We consider a vector space \(V\) of dimension \(2\) over a field \(\mathbb{K}\). The matrix:
\[A=\left( \begin{array}{cc}
0 & 1 \\
0 & 0 \end{array} \right)\] has several wonderful properties!

Only zero as eigenvalue, but minimal polynomial of degree \(2\)

Zero is the only eigenvalue. The corresponding characteristic space is \(\mathbb{K} . e_1\) where \((e_1,e_2)\) is the standard basis. The minimal polynomial of \(A\) is \(\mu_A(X)=X^2\).

\(A\) is not diagonalizable

As zero is the only eigenvalue of \(A\), if \(A\) was diagonalizable, \(A\) would be similar to the null matrix. That cannot be as the null matrix commutes with all matrices.

\(A\) is not divisible

Suppose that \(A=B^n\) where \(n \ge 2\) is an integer and \(B\) is a 2×2 matrix. As \(A^2=0\) we have \(B^{2n}=0\). Hence \(B\) minimal polynomial \(\mu_B\) divides \(X^{2n}\). The degree of \(\mu_B\) is less or equal to \(2\). Therefore \(\mu_B=X\) or \(\mu_B=X^2\). \(\mu_B=X\) is impossible as \(B\) cannot be the null matrix. If \(\mu_B=X^2\), \(B\) would be similar to \(A\). For the proof, consider a vector \(u\) such that \(B.u \neq 0\). \((B.u,u)\) is a base in which the matrix of \(B\) is \(A\). This leads to the contradiction \(0=B^2 \cdot B^{n-2}=A \neq 0\). Finally, \(A\) is not divisible.

\(A\) is nilpotent but not null

We have \(A^2=0\) while \(A \neq 0\). Therefore \(A\) is nilpotent.

\(A\) can be used to prove non commutativity of matrices

Indeed we have:
\[\left( \begin{array}{cc}
0 & 1 \\
0 & 0 \end{array} \right)=
\left( \begin{array}{cc}
1 & 1 \\
0 & 0 \end{array} \right)A \neq A
\left( \begin{array}{cc}
1 & 1 \\
0 & 0 \end{array} \right)=
\left( \begin{array}{cc}
0 & 0 \\
0 & 0 \end{array} \right)
\]