MTH 102: Linear Algebra

Department of Mathematics and Statistics Indian Institute of Technology - Kanpur

Problem Set 3

Problems marked (T) are for discussions in Tutorial sessions.

1. Draw and illustrate in R2 .

(a) e1 + {ne2 |n N}.

(b) e1 + {e2 | R}.
 
1
2. In R2 , Is {e1 | R} + {e2 | R} = R2 ? What about {e1 | R} + { | R} = R2 ?
1

2 1 0
( " # ) ( " # ) ( " # )
3. In R3 prove that 1 | R + 1 | R + 1 | R = R3 . Do you use Gauss-
1 0 1
Jordan Elimination (GJE) method somewhere?

2 1 0
" # " # " #
Solution: Put A = { 1 | R}, B = { 1 | R}, C = { 1 | R}. Then A + B + C =
1 0 1
{a + b + c|a A, b B, c C} R3 .
x1 2 1 0 x1
" # " # " # " # " #
Let x = x2 R3 . We want to find , , s.t. 1 + 1 + 1 = x2 . That is, need
x 1 0 1 x3
" 3
2 1 0 x1
#" # " #
x1 x2 +x3
to solve 1 1 1 = x2 . We may use GJE to find the values of = 2 , =
1 0 1 x3
x2 x3 , = x1 +x
2
2 +x3
. But without doing so, we may find the determinant and conclude that
the system has a unique solution. But, we will need GJE for higher order vectors.

4. Let L1 and L2 be two nonparallel lines passing through origin in R3 . What is L1 + L2 ?

5. (T) Let L1 and L2 be two skewed (non parallel, nonintersecting) lines in R3 ? What is L1 + L2 ?

Solution:
A plane. Take a L1 , b L2 . Then L1h = L1 a and L2h = L2 b both pass through 0.
Thus L1 + L2 = a + b + L1h + L2h . As L1h + L2h is a plane, we are done.
Alternately: Put L1 = L1 + (b a). This is the line parallel to L1 passing through b. Then
L1 + L2 is a plane parallel to both L1 and L2 passing through 2b (be clear, not b, for example
L1 := (1, y, 0) and L2 = (1, 0, z)). So adding a b to it (that is, making the plane trace back
(b a)) will give us the plane through a + b. So our answer is L1 + L2 + a b which is a plane.

6. (T) Fix a non-negative integer n and let R[x; n] be the set of polynomials with real coefficients
P n i

and degree less than or equal to n. That is, R[x; n] = i=0 ci x : c0 , c1 , , cn R . Show
that R[x; n] is a vector space over R with respect to the usual addition and scalar multiplication.

Solution: For p(x) = ni=0 ai xi , q(x) = ni=0 bi xi , r(x) = ni=0 ci xi , we define the following:
P P P
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[Vector Addition:]
n
X
(p + q)(x) = (ai + bi )xi R[x; n]. (1)
i=0

[Scalar Multiplication:] for R,

n
X
(p)(x) = (ai )xi R[x; n]. (2)
i=0

Verify all vector space requirements:

i. Clearly, p + q = q + p as
n
X n
X
(p + q)(x) = (ai + bi )xi = (bi + ai )xi = (q + p)(x).
i=0 i=0

ii. (p + q) + r = p + (q + r) as
n
X n
X n
X
i i
(p + q)(x) + r(x) = (ai + bi )x + ci x = ((ai + bi ) + ci )xi =
i=0 i=0 i=0
n
X n
X n
X
i i
(ai + (bi + ci ))x = ai x + (bi + ci )xi = p(x) + (q + r)(x).
i=0 i=0 i=0

iii. The zero polynomial, z(x) = 0, satisfies p + z = p as

n
X n
X
(p + z)(x) = (ai + 0)xi = ai xi .
i=0 i=0

Pn i
iv. For all p(x) R[x; n], there is (p)(x) := i=0 (ai )x such that
n
X n
X
i
(p + (p))(x) = (ai + (ai ))x = 0xi = 0 = z(x)
i=0 i=0

