Linear Algebra

260 Chapter Three. Maps Between SpacesWith the matrix, calculating the orthogonal projection of any vector into P iseasy.⎛⎞ ⎛1/2 0 −1/2proj P (⃗v) = ⎝ 0 1 0 ⎠ ⎝ 1⎞ ⎛−1⎠ = ⎝ 0 ⎞−1⎠−1/2 0 1/2 1 0Exerciseš 3.10 Project ( ) the vectors ( into ) M along N. ( )3x ∣∣ x ∣∣(a) , M = { x + y = 0}, N = { −x − 2y = 0}−2yy( ( ) ( )1 x ∣∣ x ∣∣(b) , M = { x − y = 0}, N = { 2x + y = 0}2)yy( )) ( ) 3 x ∣∣ 1 ∣∣(c) 0 , M = {(y x + y = 0}, N = {c · 0 c ∈ R}1z1̌ 3.11 Find M ⊥ .( ) ( )x ∣∣ x ∣∣(a) M = { x + y = 0} (b) M = { −2x + 3y = 0}yy( ) ( )x ∣∣ x ∣∣(c) M = { x − y = 0} (d) M = { ⃗0 } (e) M = { x = 0}yy) ( ) x ∣∣ x ∣∣(f) M = {(y −x + 3y + z = 0} (g) M = { y x = 0 and y + z = 0}zz3.12 This subsection shows how to project orthogonally in two ways, the method ofExample 3.2 and 3.3, and the method of Theorem 3.8. To compare them, considerthe plane P specified by 3x + 2y − z = 0 in R 3 .(a) Find a basis for P .(b) Find P ⊥ and a basis for P ⊥ .(c) Represent this vector with respect to the concatenation of the two bases fromthe prior item.)⃗v =( 112(d) Find the orthogonal projection of ⃗v into P by keeping only the P part fromthe prior item.(e) Check that against the result from applying Theorem 3.8.̌ 3.13 We have three ways to find the orthogonal projection of a vector into a line,the Definition 1.1 way from the first subsection of this section, the Example 3.2and 3.3 way of representing the vector with respect to a basis for the space andthen keeping the M part, and the way of Theorem 3.8. For these cases, do allthree ways. ( ) ( )1x ∣∣(a) ⃗v = , M = { x + y = 0}−3y( )) 0 x ∣∣(b) ⃗v = 1 , M = {(y x + z = 0 and y = 0}2z3.14 Check that the operation of Definition 3.1 is well-defined. That is, in Example3.2 and 3.3, doesn’t the answer depend on the choice of bases?