Download Solutions to Homework 6

Homework 5 M 373K
Mark Lindberg and Travis Schedler
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1. Artin, Chapter 3, Exercise 2.1. Prove that the numbers of the form a + b 2, where a and
b are rational numbers, form a subfield of C.
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Let F be the numbers of the form a + b 2, where a and b are rational numbers. Then
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• Let a + b 2 and c + d 2 be in F . Then a + c + (b + d) 2 ∈ F because the rationals
are closed under addition.
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• Let a + b 2 ∈ F . Then −a − b 2 ∈ F , and a + b 2 + (−a − b 2) = 0.
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• Let a+b 2 and c+d 2 be in F . Then (a+b 2)(c+d 2) = ac+2bd+(ad+bc) 2 ∈ F ,
because the rationals are closed under addition and multiplication.
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√
√
√
2
a
b
2 ∈ F , because
• Let a+b 2 ∈ F , a+b 2 6= 0. Then a+b1√2 = aa−b
2 −2b2 = a2 −2b2 − a2 −2b2
the rationals are closed under multiplication, subtraction, and division.a2 − 2b2 6= 0,
±a
because if it were, b = √
, which is not rational.
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• a = 1, b = 0 ⇒ 1 ∈ F .
Thus, since all necessary properties hold, F is a subfield of C. 2. Artin, Chapter 3, Exercise 2.3. Compute the product polynomial (x3 + 3x2 + 3x + 1)(x4 +
4x3 + 6x2 + 4x + 1) when the coefficients are regarded as elements of the field F7 . Explain
your answer.
(x3 + 3x2 + 3x + 1)(x4 + 4x3 + 6x2 + 4x + 1) = (x + 1)3 (x + 1)4 = (x + 1)7 = x7 + 7x6 +
21x5 + 35x4 + 35x3 + 21x2 + 7x + 1 = x7 + 1, because 0 = 7 = 21 = 35 in F7 .
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Alternative answer: This equals
(x + 1)
, and in general by the binomial theorem, for p
Pp−1
p i p−i
p
p
any
prime, (a + b) = a + b + i=1 i a b , with the last term vanishing in Fp because
p
i is a multiple of p if 1 ≤ i ≤ p − 1. Therefore the “Freshman’s Dream” holds for p-th
powers in Fp , that is, (a + b)p = ap + bp . Thus in our case, we get x7 + 17 = x7 + 1 in F7 .
3. Artin, Chapter 3, Exercise 2.5. Determine the primes p such that the matrix


