Question from a proof in Axler 2nd Ed, 'Linear Algebra Done Right'

My question is motivated by the proof of TH 5.13 on p 84 in the 2nd edition of Linear Algebra Done Right. (This proof differs from that in the 4th ed - online at: https://linear.axler.net/index.html chapter 5 )

In the proof we arrive at the following situation:
##T## is a linear operator on a finite dimensional complex vector space ##V## and ##\lambda ## is a an eigenvalue of ##T\ ##. The subspace ##U## is the range of the linear operator ##T−λI## The set of vectors ##\{ u_1,u_2,...,u_m, v_1,v_2,...v_k\}## is a basis for ##V## such that ##\{ u_1,u_2,...u_m\}## is a basis for ##U## and such that the matrix of the operator ##T - \lambda I ## restricted to ##U## is upper triangular in that basis.

We have the identity ## Tv_j = (T - \lambda I) v_j + \lambda v_j##. Since ##(T-\lambda I) v_j \in U##, this exhibits ##Tv_j## as the sum of two vectors where the first can be expressed as a linear combination of the vectors ##u_i## and the second is ##\lambda v_j##.

My question: (for example in the case Dim ##U = m = 2,\ ##, Dim ## V = 5## ) Does this show the matrix of ##T## has the form
##\begin{pmatrix} a_{1,1}&a_{1,2}& a_{1,3} & a_{1,4} & a_{1,5} \\ 0 & a_{2,2}, & a_{2,3} & a_{2,4} & a_{2,5} \\ 0 & 0 & \lambda & 0 & 0 \\ 0 & 0 &0 & \lambda &0\\ 0 & 0 & 0 & 0 & \lambda \end{pmatrix} ## ?

This is not how Axler ends the proof. He makes the less detailed observation that ##v_j \in ## Span ## \{u_1,u_2,...v_j\} ## for ##j = 1, k\ ##. That property characterizes an upper triangular matrix by a previously proved theorem, TH 5.12.

pardon me, I did not try read axler, but this theorem seems off hand to have a trivial proof. just take v1 as an eigenvector for T, then mod out V by the space spanned by v1, and take v2 to be a vector representing an eigenvector for the induced map on V/<v1>. Then take v3 to be a vector representing an eigenvector for the induced map on V/<v1,v2>, ...... I.e. at each stage, vk is an eigenvector mod the previous vectors, so T(vk)-ck.vk is a linear combination of v1,...,vk-1. Is this nonsense?

but I see your questions is otherwise. I think the answer to it is yes.

Here's my discomfort with the zeroes: If all those zeroes appear, why do we ever need generalized eigenvectors? This is just an intuitive discomfort. I haven't figured out whether a defective matrix couldn't have them.

I see why my intuition is wrong. For example, ##Tv_1 = \begin{pmatrix} a_{1,3} \\ a_{2,3} \\ \lambda,\\0,\\0 \end{pmatrix} ## , which isn't equal to ##\lambda v_1##. So ##\lambda v_1## isn't necessarily an eigenvector.