Linear Algebra Exercises-n-Answers.pdf
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Linear Algebra Exercises-n-Answers.pdf

Linear Algebra Exercises-n-Answers.pdf Linear Algebra Exercises-n-Answers.pdf

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174 Linear Algebra, by Hefferon when deleted, leave an upper triangular minor, because entry i, j of the minor is either entry i, j of M (this happens if a > i and b > j; in this case i < j implies that the entry is zero) or it is entry i, j + 1 of M (this happens if i < a and j > b; in this case, i < j implies that i < j + 1, which implies that the entry is zero), or it is entry i + 1, j + 1 of M (this last case happens when i > a and j > b; obviously here i < j implies that i + 1 < j + 1 and so the entry is zero). Thus the determinant of the minor is the product down the diagonal. Observe that the a − 1, a entry of M is the a − 1, a − 1 entry of the minor (it doesn’t get deleted because the relation a > b is strict). But this entry is zero because M is upper triangular and a − 1 < a. Therefore the cofactor is zero, and the adjoint is upper triangular. (The lower triangular case is similar.) (b) This is immediate from the prior part, by Corollary 1.11. Four.III.1.27 We will show that each determinant can be expanded along row i. The argument for column j is similar. Each term in the permutation expansion contains one and only one entry from each row. As in Example 1.1, factor out each row i entry to get |T | = t i,1 · ˆT i,1 + · · · + t i,n · ˆT i,n , where each ˆT i,j is a sum of terms not containing any elements of row i. We will show that ˆT i,j is the i, j cofactor. Consider the i, j = n, n case first: t n,n · ˆT n,n = t n,n · ∑ t 1,φ(1) t 2,φ(2) . . . t n−1,φ(n−1) sgn(φ) φ where the sum is over all n-permutations φ such that φ(n) = n. To show that ˆT i,j is the minor T i,j , we need only show that if φ is an n-permutation such that φ(n) = n and σ is an n − 1-permutation with σ(1) = φ(1), . . . , σ(n − 1) = φ(n − 1) then sgn(σ) = sgn(φ). But that’s true because φ and σ have the same number of inversions. Back to the general i, j case. Swap adjacent rows until the i-th is last and swap adjacent columns until the j-th is last. Observe that the determinant of the i, j-th minor is not affected by these adjacent swaps because inversions are preserved (since the minor has the i-th row and j-th column omitted). On the other hand, the sign of |T | and ˆT i,j is changed n − i plus n − j times. Thus ˆT i,j = (−1) n−i+n−j |T i,j | = (−1) i+j |T i,j |. Four.III.1.28 This is obvious for the 1×1 base case. For the inductive case, assume that the determinant of a matrix equals the determinant of its transpose for all 1×1, . . . , (n−1)×(n−1) matrices. Expanding on row i gives |T | = t i,1 T i,1 +. . . +t i,n T i,n and expanding on column i gives |T trans | = t 1,i (T trans ) 1,i +· · ·+t n,i (T trans ) n,i Since (−1) i+j = (−1) j+i the signs are the same in the two summations. Since the j, i minor of T trans is the transpose of the i, j minor of T , the inductive hypothesis gives |(T trans ) i,j | = |T i,j |. Four.III.1.29 This is how the answer was given in the cited source. Denoting the above determinant by D n , it is seen that D 2 = 1, D 3 = 2. It remains to show that D n = D n−1 + D n−2 , n ≥ 4. In D n subtract the (n − 3)-th column from the (n − 1)-th, the (n − 4)-th from the (n − 2)-th, . . . , the first from the third, obtaining 1 −1 0 0 0 0 . . . 1 1 −1 0 0 0 . . . F n = 0 1 1 −1 0 0 . . . . 0 0 1 1 −1 0 . . . ∣. . . . . . . . . ∣ By expanding this determinant with reference to the first row, there results the desired relation. Topic: Cramer’s Rule 1 (a) x = 1, y = −3 (b) x = −2, y = −2 2 z = 1 3 Determinants are unchanged by pivots, including column pivots, so det(B i ) = det(⃗a 1 , . . . , x 1 ⃗a 1 +· · ·+ x i ⃗a i + · · · + x n ⃗a n , . . . ,⃗a n ) is equal to det(⃗a 1 , . . . , x i ⃗a i , . . . ,⃗a n ) (use the operation of taking −x 1 times the first column and adding it to the i-th column, etc.). That is equal to x i · det(⃗a 1 , . . . ,⃗a i , . . . ,⃗a n ) = x i · det(A), as required.

