Linear Algebra Exercises-n-Answers.pdf

166 Linear Algebra, by Hefferon
Four.I.3.30
Let T be n×n, let J be p×p, and let K be q×q. Apply the permutation expansion formula
∑
|T | =
t 1,φ(1) t 2,φ(2) . . . t n,φ(n) |P φ |
permutations φ
Because the upper right of T is all zeroes, if a φ has at least one of p+1, . . . , n among its first p column
numbers φ(1), . . . , φ(p) then the term arising from φ is 0 (e.g., if φ(1) = n then t 1,φ(1) t 2,φ(2) . . . t n,φ(n)
is 0). So the above formula reduces to a sum over all permutations with two halves: first 1, . . . , p are
rearranged, and after that comes a permutation of p+1, . . . , p+q. To see this gives |J|·|K|, distribute.
[
∑
] [
∑
]
t 1,φ1(1) · · · t p,φ1(p) |P φ1 | ·
t p+1,φ2(p+1) · · · t p+q,φ2(p+q) |P φ2 |
perms φ 1
of 1,...,p
perms φ 2
of p+1,...,p+q
Four.I.3.31 The n = 3 case shows what happens.
t 1,1 − x t 1,2 t 1,3
|T − rI| =
t 2,1 t 2,2 − x t 2,3
∣ t 3,1 t 3,2 t 3,3 − x∣
Each term in the permutation expansion has three factors drawn from entries in the matrix (e.g.,
(t 1,1 − x)(t 2,2 − x)(t 3,3 − x) and (t 1,1 − x)(t 2,3 )(t 3,2 )), and so the determinant is expressible as a
polynomial in x of degree 3. Such a polynomial has at most 3 roots.
In general, the permutation expansion shows that the determinant can be written as a sum of
terms, each with n factors, giving a polynomial of degree n. A polynomial of degree n has at most n
roots.
Four.I.3.32 This is how the answer was given in the cited source. When two rows of a determinant are
interchanged, the sign of the determinant is changed. When the rows of a three-by-three determinant
are permuted, 3 positive and 3 negative determinants equal in absolute value are obtained. Hence the
9! determinants fall into 9!/6 groups, each of which sums to zero.
Four.I.3.33 This is how the answer was given in the cited source. When the elements of any column
are subtracted from the elements of each of the other two, the elements in two of the columns of the
derived determinant are proportional, so
∣
the determinant ∣
vanishes. That is,
2 1 x − 4
∣∣∣∣∣ 1 x − 3 −1
∣∣∣∣∣ 4 2 x − 3
∣6 3 x − 10∣ x − 2 −1 −2
2 x − 1 −2
3 x − 7 −3∣ x + 1 −2 −4
x − 4 −3 −6∣ Four.I.3.34 This is how the answer was given in the cited source. Let
a b c
d e f
g h i
have magic sum N = S/3. Then
N = (a + e + i) + (d + e + f) + (g + e + c)
− (a + d + g) − (c + f + i) = 3e
and S = 9e. Hence, adding rows and columns,
∣ ∣ ∣ a b c
∣∣∣∣∣ a b c
∣∣∣∣∣
D =
d e f
∣g h i ∣ = a b 3e
∣∣∣∣∣
d e f
3e 3e 3e∣ = a b e
d e 3e
3e 3e 9e∣ = d e e
1 1 1∣ S.
Four.I.3.35 This is how the answer was given in the cited source. Denote by D n the determinant in
question and by a i,j the element in the i-th row and j-th column. Then from the law of formation of
the elements we have
a i,j = a i,j−1 + a i−1,j , a 1,j = a i,1 = 1.
Subtract each row of D n from the row following it, beginning the process with the last pair of rows.
After the n − 1 subtractions the above equality shows that the element a i,j is replaced by the element
a i,j−1 , and all the elements in the first column, except a 1,1 = 1, become zeroes. Now subtract each
column from the one following it, beginning with the last pair. After this process the element a i,j−1
is replaced by a i−1,j−1 , as shown in the above relation. The result of the two operations is to replace
a i,j by a i−1,j−1 , and to reduce each element in the first row and in the first column to zero. Hence
D n = D n+i and consequently
D n = D n−1 = D n−2 = · · · = D 2 = 1.