Math 5327 Homework 6 Fall 2018 Solutions
1. Find the sign for each of the following permutations. (a) (5, 4, 3, 2, 1)
Solution: The sign is (−1)k, where the permutation can be put in order in k transposition steps, or it is the determinant of the corresponding permutation matrix. In this case, it is easy to see that the permutation can be put in proper order with just two transpositions: interchange 1 ↔ 5, and interchange 2 ↔ 4. Often it is hard to count so we use the involution number: for each i we count how many things to the right of i are smaller than i. This tells how many transpositions of adjacent numbers are needed to put the permutation in order. In this case, the inversion number is 4 + 3 + 2 + 1 = 10, which is again even, so the sign of the permutation is +1.
(b) (3, 7, 5, 2, 1, 8, 4, 6)
Solution: The inversion number is 2 + 5 + 3 + 1 + 0 + 2 + 0 = 13 so the sign is - 1.
(c) (2, 4, 6, 8, 7, 5, 3, 1)
Solution: The inversion number is 1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 so the sign is + 1.
2. Let U be the space of all functions from F n×n to F,V the set of all n-linear functions in U and W the set of all n-linear, alternating functions in U. (a) Show that V is a subspace of U.
Solution: I think I proved in class that a combination of n-linear functions is n-linear. Here is a proof anyway. Suppose that D and E are both n-linear. Let Di and Ei be the functions you get when you fix every row but the i’th row. If we show Di + kEi is linear, then we will have verified n-linearity. We have (Di + kEi)(u + cv) = Di(u + cv) + kEi(u + cv) = Di(u) + cDi(v) + kEi(u) + kcEi(v) = (Di + kEi)(u) + c(Di + kEi)(v), as desired, so V is a subspace.
(b) Show that W is a subspace of V.
Solution: As mentioned in class, an n-linear function D is alternating if D(A) = 0 whenever A has two equal rows. So let D and E be alternating. Given a ma- trix A with two equal rows, (D + kE)(A) = D(A) + kE(A) = 0, so D + kE is also alternating.
(c) What is dim(V )?
Solution: Given any n-linear function D, we can use linearity in the first row to write D(A) as a sum of n terms. We repeat this for each of the rows to
get D(A) as a sum of terms of the form a1,i1 ··· an,in D(M) where M has rows n ei1 , ··· , ein . The function D depends on the values of each of these n expres- sions D(M), so the vector space has dimension nn.
To be more formal, let f be a function from {1, 2, . . . , n} to {1, 2, . . . , n}. This function need not be one-to-one or onto. Given f we can define Df (A) = a1,f(1) ··· an,f(n), and this is n-linear. The above showed that the set of all Df form a spanning set for the set of all n-linear functions. That is, if D is n-linear then X D(A) = xf Df (A) f
where xf is D(Mf ), what D does to Mf , the matrix with rows ef(1), ··· , ef(n). n There are n such functions, Df , one for each possible f. These functions are P also independent. To show this, suppose that f cf Df = O, the 0-operator. If g is some function from {1, 2, . . . , n} to {1, 2, . . . , n} then consider Df (Mg). This takes the f(1) entry from the first row of Mg, the f(2) entry from the second row, and so on, and then multiplies all these together. If any one of the entries is 0, then Df (Mg) = 0. But the f(i) element of eg(i) is 0 unless f(i) = g(i). This means that Df (Mg) = 0 if f 6= g and Dg(Mg) = 1. As a P consequence, 0 = O(Mg) = cf Df (Mg) = cg. That is, every scalar in the P f linear combination f cf Df must be 0 if this combination gives the 0-operator, so all the Df are independent. Thus, the Df form a basis so the dimension of V is nn.
(d) What is dim(W )?
Solution: Once the alternating condition is added, the dimension drops greatly. Our theorem on determinants says that if D is any n-linear, alternating function, then for all A, D(A) = det(A)D(I). This means D is a scalar multiple of the determinant function (with D(I) as the scalar) so the determinant function forms a basis for W . That is, W is 1-dimensional.
Page 2 3. An n × n matrix A is called orthogonal if AAt = I. Show that if A is orthogonal then det(A) = ±1. Give an example of an orthogonal matrix with determinant -1.