v. For all , R and p(x) R[x; n], (p) = ()p as

n
X n
X
((p))(x) = (ai )xi = ()ai xi = (()p)(x).
i=0 i=0

vi. For all R, (p + q) = p + q as

n
X n
X n
X n
X
((p + q))(x) = (ai + bi )xi = (ai + bi )xi = ai xi + bi xi
i=0 i=0 i=0 i=0
= (p)(x) + (q)(x) = ((p) + (q))(x)
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Pn i
vii. For all , R and p(x) = i=0 ai x R[x; n], ( + )p = p + p as
n
X n
X
(( + )p)(x) = ( + )ai xi = (ai + ai )xi = (p)(x) + (p)(x) = ((p) + (p))(x).
i=0 i=0

viii. For all p(x) R[x; n], 1(p) = p as

n
X n
X
i
(1p)(x) = (1ai )x = ai xi = p(x).
i=0 i=0

7. Show that the space of all real m n matrices is a vector space over R with respect to the usual
addition and scalar multiplication.

Solution: Similar to Problem 4; a straightforward verification of all vector space requirements.

8. Let Mn (R) be the set of all n n real matrices. Then, from above we see that Mn (R) is a real
vector space. Now, prove the following:

(a) S = {A Mn (R) : At = A is a subspace of Mn (R).

(b) Fix A Mn (R). Define U = {B Mn (R) : AB = BA}. Then, U is a subspace of Mn (R).
(c) Let W = {a0 I + a1 A + + am Am : m is a non-negative integer, ai R}. Then, W is a
subspace of U.

9. In R, consider the addition x y = x + y 1 and a.x = a(x 1) + 1. Show that R is a real

vector space with respect to these operations with additive identity 1 (note that 0 is NOT the
additive identity).

Solution: Again, an easy verification of all vector space requirements.

10. (T) Which of the following are subspaces of R3 :

(a) {(x, y, z) | x 0}, (b) {(x, y, z) | x + y = z}, (c) {(x, y, z) | x = y 2 }.

Solution:

(a) Not a subspace : 1(1, 0, 0) does not belong to the set.

(b) Is a subspace.
(c) Not a subspace : (1, 1, 0) + (4, 2, 0) is not in the set. Since the relation is non-linear, closure
is a problem.

11. Find the condition on real numbers a, b, c, d so that the set {(x, y, z) | ax + by + cz = d} is a
subspace of R3 .

Solution: If d = 0, then this is a subspace. For it to be a subspace, (0, 0, 0) had to be in the

space and hence d = 0.
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12. (T) Let W1 and W2 be subspaces of a vector space V such that W1 W2 is also a subspace.
Prove that one of the spaces Wi , i = 1, 2 is contained in the other.

Solution: Suppose W1 is not a subset of W2 . Then to prove the result, we have to show that
W2 is a subset of W1 .
Let w2 W2 . To show that W2 is contained in W1 , we need to show that w2 W1 . Since
W1 6 W2 , we can choose w1 W1 such that w1 6 W2 . Then w2 w1 W1 W2 as it is a
subspace but w2 w1 6 W2 because then w1 = w2 (w2 w1 ) W2 . So, w2 w1 W1
w2 = (w2 w1 ) + w1 W1 .

13. Let v1 , v2 , . . . , vn be n vectors from a vector space V over R. Define linear span of this set of
vectors as

LS({v1 , v2 , . . . , vn }) = {c1 v1 + c2 v2 + cn vn : c1 , c2 , . . . , cn R},

that is, the set of all linear combinations of vectors v1 , v2 , . . . , vn . Show that
LS({v1 , v2 , . . . , vn }) is a subspace of V .

Solution: If u = c1 v1 + c2 v2 + cn vn and w = d1 v1 + d2 v2 + dn vn , then

u + w = (c1 + d1 )v1 + (c2 + d2 )v2 + (cn + dn )vn span({v1 , v2 , . . . , vn })

and
u = (c1 )v1 + (c2 )v2 + (cn )vn span({v1 , v2 , . . . , vn })
for R. Rest is straightforward.

14. (T) Show that {(x1 , x2 , x3 , x4 ) : x4 x3 = x2 x1 } = LS({(1, 0, 0, 1), (0, 1, 0, 1), (0, 0, 1, 1)}
and hence is a subspace of R4 .

