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

[Stephen Tashi]

TL;DR

A question about the final step in a proof (by induction) that each linear transformation in a finite dimensional complex vector space has a basis in which its matrix is upper triangular.

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.

 

[mathwonk]

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.

 

[Stephen Tashi]

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.

 

[Stephen Tashi]

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.