Minimal polynomial/Diagonal matrix/Different entries/Example

For a diagonal matrix

${\displaystyle {}M={\begin{pmatrix}d_{1}&0&\cdots &0\\0&d_{2}&\ddots &\vdots \\\vdots &\ddots &\ddots &0\\0&\cdots &0&d_{n}\end{pmatrix}}\,}$

with different entries ${\displaystyle {}d_{i}}$, the minimal polynomial is

${\displaystyle {}P=(X-d_{1})(X-d_{2})\cdots (X-d_{n})\,.}$

This polynomial is sent under the substitution to

${\displaystyle (M-d_{1}E_{n})\circ (M-d_{2}E_{n})\circ (M-d_{n}E_{n}).}$

We apply this to a standard vector ${\displaystyle {}e_{i}}$. Then factor ${\displaystyle {}(M-d_{j}E_{n})}$ sends ${\displaystyle {}e_{i}}$to ${\displaystyle {}(d_{i}-d_{j})e_{i}}$. Therefore, the ${\displaystyle {}i}$-th factor ensures that ${\displaystyle {}e_{i}}$ is annihilated. Since a basis is mapped under ${\displaystyle {}P(M)}$ to ${\displaystyle {}0}$, it must be the zero mapping.

Assume now that ${\displaystyle {}P}$ is not the minimal polynomial ${\displaystyle {}\mu }$. Then there exists, due to fact, a polynomial ${\displaystyle {}Q}$ with

${\displaystyle {}P=Q\mu \,,}$

and, because of fact, ${\displaystyle {}\mu }$ is a partial product of the linear factors of ${\displaystyle {}P}$. But as soon as one factor of ${\displaystyle {}P}$ is removed, say we remove ${\displaystyle {}X-d_{i}}$, then ${\displaystyle {}e_{i}}$ is not annihilated by the corresponding mapping.