Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 29
One.III.2.13 (a) They have the form (
a
)
0
b 0
where a, b ∈ R.
(b) They have this form (for a, b ∈ R). (
1a
)
2a
1b 2b
(c) They have the form (
a
)
b
c d
(for a, b, c, d ∈ R) where ad − bc ≠ 0. (This is the formula that determines when a 2×2 matrix is
nonsingular.)
One.III.2.14 Infinitely many. For instance, in (1 ) k
0 0
each k ∈ R gives a different class.
One.III.2.15
No. Row operations do not change the size of a matrix.
One.III.2.16 (a) A row operation on a zero matrix has no effect. Thus each zero matrix is alone in
its row equivalence class.
(b) No. Any nonzero entry can be rescaled.
One.III.2.17 Here are two. ( ) 1 1 0
and
0 0 1
One.III.2.18
matrix with all 0’s except for 1’s down the diagonal.
⎛
( 1 0
) 0
0 0 1
Any two n×n nonsingular matrices have the same reduced echelon form, namely the
⎞
1 0 0
0 1 0
⎜
⎝
. ..
⎟
⎠
0 0 1
Two 2×2 singular matrices need not
(
row
)
equivalent.
1 1
and
0 0
(
1 0
0 0
One.III.2.19 Since there is one and only one reduced echelon form matrix in each class, we can just
list the possible reduced echelon form matrices.
For that list, see the answer for Exercise 11.
One.III.2.20 (a) If there is a linear relationship where c 0 is not zero then we can subtract c 0β0 ⃗ and
divide both sides by c 0 to get β ⃗ 0 as a linear combination of the others. (Remark. If there are no
others — if the relationship is, say, ⃗0 = 3 · ⃗0 — then the statement is still true because zero is by
definition the sum of the empty set of vectors.)
If β ⃗ 0 is a combination of the others β ⃗ 0 = c 1β1 ⃗ + · · · + c nβn ⃗ then subtracting β ⃗ 0 from both sides
gives a relationship where one of the coefficients is nonzero, specifically, the coefficient is −1.
(b) The first row is not a linear combination of the others for the reason given in the proof: in the
equation of components from the column containing the leading entry of the first row, the only
nonzero entry is the leading entry from the first row, so its coefficient must be zero. Thus, from the
prior part of this question, the first row is in no linear relationship with the other rows. Hence, to
see if the second row can be in a linear relationship with the other rows, we can leave the first row
out of the equation. But now the argument just applied to the first row will apply to the second
row. (Technically, we are arguing by induction here.)
One.III.2.21 (a) As in the base case we will argue that l 2 isn’t less than k 2 and that it also isn’t
greater. To obtain a contradiction, assume that l 2 ≤ k 2 (the k 2 ≤ l 2 case, and the possibility that
either or both is a zero row, are left to the reader). Consider the i = 2 version of the equation
that gives each row of B as a linear combination of the rows of D. Focus on the l 1 -th and l 2 -th
component equations.
b 2,l1 = c 2,1 d 1,l1 + c 2,2 d 2,l1 + · · · + c 2,m d m,l1 b 2,l2 = c 2,1 d 1,l2 + c 2,2 d 2,l2 + · · · + c 2,m d m,l2
)