Show that a nonzero nilpotent matrix \(A\) is not diagonalizable over any field. Some useful facts:
\(\spec A = \ts{0}\): if \(Av=\lambda v\) and \(A^N=0\), then \(0=A^Nv=\lambda^N v\), so \(\lambda=0\). Thus every Jordan block is nilpotent.
If \(r\) is the nilpotency index of \(A\), then \(\min_A(x)=x^r\).
If \(A\) were diagonalizable, its diagonal form would be \(0\).
Solution
1. The only eigenvalue of a nilpotent matrix is \(0\).
Choose \(N\ge1\) with \(A^N=0\). If \(Av=\lambda v\) for some nonzero vector \(v\), then \[ 0=A^Nv=\lambda^N v. \] Since \(v\neq0\) and the scalars form a field, \(\lambda^N=0\), hence \(\lambda=0\).
2. If a nilpotent matrix \(A\) is diagonalizable, then \(A=0\).
If \(A\) is diagonalizable, there is an invertible matrix \(P\) and a diagonal matrix \(D\) such that \[ A=PDP^{-1}. \] The diagonal entries of \(D\) are the eigenvalues of \(A\). By 1 every eigenvalue is \(0\), so \(D=0\). Therefore \[ A=P0P^{-1}=0. \]
3. Hence every nonzero nilpotent matrix is not diagonalizable.
If a nonzero nilpotent matrix were diagonalizable, 2 would force it to equal \(0\), a contradiction.
4. Equivalently, if \(r\) is the nilpotency index of a nonzero nilpotent matrix, then its minimal polynomial is \(x^r\) with \(r\ge2\), so it cannot have the distinct linear factors required for diagonalizability.
By definition of the nilpotency index, \(A^r=0\) and \(A^{r-1}\neq0\), so the monic polynomial of least degree annihilating \(A\) is \(x^r\). Since \(A\neq0\), one has \(r\ge2\), and \(x^r\) has a repeated root. A matrix is diagonalizable over a field containing its eigenvalues exactly when its minimal polynomial splits into distinct linear factors; here the only factor is \(x\), repeated \(r\) times.