Linear Algebra

24 Chapter One. Linear Systemsleading variable can be expressed in terms of free variables. That will finish theproof because we can then use the free variables as the parameters and the ⃗β’sare the vectors of coefficients of those free variables.For the base step, consider the bottommost equation that is not 0 = 0. Callit equation m so we havea m,lm x lm + a m,lm +1x lm +1 + · · · + a m,n x n = 0where a m,lm ≠ 0. (The ‘l’ means “leading” so that x lm is the leading variablein row m.) This is the bottom row so any variables x lm +1, . . . after the leadingvariable in this equation must be free variables. Move these to the right sideand divide by a m,lmx lm= (−a m,lm +1/a m,lm )x lm +1 + · · · + (−a m,n /a m,lm )x nto express the leading variable in terms of free variables. (If there are no variablesto the right of x lm then x lm = 0; see the “tricky point” following this proof.)For the inductive step assume that for the m-th equation, and the (m − 1)-thequation, etc., up to and including the (m − t)-th equation (where 0 t < m),we can express the leading variable in terms of free variables. We must verify thatthis statement also holds for the next equation up, the (m − (t + 1))-th equation.As in the earlier sketch, take each variable that leads in a lower-down equationx lm , . . . , x lm−t and substitute its expression in terms of free variables. (We onlyneed do this for the leading variables from lower-down equations because thesystem is in echelon form and so in this equation none of the variables leadinghigher up equations appear.) The result has the forma m−(t+1),lm−(t+1) x lm−(t+1) + a linear combination of free variables = 0with a m−(t+1),lm−(t+1) ≠ 0. Move the free variables to the right side and divideby a m−(t+1),lm−(t+1) to end with x lm−(t+1) expressed in terms of free variables.Because we have shown both the base step and the inductive step, by theprinciple of mathematical induction the proposition is true.QEDWe say that the set {c 1⃗β 1 + · · · + c k⃗β k∣ ∣ c1 , . . . , c k ∈ R} is generated by orspanned by the set of vectors {⃗β 1 , . . . , ⃗β k }.There is a tricky point to this. We rely on the convention that the sum of anempty set of vectors is the zero vector. In particular, we need this in the casewhere a homogeneous system has a unique solution. Then the homogeneouscase fits the pattern of the other solution sets: in the proof above, we derive thesolution set by taking the c’s to be the free variables and if there is a uniquesolution then there are no free variables.Note that the proof shows, as discussed after Example 2.4, that we can alwaysparametrize solution sets using the free variables.The next lemma finishes the proof of Theorem 3.1 by considering the particularsolution part of the solution set’s description.