Linear Algebra Exercises-n-Answers.pdf

168 Linear Algebra, by Hefferon
The second matrix has one fewer inversion because there is one fewer inversion in the interval (s vs.
s + 1) and inversions involving rows outside the interval are not affected.
Proceed in this way, at each step reducing the number of inversions by one with each row swap.
When no inversions remain the result is the identity.
The contrast with Corollary 4.6 is that the statement of this exercise is a ‘there exists’ statement:
there exists a way to swap to the identity in exactly m steps. But the corollary is a ‘for all’ statement:
for all ways to swap to the identity, the parity (evenness or oddness) is the same.
Four.I.4.17 (a) First, g(φ 1 ) is the product of the single factor 2 − 1 and so g(φ 1 ) = 1. Second, g(φ 2 )
is the product of the single factor 1 − 2 and so g(φ 2 ) = −1.
(b) permutation φ φ 1 φ 2 φ 3 φ 4 φ 5 φ 6
g(φ) 2 −2 −2 2 2 −2
(c) Note that φ(j) − φ(i) is negative if and only if ι φ(j) and ι φ(i) are in an inversion of their usual
order.
Subsection Four.II.1: Determinants as Size Functions
Four.II.1.8 For each, find the determinant and take the absolute value.
(a) 7 (b) 0 (c) 58
Four.II.1.9
Solving
⎛
c 1
⎝ 3 ⎞ ⎛
3⎠ + c 2
⎝ 2 ⎞ ⎛
6⎠ + c 3
⎝ 1 ⎞ ⎛
0⎠ = ⎝ 4 ⎞
1⎠
1 1 5 2
gives the unique solution c 3 = 11/57, c 2 = −40/57 and c 1 = 99/57. Because c 1 > 1, the vector is not
in the box.
Four.II.1.10
Move the parallelepiped to start at the origin, so that it becomes the box formed by
( (
3 2
〈 , 〉
0)
1)
and now the absolute value of this determinant is easily computed as 3.
∣ 3 2
0 1∣ = 3
Four.II.1.11 (a) 3 (b) 9 (c) 1/9
Four.II.1.12 Express each transformation with respect to the standard bases and find the determinant.
(a) 6 (b) −1 (c) −5
Four.II.1.13
is 84.
The starting area is 6 and the matrix changes sizes by −14. Thus the area of the image
Four.II.1.14 By a factor of 21/2.
Four.II.1.15 For a box we take a sequence of vectors (as described in the remark, the order in which
the vectors are taken matters), while for a span we take a set of vectors. Also, for a box subset of R n
there must be n vectors; of course for a span there can be any number of vectors. Finally, for a box
the coefficients t 1 , . . . , t n are restricted to the interval [0..1], while for a span the coefficients are free
to range over all of R.
Four.II.1.16 That picture is drawn to mislead. The picture on the left is not the box formed by two
vectors. If we slide it to the origin then it becomes the box formed by this sequence.
( (
0 2
〈 , 〉
1)
0)
Then the image under the action of the matrix is the box formed by this sequence.
( (
1 4
〈 , 〉
1)
0)
which has an area of 4.
Four.II.1.17 Yes to both. For instance, the first is |T S| = |T | · |S| = |S| · |T | = |ST |.