A theorem in Linear Algebra; linear dependence - Axler 


I really am having trouble understanding the statement and the proof. Why does the theorem pick $v_1 \neq 0 $? Why not $v_2$? Also in proving (a), why do we consider the largest $j$?
I do not understand the statement Note all of $a_2, a_3, \dots, a_m$ can be $0$ (because $v_1 = 0 $) Can someone teach me by example?

First, there's nothing special about $v_1$ that we set it not equal to $0$.  We're putting an arbitrary order to our list of basis vectors: the theorem would work the same if we switched the places of $v_1$ and $v_2$ (and changed the statement accordingly so that we could potentially express $v_2$ in terms of $v_1$ and require that $v_2 \ne 0$).
We have to require that $v_1 \ne 0$ because otherwise the singleton $\{ v_1 \}$ is linearly dependent, but it is nonsensical to express $v_1$ in terms of previous basis vectors (unless you consider $v_1 = 0$ an example of that).
We consider the largest $j$ with nonzero coefficient simply because we're expressing $v_j$ in terms of previous basis vectors.  If the coefficient of $v_k$ is $0$, then the statement says nothing about the vector $v_k$.  We want to be able to divide out the coefficient, and taking the largest $j$ vector allows us to do that.
Now to answer your last question, suppose all of $a_2 \dots a_m$ are $0$.  We're assuming that $a_1$ and $v_1$ are not zero, so $a_1 v_1 \ne 0$.  But $0 = a_1 v_1 + \dots a_m v_m = a_1 v_1$ if the other coefficients are all $0$.  This is a contradiction.
Hope that helped.
