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Math 415 - Assignment 3 Solutions
Problems: 2.1.7, 2.2.1, 2.2.7, 2.2.9, 2.3.2, 2.3.6, 2.3.7, 2.3.8, 2.3.22, 2.3.23, 2.3.
Problem 2.1.
(a) Not (i) since it is a scalar, not (v) since it is in M 2 x 2 , not (vi) since it is in R 3 .
(b) − 5
x − y
xy
e x
cos y
− 5 x + 5y − 5 e x − 5
− 5 xy − 5 cos y − 15
(c) The zero element is the zero function 0 (x, y) =
Problem 2.2.
(a) Let v =
x
y
z
(^) and w =
X
Y
Z
(^) both satisfy the equation, i.e. x − y + 4z = 0, X − Y + 4Z = 0.
For scalars a and b we have av + bw =
ax + bX
ay + bY
az + bZ
. We need to show that this vector also satisfies
the equation. Here it is: (ax + bX) − (ay + bY ) + 4(az + bZ) = a(x − y + 4z) + b(X − Y + 4Z) =
a × 0 + b × 0 = 0.
(b) the zero vector must lie in a subspace, but x = 0, y = 0, z = 0 does not satisfy x − y + 4z = 1.
Problem 2.2.
(a) Yes: a
x
...
x
y
...
y
ax + by
...
ax + by
which again has all equal entries
(b) No: − 2
(^) which has negative entries
(c) Yes: a
... (anything)
0
(^) + b
... (anything)
0
... (anything)
0
(^) which again has first
and last entries 0
(d) Yes: proof is similar to 2.2.1 above
(e) No: the zero vector does not satisfy this condition. Alternately, if x satisfies it, then the scalar
multiple −x does not.
Problem 2.2.
Convince your self in the case of general 3x3 or 4x4 matrices:
a
c 0 0 0
d e 0 0
f g h 0
+b
C 0 0 0
D E 0 0
F G H 0
ac + bC 0 0 0
ad + bD ae + bE 0 0
af + bF ag + bG ah + bH 0
, again strictly
lower triangular.
Problem 2.3.
We need to find real numbers c 1 , c 2 , c 3 such that
= c 1
c 1
c 2
c 3
so Gaussian eliminate the augmented matrix:
⇒ c 3 = 1, c 2 = 2c 3 = 2, c 1 = −3 + 2c 2 + 2c 3 = 3
Check: 3
Problem 2.3.
V will be a subspace of U if v 1 and v 2 are both in U (since U is a vector space itself). So we need
to find constants c 1 , c 2 and d 1 , d 2 such that
v 1 =
= c 1
and v 2 =
= d 1
, i.e.
c 1
c 2
(^) and
d 1
d 2
. Let’s solve these together by
Gaussian reducing a double augmented matrix:
. Since there are two piv-
ots, we can uniquely solve for c 1 , ..., d 2 (we don’t need to solve explicitly. It is enough to know that
we can to say that v 1 and v 2 are in U ). The next question is whether U and V are the same. It is
enough then to show that U is a subspace of V. We can do this in two ways. One is to repeat the
above calculation with U and V interchanged. This leads to the calculation
6 5
12 5
6 5
(^) and again we have two piv-
ots. Thus u 1 and u 2 are in V and so U = V. A second approach is simply to show that v 1 and
v 2 are linearly independent. The calculation just above shows that the matrix with v 1 and v 2 as
columns has two pivots and this demonstrates the linear independence.
Problem 2.3.
(a)
a b
b c
= a
. Done!
(b) Since
a b c
b d e
c e f
= a
d
(^) + e
(^) + f
, the six special matrices shown here span the
set of 3x3 symmetric matrices.
Problem 2.3.
(a) Yes: Since we are given 3 polynomials of degree 2, if they are linearly independent, then they will
span P (2) (since this space has dimension 3). Alternately, we need to show that a general polynomial
of degree 2 can be written as a linear combo of these three. This leads to the problem
p(x) = a 0 +a 1 x+a 2 x 2 = A(x 2 +1)+B(x 2 −1)+C(x 2 +x+1) = (A−B +C)+Cx+(A+B +C)x 2
. We
conclude that A−B +C = a 0 , C = a 1 , A+B +C = a 2. We solve these to get A =
1 2
(a 0 +a 2 )−a 1 , B = 1 2
(a 2 − a 0 ), C = a 1. Check:.
1 2
(a 0 + a 2 ) − a 1
(x 2
1 2
(a 2 − a 0 )(x 2 − 1) + a 1 (x 2
a 0 + a 2 x 2
(b) Yes: Here we need to solve
p(x) = a 0 +a 1 x+a 2 x
2 +a 3 x
3 = A(x
3 −1)+B(x
2 +1)+C(x−1)+D = (D−A+B−C)+Cx+Bx
2 +Ax
3 .
This is satified by taking A = a 3 , B = a 2 , C = a 1 , D = a 0 + a 3 − a 2 + a 1 , and so P
(3) is spanned by
1 2
Since the initial 4x4 matrix has 4 pivots, the vectors are linearly
independent.
(b) Since R
4 has dimension 4, any set of 4 linearly independent vectors from R
4 has to span R
4 .
(c) By part (a) if
= c 1
then
c 1 , c 2 , c 3 , c 4 must satisfy − 2 c 4 = 1 2
, − 2 c 3 − 2 c 4 = − 1 , − 2 c 2 − c 3 − c 4 = − 1 , c 1 + c 2 + c 3 + c 4 = 1. These
have solution c 4 = −
1 4
, c 3 = −
1 2
1 4
3 4
, c 2 = −
1 2
3 4
1 4
1 4
, c 1 = 1 −
1 4
3 4
1 4
1 4
Check:
1 4
1 4
3 4
1 4
Problems 2.3.
False: Write c 1
(^) + c 2
(^) + c 3
(^) + c 4
+c 5
(^) + c 6
c 1 + c 6 c 2 + c 4 c 3 + c 5
c 3 + c 4 c 1 + c 5 c 2 + c 6
c 2 + c 5 c 3 + c 6 c 1 + c 4
This leads to to nine equations c 1 + c 6 = 0, c 3 + c 4 = 0, c 2 + c 5 = 0, c 2 + c 4 = 0, c 1 + c 5 = 0, c 3 + c 6 =
0 , c 3 + c 5 = 0, c 2 + c 6 = 0, c 1 + c 4 = 0. These can be written as the 9x6 system
1 0 0 0 0 1
0 0 1 1 0 0
0 1 0 0 1 0
0 1 0 1 0 0
1 0 0 0 1 0
0 0 1 0 0 1
0 0 1 0 1 0
0 1 0 0 0 1
1 0 0 1 0 0
c 1
c 2
c 3
c 4
c 5
c 6
and so we need to reduce this system. Here it is!
Since there are only 5 pivots, there is one free variable and hence many nontrivial solutions for the
ci’s. Here is an example:
