Let \(V\) be \(6\)-dimensional and let \(T\in\operatorname{End}(V)\) satisfy \[ T^6=0, \qquad T^5\ne0. \] Show that \(T\) has a single Jordan block of size \(6\).
Solution
Since \(T^6=0\), the minimal polynomial is a power of \(x\): \[ m_T(x)=x^r \] for some \(r\le6\). The condition \(T^5\ne0\) forces \(r>5\), so \[ m_T(x)=x^6. \]
Because \(\dim V=6\), the characteristic polynomial has degree \(6\). Since the only eigenvalue is \(0\), \[ \chi_T(x)=x^6. \]
In Jordan form, the exponent of \(x\) in the minimal polynomial is the size of the largest Jordan block. Thus the largest block has size \(6\). But the sum of all Jordan-block sizes is also \(6\). Therefore there can be only one block: \[ J(T)=J_6(0) = \begin{pmatrix} 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1\\ 0&0&0&0&0&0 \end{pmatrix}. \]