**MATH1025
AF**

**FW00F**

**Answers
to Tutorial 7 **

Anton and Rorres

Section 5.1

5.
The set of vectors from R^{2} of the form (x, 0) is a vector
space. Let

**u**
= (x, 0) and **v** = (x´, 0). Then **u** + **v** = (x
+ x´, 0) which is of the required form. Similarly k**u** =
(kx, 0) which is also of the required form. Moreover, the zero
vector (0, 0) and -**u** = (-x, 0) are both of this form. The
remaining axioms hold since they are valid for any vector in R^{2}.

6.
The set of vectors from R^{2} of the form **u** = (x, y)
with x greater or equal to zero is not a vector space since k**u**
with k negative and -**u** are not of the required form as the
first element is negative in these cases.

9. The set of 2 by 2 matrices of the form M = is not a vector space since if

N = then M + N =which is not of the required form. Also the zero matix is not of this form. Moreover, -M = and kM =are not of the required form if k is different from one.

10. The set of 2 by 2 matrices of the form M = is a vector space since if

N = then M + N =which is of the required form. Also the zero matix is of this form. Moreover, -M = and kM =are also of this form. The remaining axioms hold for any 2 by 2 matrix.

15.
Here the vectors are the positive real numbers but `addition' is
defined as multiplication and `multiplication by a scalar' is defined
as exponentiation. Thus x + y = xy which is positive if x and y are
and kx = x^{k} which is also a positive number if x is.
Since multiplication of scalars is commutative and associative,
axioms 2 and 3 hold. The 'zero' vector is 1 since 1+x = 1x = x and
-x = 1/x since x - x = x(1/x) = 1 which is the 'zero' vector. Now
k(x+y) = (xy)^{k} = x^{k}y^{k} = kx + ky so
axiom 7 holds. Also (k+j)x = x^{k+j} = x^{k}x^{j}
= kx + jx so axiom 8 holds. As well k(jx) = (x^{j})^{k }=
x^{kj} = (kj)x and 1x = x^{1} = x which shows that
axioms 9 and 10 hold as well.

Section 5.2

1. To prove that a set is a subspace we only need to prove the two closure axioms.

(a)
If **u** = (a, 0, 0) and **v** = (b, 0, 0) then **u** + **v
**= (a+b, 0, 0) and k**u** = (ka, 0, 0) both of which are of the
required form. Thus it is a subspace of R^{3}.

(b)
If **u** = (a, 1, 1) and **v** = (b, 1, 1) then **u** + **v
**= (a+b, 2, 2) and k**u** = (ka, k, k) neither of which are of
the required form. Thus it is not a subspace of R^{3}.

(c)
If **u** = (a, b, c) with b = a + c and **v** = (p, q, r) with
q = p + r then **u** + **v **= (a+p, b+q, c+r) and b+q = (a +
c) + (p+r) = (a+p) + (c+r). Futhermore. k**u** = (ka, kb, kc) and
kb = k(a+c) = ka + kc. Thus both are of the required form and it is
a subspace of R^{3}.

(d)
If **u** = (a, b, c) with b = a + c + 1 and **v** = (p, q, r)
with q = p + r + 1 then **u** + **v **= (a+p, b+q, c+r) and
b+q = (a+c+1) + (p+r+1) = (a+p) + (c+r) + 2. Futhermore. k**u** =
(ka, kb, kc) and kb = k(a+c+1) = ka + kc + k. Thus neither are of
the required form and it is not a subspace of R^{3}.

3.
(a) If **u** = a_{0} + a_{1}x + a_{2}x^{2}
+ a_{3}x^{3} with a_{0} = 0 and **v** = b_{0}
+ b_{1}x + b_{2}x^{2} + b_{3}x^{3}
with b_{0} = 0 then **u** + **v** = a_{0} + a_{1}x
+ a_{2}x^{2} + a_{3}x^{3} + b_{0}
+ b_{1}x + b_{2}x^{2} + b_{3}x^{3
}= a_{0 }+ b_{0} + (a_{1} + b_{1})x
+

(a_{2}
+ b_{2})x^{2} + (a_{3} + b_{3})x^{3}
with a_{0} + b_{0} = 0. Also k**u** = ka_{0}
+ ka_{1}x + ka_{2}x^{2} + ka_{3}x^{3
}with

ka_{0}
= 0. Thus they both have the required form and it is a subspace of
P_{3}.

