Math 317: Linear Algebra
Homework 6 Solutions
Turn in the following problems.
1. For each of the following sets V , determine if V forms a subspace in Rn . Justify your
answer in each case by providing a proof or a counterexample where appropriate.
(
)
n
X
n
(a) V = x ∈ R :
xi = 0
i=1
Proof : We claim that
0=
Pn is V is a subspace. We note that 0 ∈ VPsince
n
(0, 0,P
. . . , 0) =⇒
0
=
0.
Suppose
that
x,
y
∈
V
.
Then
x
i=1 P
i=1 i = 0
Pn
Pn
n
n
and i=1 yi = 0. Thus i=1 xi + i=1 yi = 0 =⇒
i=1 (xi + yi ) = 0.
P
n
Thus
x
+
y
∈
V
.
Suppose
that
c
∈
R
and
x
∈
V
.
Then
i=1 xi = 0 =⇒
Pn
Pn
c i=1 xi = c0 = i=1 cxi = 0. Thus cx ∈ V . Since the three properties
of a subspace have been satisfied we may say that V is a subspace of Rn .
(b) V = x ∈ Rn : Ax = b, A 6= 0, b 6= 0, b ∈ Rm , A ∈ Rm×n
V is not a subspace because 0 is not an element of V . This is true since
if 0 were to be in V , then A0 = b for some b 6= 0. This of course is
impossible.
2. Determine which of the following subsets of Rn×n are in fact subspaces of Rn×n .
Justify your answer in each case by providing a proof or a counterexample where
appropriate.
(a) The symmetric matrices.
Let V = A ∈ Rn×n | A = AT . We claim that V is a subspace of Rn .
We begin by observing that 0n×n ∈ V , since 0T = 0. That is, the zero
matrix is symmetric. Let A, B ∈ V . Then A = AT and B = B T (wts.
A + B is symmetric). Then (A + B)T = AT + B T = A + B so A + B
is symmetric and hence is an element of V . Let c ∈ R and suppose that
A ∈ V . Then (cA)T = cAT = cA and so cA is symmetric and hence an
element of V . Since all three conditions of a subspace were met, we may
conclude that V is a subspace of Rn×n .
(b) The nonsingular matrices.
The set of all nonsingular matrices do not form a subspace of Rn×n .
This is true since 0n×n is a singular matrix (0x = 0 has infinitely many
solutions.). Since the zero matrix is not in the set of all nonsingular
matrices, the set does not form a subspace of Rn×n .
(c) The singular matrices.
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Math 317: Linear Algebra
Homework 6 Solutions
The set of singular matrices do not form a subspace in Rn×n . This is true
because the set is not closed under matrix addition. To see why, consider
the singular
which have a row ofzerosin its echelon form)
matrices (all
of 1 0
0 0
1 0
A=
and B =
. Then A + B =
which is the 2 × 2
0 0
0 1
0 1
identity matrix and is nonsingular. So A+B is not in the set, even though
A and B are.
3. Section 3.1, Problem 6
Proof: Suppose that U and V are subspaces of Rn . We prove that U ∩ V (the
intersection of U and V ) is a subspace of Rn . Since U and V are subspaces
of mathbbRn , we know that 0 ∈ U and 0 ∈ V . Thus 0 ∈ U ∩ V . Suppose
that x, y ∈ U ∩ V . Then x, y ∈ U and x, y ∈ V . Since U, V are subspaces, we
know that x + y ∈ U and x + y ∈ V . Thus x + y ∈ U ∩ V . Now suppose that
c ∈ R and let x ∈ U ∩ V . Then U, V being subspaces =⇒ cx ∈ U, cx ∈ V .
Thus, cx ∈ U ∩ V and so U ∩ V is a subspace of Rn .
For examples of this consider U = Rn and V = {0}, which are both subspaces
of Rn . Then U ∩V = {0} which is still a subspace of Rn . For another example,
consider the x-axis and y-axis in R2 which can be interpretated as lines passing
through the origin. Thus, these are subspaces of R2 . In set notation, we may
