Let \(T\) be the linear operator on \(\RR^4\) which is represented in the standard ordered basis by the matrix \[ \begin{bmatrix} 0 & 0 & 0 & 0 \\ a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0 \end{bmatrix}. \] Under what conditions on \(a\), \(b\), and \(c\) is \(T\) diagonalizable?
Solution
1. The matrix of \(T\) is strictly lower triangular, so its only eigenvalue is \(0\) (with algebraic multiplicity \(4\)).
the characteristic polynomial is \(x^4\) (the matrix is nilpotent).
2. \(T\) is diagonalizable iff it is similar to the zero matrix, i.e. iff \(T = 0\).
a diagonalizable matrix with a single eigenvalue \(\lambda\) equals \(\lambda I\); here \(\lambda = 0\), so diagonalizable iff \(T = 0\).
3. \(T = 0\) iff \(a = b = c = 0\).
the matrix is zero exactly when all its nonzero entries vanish.
4. Hence \(T\) is diagonalizable iff \(a = b = c = 0\).
2 and 3.
5. Q.E.D.
4.