hence the $i$-th row of the matrix representation contains $\lambda$ in the $i$-th columns and $1$ in column $i+1$. Pretty much the matrix you asked about.
For $i=d-1$, we have $Av_{d-1} = (A-\lambda)v_{d-1} + \lambda v_{d-1}=\lambda v_{d-1}$, hence the $\lambda$ entry in the lower-right corner.