Answers to Exercises 175 4 Because the determinant of A is nonzero, Cramer’s Rule applies and shows that x i = |B i |/1. Since B i is a matrix of integers, its determinant is an integer. 5 The solution of ax +by = e cx +dy = f is ed − fb x = y = ad − bc provided of course that the denominators are not zero. af − ec ad − bc 6 Of course, singular systems have |A| equal to zero, but the infinitely many solutions case is characterized by the fact that all of the |B i | are zero as well. 7 We can consider the two nonsingular cases together with this system x 1 + 2x 2 = 6 x 1 + 2x 2 = c where c = 6 of course yields infinitely many solutions, and any other value for c yields no solutions. The corresponding vector equation ( ( ( 1 2 6 x 1 · + x 1) 2 · = 2) c) gives a picture of two overlapping vectors. Both lie on the line y = x. In the c = 6 case the vector on the right side also lies on the line y = x but in any other case it does not. Topic: Speed of Calculating Determinants 1 (a) Under Octave, rank(rand(5)) finds the rank of a 5×5 matrix whose entries are (uniformily distributed) in the interval [0..1). This loop which runs the test 5000 times octave:1> for i=1:5000 > if rank(rand(5)) endfor produces (after a few seconds) returns the prompt, with no output. The Octave script function elapsed_time = detspeed (size) a=rand(size); tic(); for i=1:10 det(a); endfor elapsed_time=toc(); endfunction lead to this session. octave:1> detspeed(5) ans = 0.019505 octave:2> detspeed(15) ans = 0.0054691 octave:3> detspeed(25) ans = 0.0097431 octave:4> detspeed(35) ans = 0.017398 (b) Here is the data (rounded a bit), and the graph. matrix rows 15 25 35 45 55 65 75 85 95 time per ten 0.0034 0.0098 0.0675 0.0285 0.0443 0.0663 0.1428 0.2282 0.1686 (This data is from an average of twenty runs of the above script, because of the possibility that the randomly chosen matrix happens to take an unusually long or short time. Even so, the timing cannot be relied on too heavily; this is just an experiment.)

<strong>Answers</strong> to <strong>Exercises</strong> 175<br />

4 Because the determinant of A is nonzero, Cramer’s Rule applies and shows that x i = |B i |/1. Since<br />

B i is a matrix of integers, its determinant is an integer.<br />

5 The solution of<br />

ax +by = e<br />

cx +dy = f<br />

is<br />

ed − fb<br />

x = y =<br />

ad − bc<br />

provided of course that the denominators are not zero.<br />

af − ec<br />

ad − bc<br />

6 Of course, singular systems have |A| equal to zero, but the infinitely many solutions case is characterized<br />

by the fact that all of the |B i | are zero as well.<br />

7 We can consider the two nonsingular cases together with this system<br />

x 1 + 2x 2 = 6<br />

x 1 + 2x 2 = c<br />

where c = 6 of course yields infinitely many solutions, and any other value for c yields no solutions.<br />

The corresponding vector equation<br />

( ( ( 1 2 6<br />

x 1 · + x<br />

1)<br />

2 · =<br />

2)<br />

c)<br />

gives a picture of two overlapping vectors. Both lie on the line y = x. In the c = 6 case the vector on<br />

the right side also lies on the line y = x but in any other case it does not.<br />

Topic: Speed of Calculating Determinants<br />

1 (a) Under Octave, rank(rand(5)) finds the rank of a 5×5 matrix whose entries are (uniformily<br />

distributed) in the interval [0..1). This loop which runs the test 5000 times<br />

octave:1> for i=1:5000<br />

> if rank(rand(5)) endfor<br />

produces (after a few seconds) returns the prompt, with no output.<br />

The Octave script<br />

function elapsed_time = detspeed (size)<br />

a=rand(size);<br />

tic();<br />

for i=1:10<br />

det(a);<br />

endfor<br />

elapsed_time=toc();<br />

endfunction<br />

lead to this session.<br />

octave:1> detspeed(5)<br />

ans = 0.019505<br />

octave:2> detspeed(15)<br />

ans = 0.0054691<br />

octave:3> detspeed(25)<br />

ans = 0.0097431<br />

octave:4> detspeed(35)<br />

ans = 0.017398<br />

(b) Here is the data (rounded a bit), and the graph.<br />

matrix rows 15 25 35 45 55 65 75 85 95<br />

time per ten 0.0034 0.0098 0.0675 0.0285 0.0443 0.0663 0.1428 0.2282 0.1686<br />

(This data is from an average of twenty runs of the above script, because of the possibility that<br />

the randomly chosen matrix happens to take an unusually long or short time. Even so, the timing<br />

cannot be relied on too heavily; this is just an experiment.)

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