Adv Alegbra HW Solutions 126

Adv Alegbra HW Solutions 126 - 126 if and only if all of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 126 if and only if all of its coefficients are 0; therefore, X is linearly independent. (ii) Define Vn = 1, x , x 2 , . . . , x n . Prove that 1, x , x 2 , . . . , x n is a basis of Vn , and conclude that dim(Vn ) = n + 1. Solution. Absent. 4.9 It is shown in analytic geometry that if slopes m 1 and m 2 , respectively, then only if m 1 m 2 = −1. If i 1 1 and and 2 2 are nonvertical lines with are perpendicular if and = {αvi + u i : α ∈ R}, for i = 1, 2, prove that m 1 m 2 = −1 if and only if the dot product v1 · v2 = 0. Solution. The lines 1 and 2 are perpendicular if and only if v1 and v2 are perpendicular if and only if v1 · v2 = 0. If v1 = (a , b), then m 1 = b/a (the slope of the line joining the origin O with (a , b)), and if v2 = (c, d ), then m 2 = d /c. Hence, m 1 m 2 = −1 ⇔ −1 = (b/a )(d /c) ⇔ b/a = −c/d ⇔ ac = −bd ⇔ ac + bd = 0 ⇔ v1 · v2 = 0 ⇔ 1 and 2 are perpendicular. 4.10 (i) A line in space passing through a point u is defined as {u + αw : α ∈ R} ⊆ R3 , where w is a fixed nonzero vector. Show that every line through u is a coset of a one-dimensional subspace of R3 . Solution. If the origin (0, 0, 0) lies on a line , then u = 0 and = {αw : α ∈ R}, where w is some fixed nonzero vector. In vector space notation, = w , a one-dimensional subspace of R3 . (The converse is true, and it is easy to prove.) (ii) A plane in space passing through a point u is defined as the subset {v ∈ R3 : (v − u ) · n = 0} ⊆ R3 , where n = 0 is a fixed normal vector and (v − u ) · n is a dot product. Prove that a plane through u is a coset of a two-dimensional subspace of R3 . ...
View Full Document

Ask a homework question - tutors are online