Linear Algebra

Section I. Solving Linear Systems 23
3.7 Lemma For any homogeneous linear system there exist vectors � β1, ... ,
�βk such that the solution set of the system is
{c1 � β1 + ···+ ck � �
βk � c1,... ,ck ∈ R}
where k is the number of free variables in an echelon form version of the system.
Before the proof, we will recall the back substitution calculations that were
done in the prior subsection. Imagine that we have brought a system to this
echelon form.
x + 2y − z +2w =0
−3y + z =0
−w =0
We next perform back-substitution to express each variable in terms of the
free variable z. Working from the bottom up, we get first that w is 0 · z,
next that y is (1/3) · z, and then substituting those two into the top equation
x + 2((1/3)z) − z + 2(0) = 0 gives x =(1/3) · z. So, back substitution gives
a paramatrization of the solution set by starting at the bottom equation and
using the free variables as the parameters to work row-by-row to the top. The
proof below follows this pattern.
Comment: That is, this proof just does a verification of the bookkeeping in
back substitution to show that we haven’t overlooked any obscure cases where
this procedure fails, say, by leading to a division by zero. So this argument,
while quite detailed, doesn’t give us any new insights. Nevertheless, we have
written it out for two reasons. The first reason is that we need the result — the
computational procedure that we employ must be verified to work as promised.
The second reason is that the row-by-row nature of back substitution leads to a
proof that uses the technique of mathematical induction. ∗ This is an important,
and non-obvious, proof technique that we shall use a number of times in this
book. Doing an induction argument here gives us a chance to see one in a setting
where the proof material is easy to follow, and so the technique can be studied.
Readers who are unfamiliar with induction arguments should be sure to master
this one and the ones later in this chapter before going on to the second chapter.
Proof. First use Gauss’ method to reduce the homogeneous system to echelon
form. We will show that each leading variable can be expressed in terms of free
variables. That will finish the argument because then we can use those free
variables as the parameters. That is, the � β’s are the vectors of coefficients of
the free variables (as in Example 3.6, where the solution is x =(1/3)w, y = w,
z =(1/3)w, andw = w).
We will proceed by mathematical induction, which has two steps. The base
step of the argument will be to focus on the bottom-most non-‘0 = 0’ equation
and write its leading variable in terms of the free variables. The inductive step
of the argument will be to argue that if we can express the leading variables from
∗ More information on mathematical induction is in the appendix.