Linear Algebra Exercises-n-Answers.pdf

136 Linear Algebra, by Hefferon
(a) c 1 = 4 (b) c 1 = 4, c 2 = 3 (c) c 1 = 4, c 2 = 3, c 3 = 2, c 4 = 1
For the proof, we will do only the k = 2 case because the completely general case is messier but no
more enlightening. We follow the hint (recall that for any vector ⃗w we have ‖ ⃗w ‖ 2 = ⃗w ⃗w).
(
0 ≤ ⃗v − ( ⃗v ⃗κ 1
· ⃗κ 1 + ⃗v ⃗κ 2
) ) (
· ⃗κ 2 ⃗v − ( ⃗v ⃗κ 1
· ⃗κ 1 + ⃗v ⃗κ 2
) )
· ⃗κ 2
⃗κ 1 ⃗κ 1 ⃗κ 2 ⃗κ 2 ⃗κ 1 ⃗κ 1 ⃗κ 2 ⃗κ 2
( ⃗v ⃗κ1
= ⃗v ⃗v − 2 · ⃗v · ⃗κ 1 + ⃗v ⃗κ )
2
· ⃗κ 2
⃗κ 1 ⃗κ 1 ⃗κ 2 ⃗κ 2
( ⃗v ⃗κ1
+ · ⃗κ 1 + ⃗v ⃗κ ) (
2 ⃗v ⃗κ1
· ⃗κ 2 · ⃗κ 1 + ⃗v ⃗κ )
2
· ⃗κ 2
⃗κ 1 ⃗κ 1 ⃗κ 2 ⃗κ 2 ⃗κ 1 ⃗κ 1 ⃗κ 2 ⃗κ 2
( ⃗v ⃗κ1
= ⃗v ⃗v − 2 · · (⃗v ⃗κ 1 ) + ⃗v ⃗κ ) (
2
· (⃗v ⃗κ 2 ) + ( ⃗v ⃗κ 1
) 2 · (⃗κ 1 ⃗κ 1 ) + ( ⃗v ⃗κ )
2
) 2 · (⃗κ 2 ⃗κ 2 )
⃗κ 1 ⃗κ 1 ⃗κ 2 ⃗κ 2 ⃗κ 1 ⃗κ 1 ⃗κ 2 ⃗κ 2
(The two mixed terms in the third part of the third line are zero because ⃗κ 1 and ⃗κ 2 are orthogonal.)
The result now follows on gathering like terms and on recognizing that ⃗κ 1 ⃗κ 1 = 1 and ⃗κ 2 ⃗κ 2 = 1
because these vectors are given as members of an orthonormal set.
Three.VI.2.19 It is true, except for the zero vector. Every vector in R n except the zero vector is in a
basis, and that basis can be orthogonalized.
Three.VI.2.20
The 3×3 case gives the idea. The set
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
{ ⎝ a d⎠ , ⎝ b e
g h
⎠ ,
⎝ c f⎠}
i
is orthonormal if and only if these nine conditions all hold
⎛ ⎞
⎛ ⎞
( a d
) g ⎝ a d⎠ = 1
( a d
) g ⎝ b e⎠ = 0
⎛g⎞
⎛h⎞
(
b e
)
h ⎝ a d⎠ = 0
(
b e
)
h ⎝ b e⎠ = 1
⎛
g
⎞
⎛
h
⎞
(
c f
)
i ⎝ a d⎠ = 0
(
c f
)
i ⎝ b e⎠ = 0
g
h
( a d g
)
(
b e h
)
(
c f i
)
⎛
⎞
⎝ c f⎠ = 0
⎛ i⎞
⎝ c f⎠ = 0
⎛
i
⎞
⎝ c f⎠ = 1
i
(the three conditions in the lower left are redundant but nonetheless correct). Those, in turn, hold if
and only if
⎛
⎝ a d g
⎞ ⎛
b e h⎠
⎝ a b c
⎞ ⎛
d e f⎠ = ⎝ 1 0 0
⎞
0 1 0⎠
c f i g h i 0 0 1
as required.
This is an example, the inverse of this matrix is its transpose.
⎛
⎝ 1/√ 2 1/ √ ⎞
2 0
−1/ √ 2 1/ √ 2 0⎠
0 0 1
Three.VI.2.21 If the set is empty then the summation on the left side is the linear combination of the
empty set of vectors, which by definition adds to the zero vector. In the second sentence, there is not
such i, so the ‘if . . . then . . . ’ implication is vacuously true.
Three.VI.2.22 (a) Part of the induction argument proving Theorem 2.7 checks that ⃗κ i is in the
span of 〈 β ⃗ 1 , . . . , β ⃗ i 〉. (The i = 3 case in the proof illustrates.) Thus, in the change of basis matrix
Rep K,B (id), the i-th column Rep B (⃗κ i ) has components i + 1 through k that are zero.
(b) One way to see this is to recall the computational procedure that we use to find the inverse. We
write the matrix, write the identity matrix next to it, and then we do Gauss-Jordan reduction. If the
matrix starts out upper triangular then the Gauss-Jordan reduction involves only the Jordan half
and these steps, when performed on the identity, will result in an upper triangular inverse matrix.
Three.VI.2.23 For the inductive step, we assume that for all j in [1..i], these three conditions are true
of each ⃗κ j : (i) each ⃗κ j is nonzero, (ii) each ⃗κ j is a linear combination of the vectors β ⃗ 1 , . . . , β ⃗ j , and