Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 137
(iii) each ⃗κ j is orthogonal to all of the ⃗κ m ’s prior to it (that is, with m < j). With those inductive
hypotheses, consider ⃗κ i+1 .
⃗κ i+1 = ⃗ β i+1 − proj [⃗κ1 ](β i+1 ) − proj [⃗κ2 ](β i+1 ) − · · · − proj [⃗κi ](β i+1 )
= ⃗ β i+1 − β i+1 ⃗κ 1
⃗κ 1 ⃗κ 1
· ⃗κ 1 − β i+1 ⃗κ 2
⃗κ 2 ⃗κ 2
· ⃗κ 2 − · · · − β i+1 ⃗κ i
⃗κ i ⃗κ i
· ⃗κ i
By the inductive assumption (ii) we can expand each ⃗κ j into a linear combination of ⃗ β 1 , . . . , ⃗ β j
= β ⃗ β
i+1 − ⃗ i+1 ⃗κ 1
·
⃗κ 1 ⃗κ ⃗β 1
1
β
− ⃗ i+1 ⃗κ
(
2
· linear combination of β
⃗κ 2 ⃗κ ⃗ 1 , ⃗ )
β
β 2 − · · · − ⃗ i+1 ⃗κ
(
i
· linear combination of β
2 ⃗κ i ⃗κ ⃗ 1 , . . . , ⃗ )
β i
i
The fractions are scalars so this is a linear combination of linear combinations of β ⃗ 1 , . . . , β ⃗ i+1 . It is
therefore just a linear combination of β ⃗ 1 , . . . , β ⃗ i+1 . Now, (i) it cannot sum to the zero vector because
the equation would then describe a nontrivial linear relationship among the β’s ⃗ that are given as
members of a basis (the relationship is nontrivial because the coefficient of β ⃗ i+1 is 1). Also, (ii) the
equation gives ⃗κ i+1 as a combination of β ⃗ 1 , . . . , β ⃗ i+1 . Finally, for (iii), consider ⃗κ j ⃗κ i+1 ; as in the i = 3
case, the dot product of ⃗κ j with ⃗κ i+1 = β ⃗ i+1 − proj [⃗κ1 ]( β ⃗ i+1 ) − · · · − proj [⃗κi ]( β ⃗ i+1 ) can be rewritten to
(
give two kinds of terms, ⃗κ ⃗βi+1 j − proj [⃗κj]( β ⃗ )
i+1 ) (which is zero because the projection is orthogonal)
and ⃗κ j proj [⃗κm ]( β ⃗ i+1 ) with m ≠ j and m < i + 1 (which is zero because by the hypothesis (iii) the
vectors ⃗κ j and ⃗κ m are orthogonal).
Subsection Three.VI.3: Projection Into a Subspace
Three.VI.3.10
(a) When bases for the subspaces
( (
1 2
B M = 〈 〉 B
−1)
N = 〈 〉
−1)
are concatenated
( ) (
⌢ 1 2
B = B M BN = 〈 , 〉
−1 −1)
and the given vector is represented (
3
−2
)
( ) (
1 2
= 1 · + 1 ·
−1 −1)
then the answer comes from retaining the M part and dropping the N part.
( (
3 1
proj M,N ( ) =
−2)
−1)
(b) When the bases
( (
1 1
B M = 〈 〉 B
1)
N 〈 〉
−2)
are concatenated, and the vector is represented,
( ( (
1 1 1
= (4/3) · − (1/3) ·
2)
1)
−2)
then retaining only the M part gives this answer.
(
1
proj M,N ( ) =
2)
(c) With these bases
⎛
⎞
⎛
⎞
( )
4/3
4/3
⎛
⎞
B M = 〈 ⎝ 1 −1⎠ , ⎝ 0 0⎠〉 B N = 〈 ⎝ 1 0⎠〉
0 1
1