write U = {(x1 , 0) | x1 ∈ R} and V = {(0, x2 ) | x2 ∈ R}. Then U ∩ V = {0}
which is a subspace of R2 . However, U ∪V is not a subspace in this example. To
see why, we note that (1, 0) ∈ U ∪V and (0, 1) ∈ U ∪V but (1, 0)+(0, 1) = (1, 1)
which is in neither U nor V , so U ∪V is not closed under vector addition. Hence,
it cannot be a subspace.
4. Section 3.1, Problem 8
Proof : As this is an if and only if statement, we must prove both directions of the claim. To begin, we assume that v ∈ span {v1 , v2 , . . . , vk }.
We show that span {v1 , v2 , . . . , vk } = span {v1 , v2 , . . . , vk , v}. To show that
two sets are equal, we must show containment on both sides. Suppose that
x ∈ span {v1 , v2 , . . . , vk } (wts. x ∈ span {v1 , v2 , . . . , vk , v}). Then x =
c1 v1 + c2 v2 + . . . + ck vk for some c1 , c2 , . . . ck ∈ R. But x = c1 v1 + c2 v2 + . . . +
ck vk + 0v and hence x ∈ span {v1 , v2 , . . . , vk , v}. So span {v1 , v2 , . . . , vk } ⊂
span {v1 , v2 , . . . , vk , v}.
We now show containment in the other direction. Suppose that x ∈ span {v1 , v2 , . . . , vk , v}.
(wts. x ∈ span {v1 , v2 , . . . , vk }). Then x = c1 v1 + c2 v2 + . . . + ck vk + ck+1 v for
c1 , . . . , ck+1 ∈ R. But, v ∈ span {v1 , v2 , . . . , vk }, and thus, v = d1 v1 + . . . +
dk vk for some d1 , . . . , dk ∈ R. Thus, we have that
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Math 317: Linear Algebra
Homework 6 Solutions
x = c1 v1 + c2 v2 + . . . + ck vk + ck+1 v
= c1 v1 + c2 v2 + . . . + ck vk + ck+1 (d1 v1 + . . . + dk vk )
= (c1 + ck+1 d1 )v1 + (c2 + ck+1 d2 )v2 + . . . + (ck + ck+1 dk )vk ,
and thus x ∈ span {v1 , v2 , . . . , vk }. So, span {v1 , v2 , . . . , vk , v} ⊂ span {v1 , v2 , . . . , vk }.
Thus span {v1 , v2 , . . . , vk } = span {v1 , v2 , . . . , vk , v}.
To go in the other direction, suppose that span {v1 , v2 , . . . , vk } = span {v1 , v2 , . . . , vk , v}.
We show that v ∈ span {v1 , v2 , . . . , vk }. We note that v = 0v1 + 0v2 +
. . . + 0vk + 1v ∈ span {v1 , v2 , . . . , vk , v} = span {v1 , v2 , . . . , vk }. Thus, v ∈
span {v1 , v2 , . . . , vk }.
5. Section 3.1, Problem 11
Proof: Suppose that U and V are orthogonal subspaces. That is u · v = 0
for any u ∈ U and v ∈ V . The since U and V are subspaces, we know that
U ∩ V is also a subspace. Hence 0 ∈ U ∩ V . In particular, {0} ⊂ U ∩ V . We
now show that U ∩ V ⊂ {0}. Suppose that x ∈ U ∩ V . Then x ∈ U and
x ∈ V and so x · x = 0. This implies that x = 0, by an earlier proposition
(i.e. x · x = x21 + x22 + . . . + x2n = 0 =⇒ x1 = x2 = . . . = xn = 0). Thus
U ∩ V ⊂ {0} and hence U ∩ V = {0}.
6. Section 3.1, Problem 18
Proof: Suppose that U and V are subspaces of Rn and let x ∈ (U + V )⊥ . Then
x · y = 0 for every y ∈ U + V . Of course, y ∈ U + V =⇒ y = u + v for
some u ∈ U and v ∈ V . Thus x · (u + v) = x · u + x · v = 0 for every u ∈ U
and v ∈ V . Now since U and V are subspaces, 0 ∈ U and 0 ∈ V . Let v = 0.
Then x · u = 0 for every u ∈ U . Thus x ∈ U ⊥ . Similarly, by letting u = 0, we
have that x ∈ V ⊥ . Thus x ∈ U ⊥ ∩ V ⊥ and so (U + V )⊥ ⊂ U ⊥ ∩ V ⊥ . Suppose
that x ∈⊂ U ⊥ ∩ V ⊥ . Then x · u = 0 for every u ∈ U and x · v = 0 for every
v ∈ V . Thus x · u + x · v = x · (u + v) = x · y = 0 for every y ∈ U + V . Thus
x ∈ (U + V )⊥ . So U ⊥ ∩ V ⊥ ⊂ (U + V )⊥ and so (U + V )⊥ = U ⊥ ∩ V ⊥ .
7. Section 3.2, Problem 3b


