linear algebra - 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 $\phi(A)=0$.

Answer

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 $\deg m_A=n= \deg c_A$

$$\implies m_A=c_A$$