Solution:

(x1 , x2 , x3 , x4 ) {(x1 , x2 , x3 , x4 ) : x4 x3 = x2 x1 }
(x1 , x2 , x3 , x4 ) = (x1 , x2 , x3 , x1 + x2 + x3 ) as x4 = x1 + x2 + x3
= x1 (1, 0, 0, 1) + x2 (0, 1, 0, 1) + x3 (0, 0, 1, 1)

Moreover, {(x1 , x2 , x3 , x4 ) : x4 x3 = x2 x1 } is a subspace of R4 because it is a linear span of

vectors in R4 .

15. Suppose S and T are two subspaces of a vector space V . Define the sum

S + T = {s + t : s S, t T }.

Show that S + T satisfies the requirements for a vector space. Moreover, LS(S T ) = S + T .

Solution: Straightforward to check all vector space requirements.

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16. (T) Find all the subspaces of R2 .

Solution: Let W be a subspace of R2 . Assume that W 6= {0}, then there exists 0 6= (w1 , w2 )
W . If the span, L({(w1 , w2 )}) = W , then W is a line through origin. If L({(w1 , w2 )}) is
a proper subset of W then we show that W = R2 .  Let (u1 , u2 ) W \ L({(w1 , w2 )}). So,
w1 u1
(u1 , u2 ) 6= (w1 , w2 ) for all R. So, A = is invertible. Therefore, we see that for
w2 u2
any (x, y) R2 , we need to find , R such that thesystem  (x, y)= (w1 , w2 ) + (u1 , u2 )
x
has a solution. Note that the above system reduces to A = . Such , exist as A is
y
invertible.

a11 a12 a1n
a21 a22 a2n
17. (T) Let A = . .. .. , with aij C. Then, we define the following 4 fundamental

.. ..
. . .
am1 am2 amn
subspaces:

(a) The column space of A is defined as

a11 a1n
a21 a2n
col(A) = {Ax : x Cn } = LS(A(:, 1), . . . , A(:, n)) = LS . , . . . , .

.
. .
.

am1 amn

(b) The column space of A is defined as

col(A ) = LS(A (1, :), . . . , A (m, :)) = {A x : x Cm }.

(c) The null space of A is defined as

Null Space(A) = N (A) = {x Cn : Ax = 0}.

(d) The null space of A is defined as

Null Space(A ) = N (A ) = {x Cm : A x = 0}.

Important: In case the matrix A has real entries, the spaces col(A ) and
Null Space(A ) are called the row-space of A and the left-null space of A, re-
spectively

Now, determine the above 4 mentioned fundamental spaces for the following matrices.

1 2 3
1 2 0

0
1 1 1
" #
2 4 6 0 1 2 0
(i) A = 2 1 1 (ii) A = (iii) B =

2 6 8 0 0 1 2
1 0 0
2 8 10 2 0 0 1
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(iv) Suppose B and C are two m n matrices and S = col(B) and T = col(C), then S + T is
a column space of what matrix M ?

1 2 3 1 2 3 1 2 3 1 2 3
2 4 6 0 0 0 0 2 2 0 2 2
Solution: (ii) 2 6 8 0 2 2 0 0 0 0 0 0

2 8 10 0 4 4 0 4 4 0 0 0

1 0 1 1 0 1
0 2 2 0 1 1
0 0 0 0 0 0 Reduced Row Echelon Form of A.

0 0 0 0 0 0

x z 1 1
So, the solutions are y = z = z 1 . Thus, N (A) = LS 1 .
z z 1 1
 
(iv) Let M = B C . In other words, M is an m (2n) matrix whose first n columns
are same as columns of B and next n columns are same as columns of C. It is easy to see that
if u col(M ) then u S + T . Similarly, if u S + T then u = s + t where s is a linear
combination of columns of B and t is a linear combination of columns of C which implies that
u col(M ).

18. Construct a matrix whose column space contains [ 1 1 1 ]T and whose null space is the line of
multiples of [ 1 1 1 1 ]T .

Solution: Clearly, the matrix we are looking for is a 3 4 matrix with rank 3. Two such
matrices are
1 0 0 1 1 1 1 3
0 1 0 1 1 2 3 6 .
0 0 1 1 1 4 9 14

19. (T) Suppose A is an m by n matrix of rank r.

