# Math 2270 Spring 2004 Homework 10 Solutions (1) General elements of P ```Math 2270 Spring 2004
Homework 10 Solutions
Section 4.1) 1, 2, 6, 10, 13, 14, 20, 25, 30, 35
(1) General elements of P2 have the form a + bt + ct2 where a, b and c are arbitrary real
numbers. The subset V consists of elements of P2 that also satisfy p(0) = a = 2, or have
the general form 2 + bt + ct2 . We first check that the zero element (0 + 0 &middot; t + 0 &middot; t2 , or
a = b = c = 0) is in the subset V . But, if a = 2, then the zero element is obviously not in
V , so V is not a subspace of P2 .
(2) General elements of P2 have the form a + bt + ct2 where a, b and c are arbitrary real
numbers. The subset V consists of elements of P2 that also satisfy p(2) = a + 2b + 4c = 0, or
a = −2b − 4c. This gives the general form (−2b − 4c) + bt + ct2 = b(−2 + t) + c(−4 + t2 ). We
first check that the zero element (0 + 0 &middot; t + 0 &middot; t2) is in the subset V . By setting b = c = 0, we
see that the zero element is in this subset. Next we must check that the subset is closed under
linear combinations. Take p(t) = b1 (−2 + t) + c1 (−4 + t2 ) and q(t) = b2 (−2 + t) + c2 (−4 + t2 )
to be elements of V , and k1 and k2 to be scalars. Then,
k1 p(t) + k2 q(t) = k1 b1 (−2 + t) + c1 (−4 + t2 ) + k2 b2 (−2 + t) + c2 (−4 + t2 )
= (k1 b1 + k2 b2 )(−2 + t) + (k1 c1 + k2 c2 )(−4 + t2 )
which is in the set V . Since V contains the zero polynomial and is closed under linear combinations, it must be a subspace of P2 . One possible basis for the subspace V is
{−2 + t, −4 + t2 }
(6) The 3 x 3 zero matrix is not invertible, and is therefore not in the subset of ℜ3x3 consisting
of all invertible matrices. Therefore, this subset is not a subspace of ℜ3x3 .
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1
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(10) Let V be the set of all 3 x 3 matrices A such that A  2  = ~0. We see that the zero
3
~
matrix must be in this set since 0~x = 0 for any vector ~x. Let A and B be elements in V ,
and let k1 and k2 be scalars. Then,












1
1
1
1
1

 


 





