Some readers of [1] might appreciate the following comments that make more explicit how Ptak’s beautiful insight there leads to a trivial proof of (the basics of) the Jordan normal form. The proof of Theorem 1 of [1] can also be based on the observation that, X being finite-dimensional, the sequence {0} ⊆ ker A ⊆ ker A2 ⊆ · · · must eventually be stationary, i.e., ker A = ker Aq+p for some q and all p > 0. For such q, let Xr and Xs be the range and the kernel, respectively, of A , hence dim X = dim Xr + dim Xs . Further, for any x ∈ Xr ∩ Xs, x = Az for some z, and so z ∈ ker A2q = ker A , hence x = 0. Therefore, X is the direct sum of the two A-invariant subspaces Xs and Xr , and A is regular on Xr (since A is) and nilpotent on Xs . In the setup and notation of Theorem 2 of [1], there must be, by duality, some y0 in Y for which 〈x0Aq−1, y0〉 / = 0, hence the q-order matrix (〈x0Aj−1, y0A∗q−i〉: i,j = 1, . . . , q) is triangular with nonzero diagonal entries, therefore invertible, and this guarantees that X is the direct sum X0 + X′, with X0 the linear span of (x0Aj−1: j = 1, . . . , q) and X′ the annihilator of {y0A∗q−i : i = 1, . . . , q}, both of which are Ainvariant. Moreover, it shows (x0Aj−1: j = 1, . . . , q) to be a basis for X0, and the matrix representation, with respect to this basis, of A restricted to X0 has the familiar form of a Jordan block (for the eigenvalue 0). Now, X being finite-dimensional, there are A-invariant direct sum decompositions X = X1 + · · · + Xm that are minimal in the sense that none of its summands is the direct sum of two nontrivial A-invariant subspaces. Take any one such. Then the matrix representation for A with respect to any basis made up from bases for the summands Xi is block diagonal, with the ith block the matrix representation of the restriction Ai of A to Xi with respect to the chosen basis for Xi . Assuming the underlying field to be algebraically closed, the restriction Ai of A to Xi has some eigenvalue, λi , and, in view of the minimality of Xi , Theorem 1 ensures
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