Linear Algebra

Section II. Linear Independence 99In the same vector space, {(40 15), (20 7.5)} is linearly dependent sincewe can satisfyc 1 (40 15) + c 2 · (20 7.5) = (0 0)with c 1 = 1 and c 2 = −2.1.5 Example The set {1 + x, 1 − x} is linearly independent in P 2 , the space ofquadratic polynomials with real coefficients, becausegives0 + 0x + 0x 2 = c 1 (1 + x) + c 2 (1 − x) = (c 1 + c 2 ) + (c 1 − c 2 )x + 0x 2c 1 + c 2 = 0c 1 − c 2 = 0−ρ 1 +ρ 2−→c 1 + c 2 = 02c 2 = 0since polynomials are equal only if their coefficients are equal. Thus, the onlylinear relationship between these two members of P 2 is the trivial one.1.6 Example The rows of this matrix⎛⎞2 3 1 0⎜⎟A = ⎝0 −1 0 −2⎠0 0 0 1form a linearly independent set. This is easy to check in this case, but also recallthat Lemma One.III.2.5 shows that the rows of any echelon form matrix form alinearly independent set.1.7 Example In R 3 , where⎛⎜3⎞ ⎛ ⎞ ⎛ ⎞24⎟ ⎜ ⎟ ⎜ ⎟⃗v 1 = ⎝4⎠ ⃗v 2 = ⎝9⎠ ⃗v 3 = ⎝18⎠524the set S = {⃗v 1 ,⃗v 2 ,⃗v 3 } is linearly dependent because this is a relationship0 · ⃗v 1 + 2 · ⃗v 2 − 1 · ⃗v 3 = ⃗0where not all of the scalars are zero (the fact that some of the scalars are zerodoesn’t matter).That example illustrates why, although Definition 1.2 is a clearer statementof what independence is, Lemma 1.3 is more useful for computations. Workingstraight from the definition, someone trying to compute whether S is linearlyindependent would start by setting ⃗v 1 = c 2 ⃗v 2 + c 3 ⃗v 3 and concluding that thereare no such c 2 and c 3 . But knowing that the first vector is not dependent on theother two is not enough. This person would have to go on to try ⃗v 2 = c 1 ⃗v 1 +c 3 ⃗v 3to find the dependence c 1 = 0, c 3 = 1/2. Lemma 1.3 gets the same conclusionwith only one computation.1.8 Example The empty subset of a vector space is linearly independent. Thereis no nontrivial linear relationship among its members as it has no members.