(k1 A + k2 B)  2  = k1 A  2  + k2 B  2  = k1 A  2  + k2 B  2  = k1~0 + k2~0 = ~0
3
3
3
3
3
so the linear combination of matrices A and B is also in the subset V . Therefore, V is a
subspace of ℜ3x3 .
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Math 2270 Spring 2004
Homework 10 Solutions
(13) Consider the following geometric sequences:
(1, 1, 1, 1, . . .) where a = 1 and r = 1, and
(1, 2, 4, 8, . . .) where a = 1 and r = 2.
Then, the sum of these two sequences is (2, 3, 5, 9, . . .). From the general form of these
sequences we know that a is the first element in the list, so a = 2. The second element is ar,
so ar = 2r = 3 implying that r = 3/2. If we use these values to calculate the third element in
the list we expect to see ar 2 = 2(9/4) = 9/2. Instead, we see a value of 5. Therefore, the sum
of the given geometric sequences does not produce another geometric sequence. The given
subset is not closed under addition and is therefore not a subspace of all infinite sequences
of real numbers.
(14) Let V be the set of all infinite sequences of real numbers that converge to zero. The
zero sequence, (0, 0, 0, 0, . . .), converges to zero so is in the set V . Next take two sequences
(x0 , x1 , x2 . . .) and (y0 , y1 , y2 , . . .) from V and two scalars k1 and k2 . Then,
k1 (x0 , x1 , x2 , . . .) + k2 (y0, y1 , y2 , . . .) = (k1 x0 , k1 x1 , k1 x2 , . . .) + (k2 y0 , k2 y1 , k2 y2 , . . .)
= (k1 x0 + k2 y0 , k1 x1 + k2 y1 , k1 x2 + k2 y2 , . . .)
We must look at the limit of this new sequence.
lim (k1 xn + k2 yn ) = k1 n→∞
lim xn + k2 n→∞
lim yn = k1 &middot; 0 + k2 &middot; 0 = 0
n→∞
The new sequence also converges to zero so is in the subset V . Therefore, V is closed under
linear combinations and is a subspace of the space of all infinite sequences of real numbers.
(20) The criteria that a + d = 0 is the same as d = −a, so general matrices are of the form
&quot;
a b
c −a
#
=a
&quot;
1 0
0 −1
#
+b
&quot;
#
+c
0 0
1 0
#)
0 1
0 0
&quot;
0 0
1 0
#
The space has dimension 3 and one possible basis is:
(&quot;
1 0
0 −1
# &quot;
,
0 1
0 0
# &quot;
,
(25) General polynomials in P2 have the form a + bt + ct2 where a, b and c are arbitrary
real numbers. If f (1) = 0, then a + b + c = 0, or a = −b − c. This gives the general form
(−b − c) + bt + ct2 = b(−1 + t) + c(−1 + t2 ). The dimension of the space is 2 and one possible
basis is {−1 + t, −1 + t2 }.
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Math 2270 Spring 2004
Homework 10 Solutions
(30) We want the set of all matrices A that satisfy
&quot;
1 2
3 6
#&quot;
a b
c d
#
=
&quot;
a + 2c b + 2d
3a + 6c 3b + 6d
#
=
&quot;
0 0
0 0
#
This reduces to a system of four equations and four unknowns:
a + 2c = 0
3a + 6c = 3(a + 2c) = 0
b + 2d = 0
3b + 6d = 3(b + 2d) = 0
which is the same as the two equations a = −2c and b = −2d. Therefore, matrices in the
set have the general form
&quot;
−2c −2d
c
d
#
=c
&quot;
−2 0
1 0
#
+d
&quot;
0 −2
0 1
#
The dimension of the space is 2 and a possible basis is
(&quot;
−2 0
1 0
# &quot;
,
0 −2
0 1
#)
(35) Yes, the set of all linear transformations from ℜn to ℜm is a subspace of the set of all
functions from ℜn to ℜm . Let ~x and ~y be two vectors in ℜn and let k be a scalar. First, we
look at the zero function f (~x) = ~0.
f (~x + ~y ) = ~0 = ~0 + ~0 = f (~x) + f (~y )
f (k~x) = ~0 = k &middot; ~0 = kf (~x)
so the zero function is a linear transformation. Take any two linear transformations from ℜn
to ℜm , say T (~x) and G(~x). If T and G are linear, then by definition we know that:
T (~x + ~y ) = T (~x) + T (~y )
T (k~x) = kT (~x)
G(~x + ~y ) = G(~x) + G(~y )
G(k~x) = kG(~x)
We must show that the linear transformations (T + G) and kT are also linear. First, we look
at (T + G):
(T + G)(~x + ~y ) = T (~x + ~y ) + G(~x + ~y ) = T (~x) + T (~y ) + G(~x) + G(~y ) = (T + G)(~x) + (T + G)(~y )
(T + G)(k~x) = T (k~x) + G(k~x) = kT (~x) + kG(~x) = k ((T + G)(~x))
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Math 2270 Spring 2004
Homework 10 Solutions
Let c be another scalar and we look at the function kT :
(kT )(~x + ~y ) = k (T (~x + ~y )) = k (T (~x) + T (~y )) = kT (~x) + kT (~y ) = (kT )(~x) + (kT )(~y )
(kT )(c~x) = k (T (c~x)) = k (cT (~x)) = c(kT )(~x)
Therefore, the set of linear transformations is closed under linear combinations and is therefore a subspace of all functions. Since we are only looking at linear transformations from
ℜn to ℜm this proof could also be done by defining matrices A and B corresponding to the
transformations T and G.
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