If you thought about Exercise 11, think about it again, after you observe what Theorem 7 tells you about the diagonalizable and nilpotent parts of \(T\).
Solution
Exercise 11 decomposes an arbitrary upper-triangular matrix \(A\) as its literal diagonal part plus its strictly upper-triangular part. The problem is that these two matrices need not commute. The actual diagonalizable and nilpotent parts of \(T\) do commute, and that extra condition is what makes the decomposition canonical.
Let \[ T=D+N, \qquad DN=ND, \] be the diagonalizable–nilpotent decomposition, with \(D\) diagonalizable and \(N\) nilpotent [@HK71].
1. The commuting operators \(D\) and \(N\) can be triangularized simultaneously.
The operator \(D\) is diagonalizable, hence triangulable. Since \(N\) is nilpotent, its minimal polynomial divides a power of \(x\), so its characteristic polynomial splits and \(N\) is triangulable. Chapter 6, Theorem 7 states that a commuting family of triangulable operators on a finite-dimensional space is simultaneously triangulable [@HK71]. Thus there is a basis in which both \(D\) and \(N\), and consequently \(T=D+N\), are upper triangular.
In such a basis the diagonal of \(N\) is zero, because every diagonal entry of an upper-triangular representation is an eigenvalue and the only eigenvalue of a nilpotent operator is zero. Hence \(T\) and \(D\) have the same diagonal entries. This observation alone does not say that the matrix of \(D\) is literally diagonal: a diagonalizable operator can have a non-diagonal triangular matrix in a poorly chosen basis. That is exactly why the argument in Exercise 11 cannot simply take the diagonal entries of an arbitrary triangular matrix for \(T\).
2. One can choose a better common triangular basis in which \(D\) is actually diagonal and \(N\) is strictly upper triangular.
Write the eigenspace decomposition of the diagonalizable operator \(D\) as \[ V=V_{\lambda_1}\oplus\cdots\oplus V_{\lambda_r}. \] Because \(DN=ND\), every eigenspace is \(N\)-invariant: if \(Dv=\lambda_i v\), then \[ D(Nv)=N(Dv)=\lambda_iNv. \] Thus \(N_i=N|_{V_{\lambda_i}}\) is a nilpotent operator on \(V_{\lambda_i}\). Its minimal polynomial is a power of \(x\), so it is triangulable. Choose a basis of each \(V_{\lambda_i}\) in which \(N_i\) is upper triangular, and concatenate these bases.
In this concatenated basis, \(D\) acts as the scalar \(\lambda_i\) on every vector of the \(i\)th block, so its matrix is diagonal. Each block of \(N\) is upper triangular and nilpotent, hence has zero diagonal; therefore the whole matrix of \(N\) is strictly upper triangular. Consequently \[ [T]=[D]+[N] \] really is a diagonal matrix plus a commuting strictly upper-triangular nilpotent matrix in this specially adapted basis.
3. This identifies the precise correction to the proof in Exercise 11.
2. In that basisthe matrix split has the desired shape and represents the actualdiagonalizable–nilpotent decomposition.