Why does T applied to basis vectors give upper triangular matrices?

[gravenewworld]

Suppose T is an element of L(V) and (v1, ..., vn) is a basis of V. Then

-the matrix of T with respect to (v1,...,vn) is upper triangular
-Tvk is an element of span(v1,...,vk) for each k=1,...,n
-span(v1,...,vk) is invariant under T for each k=1,...,n.

can some please explain why you will get an upper triangular matrix. The book doesn't show why at all because it says it is "obvious," but I just don't see why at all. Maybe I am thinking too hard. It gives a proof of the 2nd and 3rd lines by saying
Tv1 is an element of span(v1)
Tv2 is an element of span(v1,v2)
.
.
.
.
Tvk is an element of span (v1,...vK).


What i don't understand is why Tv1 is the element of just span of (v1) and Tv2 is in span(v1,v2) etc. I don't understand why you don't have to consider the entire span of the basis vector for each Tvk

 

[matt grime]

Suppose T is upper triangular, that means that T_{ij} is zero if i<j. Write out an upper tirangular matrix to "se" that is what it means.

Now, just write down any upper triangular matrix, and multiply some basis vector by it. What happens? What is the answer. Do it with actual examples to begin with.

Of course we can do it algebraically:

Te_i = T_{ij}e_j summed over the j's from 1 to n (nxn matrix)

but the entries of T are zero for j>i, so this sum stops at i=j

thus the image of v_k is in the span of v_1,...,v_k

This obviously implies the third condition by the linearity of T: pick any element in the span v_1 to v_k: sum x_iv_i, i =1 to k. each v_i is mapped into the span of the first i basis elements, and thus into the span of the first k basis elements as k is the largest of the i's.

The converse implications equally "obvious" take a matrix with those properties and write out its representation with respect to that basis. You know how to do that?

 

[gravenewworld]

Ah thank you. I was didn't read the wording of the text close enough. I was confusing the first statement with "there exists an upper triangular matrix with respect to the basis," which is what I was hung up on. I'm so tired from studying for finals.