# Linear Algebra. Answers to Exercises by Hefferson J.

By Hefferson J.

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Extra resources for Linear Algebra. Answers to Exercises

Example text

26 Each answer given is only one out of many possible. (a) We can parametrize in this way       x 1 0 { 0  x, z ∈ R} = {x 0 + z 0 x, z ∈ R} z 0 1 giving this for a spanning set.     1 0 {0 , 0} 0 1         −2/3 −1/3 −2/3 −1/3 (b) Parametrize it with {y  1  + z  0  y, z ∈ R} to get { 1  ,  0 }. 0 1 0 1     −1/2 1 −2  0     (c) {  1  ,  0 } 1 0 (d) Parametrize the description as {−a1 + a1 x + a3 x2 + a3 x3 a1 , a3 ∈ R} to get {−1 + x, x2 + x3 }.

Cn are scalars and s1 , . . , sn ∈ S. We must show that [S ∪ {v}] = [S]. Containment one way, [S] ⊆ [S ∪ {v}] is obvious. For the other direction, [S ∪ {v}] ⊆ [S], note that if a vector is in the set on the left then it has the form d0 v + d1 t1 + · · · + dm tm where the d’s are scalars and the t ’s are in S. Rewrite that as d0 (c1 s1 + · · · + cn sn ) + d1 t1 + · · · + dm tm and note that the result is a member of the span of S. The ‘only if’ is clearly true — adding v enlarges the span to include at least v.

Remark. ) (b) A set with two elements is linearly independent if and only if neither member is a multiple of the other (note that if one is the zero vector then it is a multiple of the other, so this case is covered). This is an equivalent statement: a set is linearly dependent if and only if one element is a multiple of the other. The proof is easy. A set {v1 , v2 } is linearly dependent if and only if there is a relationship c1 v1 + c2 v2 = 0 with either c1 = 0 or c2 = 0 (or both). That holds if and only if v1 = (−c2 /c1 )v2 or v2 = (−c1 /c2 )v1 (or both).