Linear Algebra

Section VI. Projection 257
We get �κ2 by starting with the given second vector � β2 and subtracting away the
part of it in the direction of �κ1.
⎛
�κ2 = ⎝ 0
⎞ ⎛
2⎠
− proj [�κ1]( ⎝
0
0
⎞ ⎛
2⎠)
= ⎝
0
0
⎞ ⎛
2⎠
− ⎝
0
2/3
⎞ ⎛
2/3⎠
= ⎝
2/3
−2/3
⎞
4/3 ⎠
−2/3
Finally, we get �κ3 by taking the third given vector and subtracting the part of
it in the direction of �κ1, and also the part of it in the direction of �κ2.
⎛
�κ3 = ⎝ 1
⎞ ⎛
0⎠
− proj [�κ1]( ⎝
3
1
⎞
⎛
0⎠)
− proj [�κ2]( ⎝
3
1
⎞ ⎛
0⎠)
= ⎝
3
−1
⎞
0 ⎠
1
Again the corollary gives that
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 −2/3 −1
〈 ⎝1⎠
, ⎝ 4/3 ⎠ , ⎝ 0 ⎠〉
1 −2/3 1
is a basis for the space.
The next result verifies that the process used in those examples works with
any basis for any subspace of an R n (we are restricted to R n only because we
have not given a definition of orthogonality for other vector spaces).
2.7 Theorem (Gram-Schmidt orthogonalization) If 〈 � β1,... � βk〉 is a basis
for a subspace of R n then, where
�κ1 = � β1
�κ2 = � β2 − proj [�κ1]( � β2)
�κ3 = � β3 − proj [�κ1]( � β3) − proj [�κ2]( � β3)
.
�κk = � βk − proj [�κ1]( � βk) −···−proj [�κk−1]( � βk)
the �κ ’s form an orthogonal basis for the same subspace.
Proof. We will use induction to check that each �κi is nonzero, is in the span of
〈 � β1,... � βi〉 and is orthogonal to all preceding vectors: �κ1 �κi = ···= �κi−1 �κi =0.
With those, and with Corollary 2.3, we will have that 〈�κ1,...�κk〉 is a basis for
the same space as 〈 � β1,... � βk〉.
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 �κ1 equal to � β1 makes it a nonzero vector
since � β1 is a member of a basis, it is obviously in the desired span, and the
‘orthogonal to all preceding vectors’ condition is vacuously met.