Linear Algebra, 2020a

Section VI. Projection 283
By the first line ⃗κ 3 ≠ ⃗0, since ⃗β 3 isn’t in the span [⃗β 1 , ⃗β 2 ] and therefore by the
inductive hypothesis it isn’t in the span [⃗κ 1 ,⃗κ 2 ]. By the second line ⃗κ 3 is in
the span of the first three ⃗β’s. Finally, the calculation below shows that ⃗κ 3 is
orthogonal to ⃗κ 1 .
(
⃗κ 1 • ⃗κ 3 = ⃗κ 1 • β ⃗ 3 − proj [⃗κ1 ](⃗β 3 )−proj [⃗κ2 ](⃗β 3 ) )
(
= ⃗κ 1 • β ⃗ 3 − proj [⃗κ1 ](⃗β 3 ) ) − ⃗κ 1 • proj [⃗κ2 ](⃗β 3 )
= 0
(Here is the difference with the i = 2 case: as happened for i = 2 the first term
is 0 because this projection is orthogonal, but here the second term in the second
line is 0 because ⃗κ 1 is orthogonal to ⃗κ 2 and so is orthogonal to any vector in
the line spanned by ⃗κ 2 .) A similar check shows that ⃗κ 3 is also orthogonal to ⃗κ 2 .
QED
In addition to having the vectors in the basis be orthogonal, we can also
normalize each vector by dividing by its length, to end with an orthonormal
basis..
2.9 Example From the orthogonal basis of Example 2.6, normalizing produces
this orthonormal basis.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎜
1/ √ 3
〈 ⎝1/ √ ⎟ ⎜
−1/ √ 6
3
1/ √ ⎠ , ⎝ 2/ √ −1/ √ 2
⎟ ⎜
6
3 −1/ √ ⎠ , ⎝ 0
6 1/ √ 2
Besides its intuitive appeal, and its analogy with the standard basis E n for
R n , an orthonormal basis also simplifies some computations. Exercise 22 is an
example.
⎟
⎠〉
Exercises
2.10 Normalize the lengths of these vectors.
⎛ ⎞
( −1 ( 1
(a) (b) ⎝ 3 ⎠ 1
(c)
2)
−1)
0
̌ 2.11 Perform Gram-Schmidt on this basis for R 2 .
( ( ) 1 −1
〈 , 〉
1)
2
Check that the resulting vectors are orthogonal.
̌ 2.12 Perform the Gram-Schmidt process on this basis for R 3 .
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
1 2 3
〈 ⎝2⎠ , ⎝ 1 ⎠ , ⎝3⎠〉
3 −3 3
̌ 2.13 Perform Gram-Schmidt on each of these bases for R 2 .