(b)
If **u** = a_{0} + a_{1}x + a_{2}x^{2}
+ a_{3}x^{3} with a_{0} + a_{1} + a_{2}
+ a_{3}^{ }= 0 and **v** = b_{0} + b_{1}x
+ b_{2}x^{2} + b_{3}x^{3} with b_{0}
+ b_{1} + b_{2} + b_{3}^{ }= 0 then **u**
+ **v** = a_{0} + a_{1}x + a_{2}x^{2}
+ a_{3}x^{3} + b_{0} + b_{1}x + b_{2}x^{2}
+ b_{3}x^{3 }= a_{0 }+ b_{0} + (a_{1}
+ b_{1})x + (a_{2} + b_{2})x^{2} +
(a_{3} + b_{3})x^{3} with a_{0} + a_{1}
+ a_{2} + a_{3}^{ } + b_{0} + b_{1}
+ b_{2} + b_{3} = 0. Also k**u** = ka_{0}
+ ka_{1}x + ka_{2}x^{2} + ka_{3}x^{3
}with ka_{0} + ka_{1} + ka_{2} + ka_{3}
= 0. Thus they both have the required form and it is a subspace of
P_{3}.

(c)
If **u** = a_{0} + a_{1}x + a_{2}x^{2}
+ a_{3}x^{3} with a_{0}, a_{1}, a_{2}
and a_{3}^{ } integers then k**u** = ka_{0}
+ ka_{1}x + ka_{2}x^{2} + ka_{3}x^{3
}and if k is not an integer the coefficients are not integers.
Thus it is not a subspace of P_{3}.

(d)
If **u** = a_{0} + a_{1}x and **v** = b_{0}
+ b_{1}x then **u** + **v** = a_{0} + a_{1}x
+ b_{0} + b_{1}x ^{ }= a_{0 }+ b_{0}
+

(a_{1}
+ b_{1})x. Also k**u** = ka_{0} + ka_{1}x.
Thus they both have the required form and it is a subspace of P_{3}.

5.
(a) If A and B are n by n matrices and tr(A) = 0 and tr(B) = 0 then
tr(A+B)^{ }= tr(A) + tr(B) = 0. Also tr(kA) = k tr(A) = 0.
Thus these matrices are a subspace of M_{nn}.

(b)
If A and B are n by n matrices and A^{T}= -A and B^{T}
= -B then (A+B)^{T} = A^{T} + B^{T} = -A + -B
= -(A + B). Also (kA)^{T} = k A^{T }= k(-A) = -kA .
Thus these matrices are a subspace of M_{nn}.

(c)
If the n by n linear system A**x** = **0** has only the trivial
system then det(A) is not zero. If det(B) is not zero this does not mean
that
det(A+B) is not zero (if B = -A for example) so these matrices are not a
subspace of M_{nn}.

(d)
If A is an n by n matrix such that AB = BA for a fixed n by n matrix
B and C is an n by n matrix such that CB = BC then (A+C)B = AB + AC =
BA + BC = B(A+C). Also (kA)B = k(AB) = k(BA) = B(kA). Thus these
matrices are a subspace of M_{nn}.

7.
We must find scalars a and b such that a**u** + b**v** is the
given vector.

(a)
We must have b = 2, -2a + 3b = 2 and 2a - b = 2. This linear system
has the solution a = 2 and b = 2 so the vector is a linear
combination of **u** and **v**.

(b)
We must have b = 3, -2a + 3b = 1 and 2a - b = 5. This linear system
has the solution a = 4 and b = 3 so the vector is a linear
combination of **u** and **v**.

(c)
We must have b = 0, -2a + 3b = 4 and 2a - b = 5. This linear system
does not have a solution so the vector is not a linear combination of
**u** and **v**.

(d)
We must have b = 0, -2a + 3b = 0 and 2a - b = 0. This linear system
has the solution a = 0 and b = 0 so the vector is a linear
combination of **u** and **v**.

8.
We must find scalars a, b and c such that a**u** + b**v** + c**w**
is the given vector.

(a)
The coefficients are solutions of the linear system 2a + b + 3c =
-9, a - b + 2c = -7 and 4a + 3b + 5c = -15. This has solution a =
-2, b = 1 and c = -2 so that (-9, -7, -15) = -2**u** + **v** -
2**w**.

(b)
The coefficients are solutions of the linear system 2a + b + 3c = 6,
a - b + 2c = 11 and 4a + 3b + 5c = 6. This has solution a = 4, b =
-5 and c = 1 so that (6, 11, 6) = 4**u** - 5**v** + **w**.

(c)
The coefficients are solutions of the linear system 2a + b + 3c = 0,
a - b + 2c = 0 and 4a + 3b + 5c = 0. This has solution a = 0, b = 0
and c = 0 so that (0, 0, 0) = 0**u** + 0**v** + 0**w**.

(d)
The coefficients are solutions of the linear system 2a + b + 3c = 7,
a - b + 2c = 8 and 4a + 3b + 5c = 9. This has solution a = 0, b = -2
and c = 3 so that (7, 8, 9) = 0**u** - 2**v** + 3**w**.

