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