1 2
0
A =  0 3 −1
−2 0
2
is invertible, when its entries are considered to be in Fp .
0 −1
3 −1
= 6 − 2(−2) = 6 + 4 = 10 = 2 · 5. 2 = 0 only in F2 and
− 2
det(A) = 1 −2
2
0
2
F5 , for p prime, so for all primes p, such that p 6= 2 and p 6= 5, det(A) 6= 0, and so A
will be invertible. Conversely, in F2 or F5 , the determinant is zero, so the matrix is not
invertible. 4. Artin, Chapter 3, Exercise
2.10.
Interpreting
matrix entries in the field F2 , prove that the
0 0
1 0
1 1
0 1
four matrices
,
,
,
form a field. Note: This gives one way to
0 0
0 1
1 0
1 1
define the (unique up to isomorphism) field of order four!
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We already know that Mat2 (F2 ) is a ring, so we only have to show that these elements give
a subring (they are closed under addition and multiplication and include 0 and 1), that
this subring is commutative, that all nonzero elements in the subring have multiplicative
inverses, and that in this ring 1 6= 0, i.e., the ring is not the zero ring.
0 0
1 0
1 1
0 1
Let 0 =
,I=
,C=
,D=
. Then
0 0
0 1
1 0
1 1
• 0 is the additive identity and I the multiplicative identity.
• Closure under addition: Note that 0 + X = X and X + X = 0 for all X. So we only
have to check I + C = D, I + D = C, C + D = I, and we are closed under addition.
• Closure under multiplication: Since 0X = X0 = 0 for all X and IX = X = XI for
all X, we only have to consider CD, DC, C 2 , and D2 . But C 2 = D and CD = I,
hence also C 3 = I, and we get DC = C 2 C = I and D2 = C 4 = C.
• Commutativity: we conclude also from the previous point that, since D = C 2 , C and
D commute, and everything always commutes with I and 0.
• Inverses: Again, we saw C and D were inverses, and I is its own iverse.
• The identity I 6= 0 is clear.
Thus, all of the properties are fulfilled, and we have formed a field of order 4.
5. Artin, Chapter 3, Exercise 3.2. Which of the following subsets is a subspace of the vector
space F n×n of n × n matrices with coefficients in F ?
(a) Symmetric matrices (A = At ).
• Let A, B be such that A = At and B = B t . Then (A + B)t = At + B t = A + B,
and so the symmetric matrices are closed under addition.
• Let A be a symmetric matrix. Then (cA)t = cAt = cA, and so the symmetric
matrices are closed under scalar multiplication.
• 0t = 0, so the zero matrix is symmetric.
Therefore, the symmetric matrices are a subspace of the vector space F n×n of n × n.
(b) Invertible matrices
By a property shown earlier, a matrix A is invertible if and only if det(A) 6= 0. But
det(0) = 0, and so 0 is not invertible, and the invertible matrices are not a subspace.
(c) Upper triangular matrices.
• Upper triangular matrices are closed under addition, by observation.
• Since c0 = 0 for any c ∈ R, again by observation, the upper triangular matrices
are closed under scalar multiplication.
• 0 is an upper triangular matrix by definition.
Therefore, the upper triangular matrices are a subspace of the vector space F n×n of
n × n.
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6. Artin, Chapter 3, Exercise 4.2. Let W
⊂ R bethe space of solutions of the system of
2 1 2 3
linear equations AX = 0, where A =
. Find a basis for W .
1 1 3 0
2
1 0 −1
3
Row reduction on A yields B =
. By an earlier result, the set of solutions
0 1
4 −3
to AX = 0 and BX = 0 must be the same, so we can now consider this new case. Let
 
x1
x2 

X=
x3  .
x4
Then x1 − x3 + 3x4 = 0 and x2 + 4x3 − 3x4 = 0, or x1 = x3 − 3x4 and x2 = −4x3 + 3x4 .
Thus, our solutions must be of the form


x3 − 3x4
−4x3 + 3x4 
.
X=


x3
x4
Letting x3 = 0, x4 = 1 and x3 = 1, x4 = 1 by turns, we get the set
   
1
−3 


   

−4
  ,  3 .
 1  0





0
1
We can see that this obviously spans the set of solutions, as
 
  

1
−3
x3 − 3x4
−4
 3 −4x3 + 3x4 

  
.
x3 
 1 = x4  0 = 

x3
0
1
x4
Then we show that they are linearly independent. Let x3 , x4 be such that
 
  