Solution: Suppose that AAt = I. Then 1 = det(I) = det(AAt) = det(A) det(At) = 0 1 det(A) det(A), so det(A)2 = 1, and det(A) can only be 1 or -1. If A = then 1 0 det(A) = −1. In fact, this is the permutation matrix for the permutation (2, 1). In general, the permutation matrix for any odd matrix is orthogonal, with determinant −3/5 4/5 -1. An example of a matrix that is not a permutation matrix is . 4/5 3/5
4. Evaluate the following determinants. In the first case, give the factored form of the determinant. 1 1 1 (a) x2 y2 z2 x4 y4 z4
Solution:
1 1 1 1 1 1 2 2 4 4 2 2 4 4 x2 y2 z2 = 0 y2 − x2 z2 − x2 = (y − x )(z − x ) − (z − x )(y − x ),
x4 y4 z4 0 y4 − x4 z4 − x4 and work is still needed to factor this. It is trickier to get to the 2 × 2 stage and factor common stuff from the columns, using y4 − x4 = (y2 − x2)(y2 + x2). We have
2 2 2 2 y − x z − x 2 2 2 2 1 1 det = = (y − x )(z − x ) , y4 − x4 z4 − x4 y2 + x2 z2 + x2 so the determinant is
(y2 − x2)(z2 − x2)(z2 − y2) = (x − y)(x + y)(x − z)(x + z)(y − z)(y + z).
Page 3 1 1 1 1 2 3 5 7 (b) 4 9 25 49 8 27 125 343
Solution: This is a Vandermonde matrix so we could just plug into the formula to get (7 - 2)(7 - 3)(7 - 5)(5 - 3)(5 - 2)(3 - 2) = 240. Alternatively,
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 3 5 7 0 1 3 5 0 1 3 5 0 1 3 5 = = = = 240. 4 9 25 49 0 5 21 45 0 0 6 20 0 0 6 20
8 27 125 343 0 19 117 335 0 0 60 240 0 0 0 40
1 2 3 4 5 6 2 3 4 5 6 7 0 0 1 2 3 4 (c) 0 0 2 3 4 5 0 0 0 0 1 2 0 0 0 0 2 3
Solution: This is a block upper triangular matrix so the determinant is the 3 1 2 product of the determinants of the blocks on the diagonal, which gives , 2 3 or (−1)3 = −1.
1 1 1 5. Use the adjoint formula to invert A = 2 0 −1 . 3 4 2
4 −7 8 Solution: The cofactor matrix is C = 2 −1 −1 and the determinant is 5 so −1 3 −2
4 2 −1 1 1 A−1 = Ct = −7 −1 3 . 5 5 8 −1 −2
Page 4 6. Recall that the rank of a matrix is the dimension of its column space. (a) Suppose that A is an n × n matrix of rank n. Prove that Rank(adj(A)) = n.
−1 1 Solution: If A has rank n then A is invertible and A = det(A) adj(A). This means adj(A) = det(A)A−1, an invertible matrix, so adj(A) must have rank n.
(b) Suppose that A is an n × n matrix and rk(A) < n − 1. Show that adj(A) is the matrix of all zeros.
Solution: The rank of a matrix is the dimension of the column space (or row space) of that matrix. If the rank is at most n − 2 then any n − 1 columns are linearly dependent. Deleting the i’th entry from each column will not change this. Consequently, every cofactor will be 0 since the determinant of a matrix with dependent columns is 0.
(c) If A is an n × n matrix and rk(A) = n − 1. Show that rk(adj(A)) = 1.
Solution: We have A · adj(A) = 0, so if v is any column of the adjoint, Av = 0. This means every column of the adjoint is in the null space of A. However, A has rank n − 1 so it has nullity 1. Since the null space is 1-dimensional every column of the adjoint is a multiple of any nonzero column. The rank will be 1 as long as the adjoint contains even a single nonzero entry. Since the rank of A is n−1,A has n−1 linearly independent columns. This means some submatrix of size n − 1 is nonsingular, so some cofactor is not 0. Consequently, at least one entry in the adjoint is nonzero, so the rank can’t be 0.
7. Find, with proof, the determinants of each of the following sequences of matrices. 2 1 0 0 2 1 0 2 1 1 2 1 0 (a) M = (2),M = ,M = 1 2 1 ,M = , ··· . 1 2 1 2 3 4 0 1 2 1 0 1 2 0 0 1 2
Solution: We calculate det(M1) = 2, det(M2) = 3, det(M3) = 4, so we should guess det(Mn) = n + 1. Now we have to show this. One way is to reduce A to R, while not using interchanges or multiplying a row by a scalar. In that case, det A = det R, and maybe det R is easier. Let’s find R for M3. We have 2 1 0 2 1 0 2 1 0 3 4 1 2 1 ⇒ 0 3/2 1 ⇒ 0 3/2 1 , so det R = 2 × 2 × 3 = 4. In 0 1 2 0 1 2 0 0 4/3 fact, in general, R for Mn will be upper triangular with 1’s on the first super
Page 5 3 4 n+1 diagonal and 2, 2 , 3 ,... n on the diagonal. I will prove this by induction on the rows of the matrix. Certainly the first row of Mn will be (2, 1, 0, ..., 0). Suppose 0 k+1 the k th row in R is (0, 0, ..., 0, k , 1, 0, ..., 0). As we do row reduction, we will reach a point where we’ve reduced the k0th row but not the one below. At this 0 0 ··· 0 k+1 1 0 ··· 0 point, these two rows look like this: k . We 0 0 ··· 0 1 2 1 ··· 0 k multiply the top row by − k+1 and add to the bottom row. The key position looks k k+2 like this: 2− k+1 = k+1 . This means that after one more row reduction step these k+1 0 0 ··· 0 k 1 0 ··· 0 two rows look like: k+2 . Consequently, when 0 0 ··· 0 0 k+1 1 ··· 0 the k0th row has the right form, the next row will also have the right form. By 3 4 n+1 induction, all rows will be as advertised and det Mn = 2× 2 × 3 ×· · ·× n = n+1.