(a) If Ax = b has a solution for every right side b, what is the column space of A?
Solution: There must be a pivot in every row, so r = m and the column space of A is all
of Rm .
(b) In part (a), what are all equations or inequalities that must hold between the numbers m,
n and r?
Solution: We always have r n. From (a), we know that r = m. From these, we deduce
that m n.
(c) Give a specific example of a 3 by 2 matrix A of rank 1 with first row [ 2 5 ]. Describe the
column space, col(A), and the null space N (A) completely.

2 5
Solution: Just use multiples of [2 5] for the other rows. For example, 4 10 . Column
0 0
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space will be the line in R3 consisting of all multiples of your first column. The null
 space
5/2
will be the line in R2 consisting of all multiples of the null space solution .
1
(d) Suppose the right side b is same as the first column in your example (part c). Find the
complete solution to Ax = b.
 
1
Solution: Adding the particular solution to the null space solution from (c), we get
0
   
1 5/2
the complete solution x = +c .
0 1
20. Suppose the matrix A has row reduced echelon form R :

1 2 1 b 1 2 0 3
A= 2 a 1 8 , R = 0 0 1 2 .
(row 3) 0 0 0 0

(a) What can you say immediately about row 3 of A?

Solution: Because row 3 of R is all zeros, row 3 of A must be a linear combination of row
1 and row 2 of A.
(b) What are the numbers a and b?
Solution: After one step of elimination, we have

1 2 1 b
0 a 4 1 8 2b .
(row 3)

From R, we see that the second column of A is not a pivot column, so a = 4. Continuing
with elimination, we get to
1 2 0 8b
0 0 1 2b 8 .
0 0 0 0
Comparing this to R, we see that b = 5.
(c) Describe all solutions of Rx = 0. Which among row spaces, column spaces and null spaces
are the same for A and for R.
Solution: Setting the free variables x2 and x4 to 1 and 0, and vice versa, and solving
Rx = 0, we get the null space solution

2 3
1 0
x = c 0 + d 2 .

0 1

The row space and the null space are always the same for A and R whereas column space
is different (row operations preserve row space but change column space).
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21. (T) Suppose that A is a 3 3 matrix. What relation is there between the null space of A and
the null space of A2 ? How about the null space of A3 ?
Solution: The null space of A is contained in the null space of A2 . The reason is that if Ax = 0,
i.e., if x is in the null space of A, then A2 x = A(Ax) = 0. Thus, x is also in the null space of
A2 . Similarly, we have
N (A) N (A2 ) N (A3 ) . . . .
Note that one can prove that if A is an n n matrix, then one has N (An ) = N (An+1 ) = . . . .
 
Ir F
22. Suppose R (an m n matrix) is in row reduced echelon form , with r non-zero rows
0 0
and first r pivot columns. Describe the column space and null space of R.
Solution: The column space is the space of all vectors whose last m r coordinates are zero.
This is clear since rank of the matrix R is r and the first r columns of R are independent.
Denote by fij the entry in the the (i, j) position in F . The null space of R is the space of all
linear combinations of the n r vectors
f1(nr)

f11 f12
f21 f22 f2(nr)
.. .. ..


. . .

fr1 fr2 fr(nr)

1 , 0 ,..., 0 .

0 1 0

0 0 0

. .. ..
..

. .
0 0 1
Clearly, these vectors are linearly independent and therefore the dimension of the null space is
n r.
n T  T o n T  T o
23. (T) Let W1 = span 1 1 0 , 1 1 0 and W2 = span 1 0 2 , 1 0 4 . Show that
W1 + W2 = R3 . Give an example of a vector v R3 such that v can be written in two different
ways in the form v = v1 + v2 , where v1 W1 , v2 W2 .

1 1 1
Solution: 1 ,
1 , 0 W1 +W2 and is linearly independent which means W1 +W2 =

0 0 2


0 1 1 1 1 0
R3 . Since 0 = 16 0 + 16 0 W2 , we have 1 = 1 + 0 W1 + W2 . Note that
1 2 4 1 0 1

0 1 1 1 1 1
1 = 1 1 + 1 1 W1 and 0 = 5 0 1 0 W2 , so
2 2 6 6
0 0 0 1 2 4

1 0 1
1 = 1 + 0 W1 + W2 .
1 0 1