Linear Algebra

Section VI. Projection 251Again the corollary gives thatis a basis for the space.⎛〈 ⎝ 1 ⎞ ⎛1⎠ , ⎝ −2/3⎞ ⎛4/3 ⎠ , ⎝ −1⎞0 ⎠〉1 −2/3 1The next result verifies that the process used in those examples works withany basis for any subspace of an R n (we are restricted to R n only because wehave not given a definition of orthogonality for other vector spaces).2.7 Theorem (Gram-Schmidt orthogonalization) If 〈 ⃗ β 1 , . . . ⃗ β k 〉 is a basisfor a subspace of R n then, where⃗κ 1 = ⃗ β 1⃗κ 2 = ⃗ β 2 − proj [⃗κ1]( ⃗ β 2 )⃗κ 3 = ⃗ β 3 − proj [⃗κ1 ]( ⃗ β 3 ) − proj [⃗κ2 ]( ⃗ β 3 )..⃗κ k = ⃗ β k − proj [⃗κ1 ]( ⃗ β k ) − · · · − proj [⃗κk−1 ]( ⃗ β k )the ⃗κ ’s form an orthogonal basis for the same subspace.Proof. We will use induction to check that each ⃗κ i is nonzero, is in the span of〈 ⃗ β 1 , . . . ⃗ β i 〉 and is orthogonal to all preceding vectors: ⃗κ 1 ⃗κ i = · · · = ⃗κ i−1 ⃗κ i = 0.With those, and with Corollary 2.3, we will have that 〈⃗κ 1 , . . . ⃗κ k 〉 is a basis forthe same space as 〈 ⃗ β 1 , . . . ⃗ β k 〉.We shall cover the cases up to i = 3, which give the sense of the argument.Completing the details is Exercise 23.The i = 1 case is trivial — setting ⃗κ 1 equal to ⃗ β 1 makes it a nonzero vectorsince ⃗ β 1 is a member of a basis, it is obviously in the desired span, and the‘orthogonal to all preceding vectors’ condition is vacuously met.For the i = 2 case, expand the definition of ⃗κ 2 .⃗κ 2 = ⃗ β 2 − proj [⃗κ1 ]( ⃗ β 2 ) = ⃗ β 2 − ⃗ β 2 ⃗κ 1⃗κ 1 ⃗κ 1· ⃗κ 1 = ⃗ β 2 − ⃗ β 2 ⃗κ 1⃗κ 1 ⃗κ 1· ⃗β 1This expansion shows that ⃗κ 2 is nonzero or else this would be a non-trivial lineardependence among the ⃗ β’s (it is nontrivial because the coefficient of ⃗ β 2 is 1) andalso shows that ⃗κ 2 is in the desired span. Finally, ⃗κ 2 is orthogonal to the onlypreceding vector⃗κ 1 ⃗κ 2 = ⃗κ 1 ( ⃗ β 2 − proj [⃗κ1]( ⃗ β 2 )) = 0because this projection is orthogonal.