Linear Algebra Extra Examples

Extra Examples, Section 4.3 31. Let T : V → W be a linear transformation from vector space V into vector space W . Show that, if {v1 , . . . , vp } is linearly dependent in V , then {T (v1 ), . . . , T (vp )} is linearly dependent in W . SOLUTION: Since {v1 , . . . , vp } is linearly dependent, there exists number c1 , c2 , . . . , ck , not all zero, so that c1 v1 + c2 v2 + · · · + cp vp = 0 Taking T of both sides and using the linearity of T , we then have: c1 T (v1 ) + c2 T (v2 ) + · · · + cp T (vp ) = 0 And since c1 , c2 , · · · , cp are not all zero, then the set of vectors {T (v1 ), . . . , T (vp )} is linearly dependent in W . 32. We want to show that if T is 1-1, then the set of images {T (v1 ), . . . , T (vp )} is linearly dependent, then the original set of vectors, {v1 , . . . , vp } is linearly dependent. SOLUTION: What is wrong with the following logic (put down a correct argument for your solution)? We are told that {T (v1 ), . . . , T (vp )} is linearly dependent in W . Therefore, c1 T (v1 ) + c2 T (v2 ) + · · · + cp T (vp ) = 0 And by the linearity of T , T (c1 v1 + c2 v2 + · · · + cp vp ) = 0 Therefore, c1 v1 + c2 v2 + · · · + cp vp = 0 where c1 , c2 , · · · , ck are not all zero. Therefore, we can conclude that the set of vectors {v1 , . . . , vp } is linearly dependent. (Hint: I did not use the fact that T was 1-1...) 33. (This problem is not 33, but is very similar to 33) Is the set of vectors p1 (t) = 2t − t2 , p2 (t) = 2 + 2t p3 (t) = 2 + 8t − 3t2 linearly independent vectors in P2 ? SOLUTION: Use the definition of linear independence. That means, we look for constants C1 , C2 , C3 , not all zero, so that C1 (2t − t2 ) + C2 (2 + 2t) + C3 (2 + 8t − 3t2 ) = 0 1 for all t I’m going to multiply this out and re-order the terms so that it is of the form: A1 t2 + A2 t + A3 = 0, for all t This will mean that A1 = 0, A2 = 0, and A3 = 0. You probably did something like this before when you solved for the coefficients in a partial fraction problem, or when you looked at power series. In this case, we have: (−C1 − 3C3 )t2 + (2C1 + 2C2 + 8C3 )t + (2C2 + 2C3 ) = 0 for all t Equating the coefficients to zero, we get three equations in three unknowns:     −C1 −C3 = 0 −1 0 −3 1 0 3 2C1 +2C2 +8C3 = 0 8 ∼ 0 1 1  ⇒  2 2 2C2 +2C3 = 0 0 2 2 0 0 0 There are an infinite number of (non-zero) solutions to our equation, so the three vectors (polynomials) are linearly dependent. In fact, you might check that p3 (t) = 3p1 (t) + p2 (t) Did you notice a relationship between the coefficients of each polynomial and the columns of our matrix? 2