Linear Algebra

262 Chapter Three. Maps Between SpacesTo check this, represent the vector as ⃗v = r 1 ⃗κ 1 + · · · + r n ⃗κ n , apply ⃗κ i to bothsides ⃗v • ⃗κ i = (r 1 ⃗κ 1 + · · · + r n ⃗κ n ) • ⃗κ i = r 1 · 0 + · · · + r i · (⃗κ i • ⃗κ i ) + · · · + r n · 0,and solve to get r i = (⃗v • ⃗κ i )/(⃗κ i • ⃗κ i ), as desired.Since obviously any member of the span of 〈⃗κ k+1 , . . . ,⃗κ n 〉 is orthogonal toany vector in M, to show that this is a basis for M ⊥ we need only show theother containment — that any ⃗w ∈ M ⊥ is in the span of this basis. The priorparagraph does this. Any ⃗w ∈ M ⊥ gives this on projections into basis vectorsfrom M: proj [⃗κ1 ](⃗w ) = ⃗0, . . . , proj [⃗κk ](⃗w ) = ⃗0. Therefore equation (∗) givesthat ⃗w is a linear combination of ⃗κ k+1 , . . . ,⃗κ n . Thus this is a basis for M ⊥ andR n is the direct sum of the two.The final sentence of the statement of this result is proved in much the sameway. Write ⃗v = proj [⃗κ1 ](⃗v ) + · · · + proj [⃗κn ](⃗v ). Then proj M (⃗v ) keeps only theM part and dropping the M ⊥ part proj M (⃗v ) = proj [⃗κk+1 ](⃗v ) + · · · + proj [⃗κk ](⃗v ).Therefore ⃗v − proj M (⃗v ) consists of a linear combination of elements of M ⊥ andso is perpendicular to every vector in M.QEDWe can find the orthogonal projection into a subspace by following the stepsof the proof but the next result gives a convenient formula.3.8 Theorem Let ⃗v be a vector in R n and let M be a subspace of R n withbasis 〈⃗β 1 , . . . , ⃗β k 〉. If A is the matrix whose columns are the ⃗β’s thenproj M (⃗v ) = c 1⃗β 1 + · · · + c k⃗β k where the coefficients c i are the entries ofthe vector (A trans A) −1 A trans · ⃗v. That is, proj M (⃗v ) = A(A trans A) −1 A trans · ⃗v.Proof The vector proj M (⃗v) is a member of M and so it is a linear combinationof basis vectors c 1 · ⃗β 1 + · · · + c k · ⃗β k . Since A’s columns are the ⃗β’s, thatcan be expressed as: there is a ⃗c ∈ R k such that proj M (⃗v ) = A⃗c. The vector⃗v − proj M (⃗v ) is perpendicular to each member of the basis so we have this.⃗0 = A trans( ⃗v − A⃗c ) = A trans ⃗v − A trans A⃗cSolving for ⃗c (showing that A trans A is invertible is an exercise)⃗c = ( A trans A ) −1Atrans · ⃗vgives the formula for the projection matrix as proj M (⃗v ) = A · ⃗c.3.9 Example To orthogonally project this vector into this subspace⎛ ⎞ ⎛ ⎞1x⎜ ⎟ ⎜ ⎟⃗v = ⎝−1⎠ P = { ⎝y⎠ ∣ x + z = 0}1zQEDfirst make a matrix whose columns are a basis for the subspace⎛⎜0 1⎞⎟A = ⎝1 0⎠0 −1