A second solution goes like this: (*) det Mn = 2 det Mn−1 −det Mn−2. For ex-
2 1 0 0 2 1 0 1 1 0 2 1 0 1 2 1 0 ample, when n = 4 we have = 2 1 2 1 + 0 2 1 = 2 1 2 1 − 0 1 2 1 0 1 2 0 1 2 0 1 2 0 0 1 2
2 1 = 2 det M3 − det M2. This works because if we do a cofactor expansion 1 2 across the top row, then the first term will be 2 det Mn−1 and the second term will be (some matrix). But if we expand down the first column of that second matrix, we get − det Mn−2. Now we can use formula (*) and given another in- ductive proof that the formula is correct. The proof goes like this: We’ve shown det Mn = n + 1 for small n (for n = 1, 2, 3.) If the formula was correct for Mn−1 and Mn−2, that is if we know det Mn−1 = n and det Mn−2 = n − 1, then 0 det Mn = 2 det Mn−1 − det Mn−2 = 2n − (n − 1) = n + 1, establishing the n th case as well. By induction, the formula holds for all n.
A third approach is to try to show directly that det(Mn) = det(Mn−1) + 1. One way to do this, for the case of M4 is to write
2 1 0 0 1 + 1 1 0 0 1 1 0 0 1 1 0 0
1 2 1 0 0 + 1 2 1 0 0 2 1 0 1 2 1 0 = = + , 0 1 2 1 0 + 0 1 2 1 0 1 2 1 0 1 2 1
0 0 1 2 0 + 0 0 1 2 0 0 1 2 0 0 1 2 by using linearity in the first column. The first matrix on the right hand side has determinant det(Mn−1), and for the second,
1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0
1 2 1 0 0 1 1 0 0 1 1 0 0 1 1 0 = = = = 1. 0 1 2 1 0 1 2 1 0 0 1 1 0 0 1 1
0 0 1 2 0 0 1 2 0 0 1 2 0 0 0 1
Page 6 To be more formal with these last two approaches, both can be done in a similar manor: If we write the first column of Mn as 2e1 + e2, and use linearity, then 0 00 det(Mn) = det(Mn) + det(Mn ), where these new matrices have 2e1 and e2 as 0 their first columns. With 2e1 as its first column, det(Mn) = 2 det(Mn−1). For 00 Mn , a cofactor expansion across the first row, and then down the first column 00 give det(Mn ) = − det(Mn−2). In the third approach, you use the fact that the first column of Mn can be written as e1 +(e1 +e2) instead. Now the new matrices have determinants det(Mn−1) and 1, respectively.
1 −1 0 0 1 −1 0 1 −1 1 1 −1 0 (b) M = (1),M = ,M = 1 1 −1 ,M = , ··· . 1 2 1 1 3 4 0 1 1 −1 0 1 1 0 0 1 1
Solution: If we calculate the first several determinants, we get 1, 2, 3, 5. This might be enough to guess that the determinants follow the Fibonacci sequence. In particular, if F0 = 0,F1 = 1,Fn = Fn−1 + Fn−2, then det Mn = Fn+1. To actually prove this, we follow the second approach used in part (a) to get det Mn = det Mn−1 + det Mn−2, so these determinants satisfy the Fibonacci relation. We get this pattern by using linearity in the first column, with that column being e1 +e2, and then doing a cofactor expansion down the first column in one case, and across the first row and down the first column for the second. Since the pattern starts correct, it must stay correct.
For extra credit:
8. Find AND PROVE a formulas for det(Mn) in the following cases. 1 2 3 4 1 2 3 1 2 −1 1 2 3 (a) M = (1),M = ,M = −1 1 2 ,M = , ··· . 1 2 −1 1 3 4 −2 −1 1 2 −2 −1 1 −3 −2 −1 1
n(n + 1) Solution: I have a proof that det(M ) = 1 + 2 + ··· + n = , but I n 2 would like a better one.
