Minimal and Characteristic Polynomial 
Find the minimal and characteristic polynomials
  of
  $$A = \begin{bmatrix}λ & 1&0&...&0\\0 & λ&1&...&0\\...&...&λ&...&1\\0&...&...&...&λ\end{bmatrix}$$

The characteristic polynomial is $(\lambda-x)^n=0$.
But how to get the minimal polynomial? The minimal polynomial should be the monic polynomial $\phi$ of least degree satisfying $\phi(A)=0$.
 A: Hint: Check matrix $A-\lambda {I}$ for nilpotency.
A: Hint The minimal polynomial divides the characteristic polynomial. In this case the latter is $c_A(x) = (x - \lambda)^n$, so the former must have the form $m_A(x) = (x - \lambda)^k$ for some $k \leq n$. Before attempting a proof for general $n$, it might be useful to work out the case $n = 3$ to see better what's going on.
A: The characteristic polynomial and the minimal polynomial are equal.
Set $0 \leq k \leq n-1. $
$(1)$ Note that the column $A^ke_1$, where $e_1$ is the first vector of the standard base, has a non-zero entry in it's $k-1$ row and zeroes above it. Can be proved by induction.
$(2)$ $A^ke_1$ is not a linear combination of $A^0e_1, Ae_1,\dots,A^{k-1}e_1$.
$(3)$ $m_A|c_A \implies \deg m_A \leq \deg c_A.$
Let $m_A(X) = \sum_{i=0}^da_iX^i$, where $a_d=1$.
So $m_A(A) = \sum_{i=0}^da_iA^i = 0$, hence $\sum_{i=0}^da_iA^ie_1 = 0$
Rearranging: $\sum_{i=0}^{d-1}-a_iA^ie_1 = A^de_1$
By $(2)$, it can't be that $d\leq n-1$, hence $\deg m_A=d \geq n.$ 
$c_A$ and $m_A$ are monic polynomials, so $\deg m_A=n= \deg c_A$
$$\implies m_A=c_A$$