1
−3
x3 − 3x4
−4
  

 = x4  3 = −4x3 + 3x4  = 0.
x3 
 1
 0 

x3
0
1
x4
Then x3 = x4 = 0 by the last two entries, and so by definition, they must be linearly
independent. Then, by a property from the book, the set is a basis for W . 7. Artin, Chapter 3, Exercise 4.3. Prove that the three functions x2 , cos(x), and ex are
linearly independent.
Let a, b, c ∈ R such that ax2 + b cos(x) + cex = 0 ∀x ∈ R. Then it also holds after
we differentiate. So differentiating twice, −b cos(x) + cex = 0 as well. Adding, we get
ax2 + 2cex = 0. Differentiating twice, we get 2cex = 0. Thus c = 0. Then a = 0 from
ax2 + 2cex = 0 and b = 0 from −b cos(x) + cex = 0. So, by the definition of linear
indepenencen, x2 , cos(x), and ex are linearly independent. 3
8. Artin, Chapter 3, Exercise 4.5. Let V = F n be the space of column vectors. Prove that
every subspace W of V is the space of solutions of some system of homogeneous linear
equations AX = 0.
Let C be an n × m matrix of whose columns span a subspace W of V . Note that rk(C) =
dim W (as we can see by taking the columns to actually be a basis, so that row reduction on
C t gives no zero rows hence rk(C t ) is the size of the basis, i.e., dim W , and rk(C) = rk(C t )).
Then consider the homogeneous system C t y = 0, where y is a column vector of length n
since C t is an m × n matrix. Let U be the vector space of solutions to this system. By
reducing C to reduced row-echelon form, we see that U is spanned by n − rk(C) vectors,
and in fact these are linearly independent. Thus dim U = n − rk(C) = n − dim W . Let B
be an n × k matrix whose columns span U . Again rk(B) = dim U . Let W 0 be the space of
solutions to the system B t x = 0. Again, we see that dim(W 0 ) = n − rk(B) = n − dim U =
n − (n − dim W ) = dim W . On the other hand it is clear that W ⊆ W 0 , and since the
dimensions are equal, we conclude W = W 0 . Therefore W is actually the vector space of
solutions to the system B t x = 0. 9. Compute the size of GLn (Fp ).
Observe from an earlier property (chapter 1, and linear algebra courses) of linear algebra
that an n × n matrix is invertible if and only if all columns are linearly independent. Then
we consider the column of GLn (Fp ). The first column must be non-zero, but there are p
possible different value for the first entry, p for the second entry, and so on, up to p for
the nth entry. Since these are independent, by a property of counting, we multiply them,
to obtain a total of pn possible column vectors, and pn − 1 non-zero column vectors. The
first column can be any of these. Then consider the second column. There are p multiples
of the first column, and still pn total possible column vectors. Since the zero vector is
included in the p multiples of the first column, there are pn − p non-zero column vectors
which are not multiples of, and thus are linearly independent from, the first column. This
process continues for every column, so let us consider the kth column in the general sense.
There are pn possible columns, and we have pk−1 previous columns which are all linearly
independent. This means that there are p multiples of each of these columns, and so
there are a total of pk−1 possible combinations of these columns–essentially, the size of
their span. Then we have pn − pk−1 possible values for column k. Since each column
is chosen separately, and we have column 1 and column 2, and so on, there are a total
n
Q
of
(pn − pk−1 ) possible matrices in which every column is linearly independent, which
k=1
means that there are the same number of matrices in GLn (Fp ).
Bonus: Do the same for GLn (Fpk ). (It is not any more work!) What about SLn (Fpk )?
All of the above statements still hold for elements of GLn (Fpk ), and so we can apply the
n
Q
same count to say that | GLn (Fpk )| =
(pkn − pk(i−1) ). Then SLn (Fpk ) is the kernel
i=1
of the determinant map det : GLn (Fpk ) → F×
, which is obviously surjective (e.g., by
pk
taking diagonal matrices with all entries but the first equal to one); thus | SLn (Fpk )| =
| GLn (Fpk )|/(pk − 1).
10. Prove that GL2 (F2 ) ∼
= S3 as follows: Observe that GL2 (F2 ) acts on the set X = F22 \ {0}
of nonzero vectors of F22 , which has size three, so we get a homomorphism GL2 (F2 ) →
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Perm(X) ∼
= S3 . Show that this is bijective (hence an isomorphism).
We know from basic matrix
properties
that I2 ∈ GL2 (F2 ), and for any column vector
1
0
1
x ∈ X = F22 \ {0} =
,
,
, I2 · x = x. We also have that for any g1 , g2 ∈
0
1
1
GL2 (F2 ), x ∈ X, by properties of matrix multiplication, g1 (g2 · x) = (g1 · g2 )x. Thus, G
acts on the set X. Then by problem 1 from homework 4, we have that f : G → Perm(X),
f (g) 7→ g · X is a homomorphism. Then we observe that the only matrix g ∈ GL2 (F2 )
such that g · xi = xi is g = I2 , and therefore the kernel of f is the identity, which means
that, by an earlier problem, f is injective. We have by the previous problem that there are
(22 − 1)(22 − 2) = 3 · 2 = 6 elements in GL2 (F2 ), and we know that since X has 3 elements,
there are 3! = 6 elements in Perm(X). Then, by Problem 1, part c, from homework 3,
since f is an invective homomorphism between two finite groups of the same size, it must
be bijective. Thus, by definition, f is an isomorphism, and so GL2 (F2 ) ∼
= Perm(X) ∼
= S3 ,
and so by the transitivity of isomorphism, GL2 (F2 ) ∼
S
.
= 3
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