Page 7 3 −2 0 0 3 −2 0 3 −2 −1 3 −2 0 (b) M = (3),M = ,M = −1 3 −2 ,M = , ··· . 1 2 −1 3 3 4 0 −1 3 −2 0 −1 3 0 0 −1 3
Solution: Here, the first several values are 3, 7, 15, 31, which might be enough n+1 for us to guess det Mn = 2 − 1. To show the guess is right, doing a cofactor expansion down the first column will tell us det Mn = 3 det Mn−1 − 2 det Mn−2, and by induction, knowing the formula is true for two preceding matrices, we n n−1 n n−1 n n have det Mn = 3(2 − 1) − 2(2 − 1) = 3 · 2 − 2 · 2 − 1 = 3 · 2 − 2 − 1 = 2 · 2n − 1 = 2n+1 − 1. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 (c) M1 = (1),M2 = 1 ,M3 = 1 2 2 ,M4 = 1 1 1 , ··· . 1 2 1 1 1 2 3 3 1 2 3 1 1 1 1 2 3 4 Solution: Row reduce as follows: subtract the (n−1)st row from the n’th row, then subtract the (n−2)nd row from the (n−1)st, and so on, finally subtracting the first row from the second. For M4, for example, we would get
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 0 2 − 1 2 − 1 2 − 1 1 1 1 ⇒ 1 1 1 1 1 2 3 3 0 0 3 − 2 3 − 2 1 1 1 1 1 1 2 3 4 0 0 0 4 − 3 We see that in general, the determinant will be
1 1 1 1 1 1 − 1 − ··· − 2 3 2 n − 1 n
1 1 −1 Now k−1 − k = (k−1)k so this becomes
−1 −1 −1 −1 (−1)n−1 × × × · · · × = . 1 · 2 2 · 3 3 · 4 (n − 1)n (n − 1)! n!
Page 8 9. For the matrices in problem 7, can you find an prove the form of the Dodgson conden- sation matrices for each Mn? That is, what do the condensed matrices look like at each stage? Can you prove your answer?
Solution: One needs to modify Dodgson’s condensation algorithm. For band ma- trices, matrices in which only the main diagonal and certain sub and super diagonals can have nonzero elements, you can just let the condensed matrix also be a band matrix, and not worry about dividing by 0’s outside the band. For example, with 2 1 0 0 0 1 2 1 0 0 M5, from 5a, M5 = 0 1 2 1 0 , there are two internal 0’s, in the (2, 4) and 0 0 1 2 1 0 0 0 1 2 (4, 2) positions. But you never have to divide by these if you assume the condensed matrices will be band matrices as well. Here is the condensation: 2 1 0 0 0 3 1 0 0 1 2 1 0 0 4 1 0 1 3 1 0 5 1 0 1 2 1 0 ⇒ ⇒ 1 4 1 ⇒ ⇒ 6. 0 1 3 1 1 5 0 0 1 2 1 0 1 4 0 0 1 3 0 0 0 1 2
This pattern, that each condensed matrix is a band matrix with k + 1 on the diag- onal if k was on the diagonal in the previous stage, and 1’s are on the first super and sub diagonals is real. For the 1’s, you will always be calculating a determinant 1 0 1 x of or , and the determinant has to be divided by the central ele- x 1 0 1 ment in the previous stage, which was also 1. For the diagonal elements, you have a k + 1 1 determinant , where the previous central element was k, leading to 1 k + 1 1 2 1 2 k ((k + 1) − 1) = k (k + 2k) = k + 2.
For the Fibonacci matrix, the condensation, say for M5 would be as follows: 1 −1 0 0 0 2 1 0 0 1 1 −1 0 0 3 −1 0 1 2 1 0 5 1 0 1 1 −1 0 → → 1 3 −1 → → 8 0 1 2 1 1 5 0 0 1 1 −1 0 1 3 0 0 1 2 0 0 0 1 1
The following appears to happen: First sub diagonal consists of all 1’s at each stage. This is easy to show. The first super diagonal consists of all entries (−1)k+1, at stage k of the condensation. Again, this is fairly easy to show. Finally, the diagonal has Fk+2 on the diagonal at stage k. Since there are n−1 stages of condensation, the final
Page 9 answer would be Fn+1, as required. Why the Fibonacci numbers on the diagonal? This would require a Fibonacci identity: when going from stage k − 1 to stage k, the appropriate 2 × 2 determinant is divided by Fk, the diagonal entry at stage k − 2. We need 2 k Fk+1 − (−1) = Fk+2. Fk 2 k This is equivalent to Fk+1 − FkFk+2 = (−1) , a standard Fibonacci identity.
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