Linear Algebra

Section VI. Projection 259Compare the projections along these (verification that R 3 = M ⊕ N and R 3 =M ⊕ ˆN is routine).⎛⎜0⎞⎛ ⎞⎟N = {k ⎝0⎠ ∣ 0⎜ ⎟k ∈ R} ˆN = {k ⎝ 1⎠ ∣ k ∈ R}1−2The projections are different because they have different effects on this vector.⎛⎜2⎞⎟⃗v = ⎝2⎠5For the first one we find a basis for N⎛⎜0⎞⎟B N = 〈 ⎝0⎠〉1⌢and represent ⃗v with respect to the concatenation B M BN .⎛⎜2⎞ ⎛⎟ ⎜1⎞ ⎛⎟ ⎜0⎞ ⎛⎟ ⎜0⎞⎟⎝2⎠ = 2 · ⎝0⎠ + 1 · ⎝2⎠ + 4 · ⎝0⎠5 0 1 1We find the projection of ⃗v into M along N by dropping the N component.⎛⎜1⎞ ⎛⎟ ⎜0⎞ ⎛⎟ ⎜2⎞⎟proj M,N (⃗v ) = 2 · ⎝0⎠ + 1 · ⎝2⎠ = ⎝2⎠0 1 1For ˆN, this basis is natural.⎛ ⎞0B ˆN = 〈 ⎜ ⎟⎝ 1⎠〉−2Representing ⃗v with respect to the concatenation⎛ ⎞ ⎛ ⎞ ⎛2 1⎜ ⎟ ⎜ ⎟ ⎜0⎞ ⎛ ⎞0⎟ ⎜ ⎟⎝2⎠ = 2 · ⎝0⎠ + (9/5) · ⎝2⎠ − (8/5) · ⎝ 1⎠5 01−2and then keeping only the M part gives this.⎛proj M, ˆN (⃗v ) = 2 · ⎜1⎞ ⎛ ⎞ ⎛ ⎞0 2⎟ ⎜ ⎟ ⎜ ⎟⎝0⎠ + (9/5) · ⎝2⎠ = ⎝18/5⎠01 9/5Therefore projection along different subspaces may yield different results.These pictures compare the two maps. Both show that the projection isindeed ‘into’ the plane and ‘along’ the line.