given is a matrix A with
$\begin{pmatrix} a & 1 & 0 & \cdots & 0 \\ 0 & a & 1 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0 & 0 & 0 & \cdots & a\end{pmatrix}$
The characteristic polynomial should by given by:
$P_A(t) = (a-t)^n$.
All Eigenvalues are $a$, so the algebraic multiplicities for each eigenvalue is $n$.
The geometrical multiplicity for each eigenvalue is then $1$, since for eigenvalue $a$ I have eigenvector $(1, 0, ..., 0)$.
Now I have to determine the minimal polynomial $m_A(t)$.
It should be $m_A(t) \in T := \{ (a-t), (a-t)^2, (a-t)^3, \cdots, (a-t)^n \}$.
At least $m_A((a-t)^n) = 0$ since $P_A(A) = (a-A)^n = 0$ right?
1) The matrix $(a\cdot E_n - A)^1$ has Rank $n-1$.
2) The matrix $(a \cdot E_n - A)^2$ has Rank $n-2$. and so on.
What I need is a matrix with rank $0 = n-n$.
This means
3) The matrix $(a \cdot E_n - A)^n = 0$. So the minimal polynomial is the characteristic polynomial.
How can I say - in a shorter and more mathematical way than I did - that with each new factor, the matrix-product $(a \cdot E_n - A)^i$ only lowers its rank down 1 number?
We have
$$B:=A-aE_n=\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \cdots & \cdots & \cdots & \cdots & 1 \\ 0 & 0 & 0 &\cdots & 0\end{pmatrix}$$ and let $\left\{v_k\right\}=\left\{(0,\ldots,0,1,0,\ldots,0)^T\right\}$ the standard basis then we see that $$Bv_k=v_{k-1}$$ hence by induction $$B^{n-1}v_n=v_1\ne0$$ hence $$B^{n-1}\ne0$$ and then the minimal polynomial of $A$ is its characteristic polynomial.