9.
We must find scalars a, b and c such that a**p**_{1} + b**p**_{2}
+ c**p**_{3} is the given polynomial.

(a)
The coefficients are solutions of the linear system 2a + b + 3c =
-9, a - b + 3c = -7 and 4a + 3b + 5c = -15. This has solution a =
-2, b = 1 and c = -2 (see question 8(a)) so that -9 - 7x - 15x^{2}
= -2**p**_{1} + **p**_{2} - 2**p**_{3}.

(b)
The coefficients are solutions of the linear system 2a + b + 3c = 6,
a - b + 2c = 11 and 4a + 3b + 5c = 6. This has solution a = 4, b =
-5 and c = 1 (see question 8(b)) so that 6 + 11x + 6x^{2} =
4**p**_{1} - 5**p**_{2} + **p**_{3}.

(c)
The coefficients are solutions of the linear system 2a + b + 3c =
0, a - b + 2c = 0 and 4a + 3b + 5c = 0. This has solution a = 0, b =
0 and c = 0 (see question 8(c)) so that ^{0} = 0**p**_{1}
+ 0**p**_{2} + 0**p**_{3}.

(d)
The coefficients are solutions of the linear system 2a + b + 3c =
7, a - b + 2c = 8 and 4a + 3b + 5c = 9. This has solution a = 0, b =
-2 and c = 3 (see question 8(b)) so that 7 + 8x + 9x^{2} =
0**p**_{1} - 2**p**_{2} + 3**p**_{3}.

11.
Given any vector **x** = (x, y, z) we must be able to find
coefficients a, b and c such that **x** = a**v**_{1} +
b**v**_{2} + c**v**_{3.}if the vectors span R^{3}.

(a)
In this case the linear system 2a = x, 2a + c = y, 2a + 3b + c = z
must have a unique solution. This is true if det(A) is not zero
where A =.
Since the determinant equals -6 the vectors span R^{3}.

(b)
In this case the linear system 2a + 4b + 8c = x, -a + b - c = y, 3a +
2b + 8c = z must have a unique solution. This is true if det(A) is
not zero where A =.
Since the determinant equals 0 the vectors do not span R^{3}.

15.
Any vector **x** = (x, y, z) in the plane can be written as a
linear combination a**u** + b**v**, i.e. x = -a + 3b, y = a +
4b, z = a + 4b. If **x** lies in the plane then its components
must be such that the linear system is consistent when solving for a
and b. The augmented matrix for this system iswhich
has row echelon form
.
For consistency, the last row mus be zero, i.e. z-y = 0 gives the
equation of the plane spanned by **u** and **v**.

Section 5.3

2.
Since the two vectors in parts (a) and (c) are not multiples of each
other,they are linearly independent. In part (d) there are 4 vectors
in R^{3} so they must be linearly dependent. To check the
vectors in part (c) we must solve the system

a(-3, 0, 4) + b(5, -1, 2) + c(1, 1, 3) = (0, 0, 0) for a, b and c. The coefficient matrix for this system is which has determinant 9, the system has only the trivial solution and the vectors are linearly independent.

5. The vectors lie in a plane only if they are linearly dependent. We proceed as in question 2(c) to check this.

(a) The coefficient matrix is which has determinant -72 so the vectors are linearly independent and do not lie in a plane.

(b) The coefficient matrix is which has determinant 0 so the vectors are linearly dependent and do lie in a plane.

7. (a) We must find scalars a, b and c such that a(0, 3, 1, -1) + b(6, 0, 5, 1) + c(4, -7, 1, 3) = (0, 0, 0, 0). One solution is a = 7, b = -2, c = 3. Thus the vectors are linearly dependent.

(b)
From part a) we can write **v**_{1}
= 2/7 **v**_{2} -
3/7 **v**_{3},
**v _{2}** = 7/2

8. Using s instead of lambda, we must find values for s such that the equation

a(s, -1/2, -1/2) + b(-1/2, s, -1/2) + c(-1/2, -1/2, s) = (0, 0, 0) has a non-trivial solution.

The
coefficient matrix is
with
determinant s^{3 } -3/4 s - 1/4 which must be zero for a
non-trivial solution. The roots are s = -1/2 (twice) and 1. The
first value should be obvious since it makes all three vectors equal
and hence linearly dependent.

14.
If **v**_{1} and **v _{2}** are linearly
independent then there are no non-=zero values for a and b such that
a

15.
It is obvious that (**u** - **v**)
+ (**v**
- **w**) + (**w**
- **u**) = **0**
and hence the vectors are
linearly dependent.