What’s wrong with the following proof of Theorem 13? Suppose that the minimal polynomial for \(T\) is a product of linear factors. Then, by Theorem 5, \(T\) is triangulable. Let \(\mathfrak{B}\) be an ordered basis such that \(A = [T]_{\mathfrak{B}}\) is upper-triangular. Let \(D\) be the diagonal matrix with diagonal entries \(a_{11}, \ldots, a_{nn}\). Then \(A = D + N\), where \(N\) is strictly upper-triangular. Evidently \(N\) is nilpotent.
Solution
1. The proof claims \(A = D + N\) with \(D\) diagonal and \(N\) nilpotent, and implicitly that \(D\) and \(N\) commute (so that this is the Jordan–Chevalley decomposition).
restate what the proof asserts.
2. The error is that \(D\) and \(N\) need not commute.
for a general upper-triangular matrix \(A\), the diagonal part \(D\) and the strictly-upper-triangular part \(N\) do not commute in general.
3. Example: \(A = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}\), so \(D = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\) and \(N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\); then \(DN = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\) but \(ND = \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}\), so \(DN \neq ND\).
explicit counterexample.
4. The Jordan decomposition requires \(D\) and \(N\) to commute (and \(D\) diagonalizable, \(N\) nilpotent); without commutativity, \(A = D + N\) is not the Jordan decomposition.
the definition of the Jordan–Chevalley decomposition.
5. Hence the proof is wrong because it does not establish (and in general cannot establish) that \(D\) and \(N\) commute.
<1>2–<1>4.
6. Q.E.D.
5.