Processing math: 100%
Back to the class
Problem #37 from pg. 61 and problem #1 from pg. 68 are graded..
#37,pg.61: True or false; if false construct a specific counterexample, if true justify. If →v1,…,→v4 are in R4×1 and {→v1,→v2,→v3} form a linearly dependent set, then {→v1,→v2,→v3,→v4} also forms a linearly dependent set.
Solution: True. Linear dependence can be characterized as "one of the vectors can be written in terms of the others" (see Theorem 7, pg. 58). Hence if {→v1,→v2,→v3} is linearly dependent, one of those can be written in terms of the others. Assume ("without loss of generality") that →v2 can be written in terms of →v1 and →v3. In this case we know there exist weights α and β so that
→v3=α→v1+β→v2.
Now consider the set {→v1,→v2,→v3,→v4}. Consider the following linear combination:
α→v1+β→v3+0→v4=α→v1+β→v3=→v2,
and we see that we can still "write one of the vectors in terms of the others". Hence the set {→v1,→v2,→v3,→v4} is always a linearly dependent set if {→v1,→v2,→v3} is, no matter what the vector →v4 is .
#1, pg.68: Let A=[2002] and define T:R2×1→R2×1 by T(→x)=A→x. Find the images under T of →u=[1−3] and →v=[ab].
Solution: We compute directly:
T(→u)=[2002][1−3]=1[20]−3[02]=[2−6,]
and
T(→v)=[2002][ab]=a[20]+b[02]=[2a2b].