1 1 0
Let A = 2 1 1. We look for matrices X and Y such that C(A) = N (X)
1 −1 2
and N (A) = C(Y ). To find such matrices, we put [A|b] in row echelon form
where b = (b1 , b2 , b3 ) to find the constraint equations for b so that Ax = b
is consistent and so that we may solve Ax = 0 to find N (A). Performing
Gaussian elimination yields
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Math 317: Linear Algebra
Homework 6 Solutions



1 1 0
b1
1 1 0 b1

b2
[A|b] =  2 1 1 b2  →  0 −1 1
1 −1 2 b3
0 0 0 3b1 − 2b2 + b3 .

So we may write C(A) = {b ∈ R3 | 3b1 − 2b2 + b3 = 0}. Letting X = [3 − 2 1]
will give a matrix whose nullspace is the column space of A.
To find the nullspace of A we use the row echelon form of A (noting that x3 is
the free variable) to obtain: x1+x2 =
0 and −x2 + x3 = 0. So, 
x2 = x3 =⇒

−1
 −1 
x1 = −x3 and so N (A) = span  1  . Thus letting Y =  1  will ensure


1
1
that the column space of Y is the nullspace of A.
8. Section 3.2, Problem 6a
   
   
0 
0 
 1
 1
Suppose that A is a matrix such that 1 , 1 ⊂ C(A) and 0 , 1 ⊂




1
0
1
1
N (A). We show that such a matrix is not possible to construct. Note the the
elements in both C(A) and N (A) will force A to be a 3 × 3 matrix. I already have two columns of A, so we just need to find the third one. Using the
nullspace as a guide, we have that

   
1 0 x1
1
0
1 1 x2  0 = 0 =⇒ x1 = x2 = x3 = −1.
1 1 x3
1
0


   
1 0 −1
0
0





So A = 1 1 −1 . However A 1 6= 0. So such an A is impossible to
1 1 −1
0
0
construct.
9. Section 3.2, Problem 10
Suppose that A is an m × n matrix and that B is an n × p matrix.
(a) We first show that N (B) ⊂ N (AB). Suppose that x ∈ N (B). Then
Bx = 0. So A(Bx) = A0 = 0 =⇒ AB(x) = 0. Thus x ∈ N (AB).
(b) Next, we show that C(AB) ⊂ C(A). Suppose that b ∈ C(AB). Then
ABx = b for some x ∈ Rp . So A(Bx) = b =⇒ Ay = b for y = Bx.
Thus, by definition, b ∈ C(A).
(c) We now show that N (B) = N (AB) when A is n × n and nonsingular. It
suffices to prove that N (AB) ⊂ N (B) since we have containment on one
side from (a). Let x ∈ N (AB). Then ABx = 0 =⇒ A(Bx) = 0. Now
4
Math 317: Linear Algebra
Homework 6 Solutions
since A is nonsingular, Ay = 0 =⇒ y = 0. Thus Bx = 0 =⇒ x ∈
N (B).
(d) Finally, we show that C(AB) = C(A) when B is n × n and nonsingular.
Once again, it suffices to only show that C(A) ⊂ C(AB) as we have
containment on one side from part (b). Suppose that b ∈ C(A). Then
Ax = b for some x ∈ Rn . Now B is nonsingular and hence B is invertible
and so Ax = ABB −1 x = AB(B −1 x) = b. So b ∈ C(AB). Thus
C(A) = C(AB).
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