MTL101 :: Linear Algebra and Di↵erential Equations
Tutorial 2
Department of Mathematics
Indian Institute of Technology Delhi
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
1 / 53
Question 1
Question 1
Suppose v1 = (1, 2), v2 = (0, 1) 2 R2 .
(a) Describe geometrically the subsets W1 := {tv1 : t 2 R},
W2 := {tv2 : t 2 R}, W3 := {sv1 + tv2 : s, t 2 R} and
W4 := {sv1 + tv2 : 0  s, t  1}.
(b) Which of W1 , W2 , W3 , W4 are subspaces of R2 ? Justify your answer
in each case.
(c) Show that {v1 , v2 } is a linearly independent subset of R2 .
(d) Suppose v3 = (2, 3). Is {v1 , v2 , v3 } linearly independent?
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
2 / 53
Question 1(a)
Question 1(a)
Suppose v1 = (1, 2), v2 = (0, 1) 2 R2 .
Describe geometrically the subsets W1 := {tv1 : t 2 R},
W2 := {tv2 : t 2 R}, W3 := {sv1 + tv2 : s, t 2 R} and
W4 := {sv1 + tv2 : 0  s, t  1}.
Solution:
W1 := {tv1 : t 2 R}
= {(t, 2t) : t 2 R}
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
3 / 53
Question 1(a) Contd.
W2 := {tv2 : t 2 R}
= {(0, t) : t 2 R}
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
4 / 53
Question 1(a) Contd.
W3 := {sv1 + tv2 : s, t 2 R}
= {(s, 2s + t) : s, t 2 R}
W4 := {sv1 + tv2 : 0  s, t  1}
= {(s, 2s + t) : 0  s, t  1}
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
5 / 53
Question 1(b)
Question 1(b)
Which of W1 , W2 , W3 , W4 are subspaces of R2 ? Justify your answer in
each case.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
6 / 53
Question 1(b)
Question 1(b)
Which of W1 , W2 , W3 , W4 are subspaces of R2 ? Justify your answer in
each case.
Recall:
A nonempty subset W is said to be the subspace of V over field F if
(a) u + v 2 W for all u, v 2 W ,
(b) au 2 W for a 2 F , u 2 W .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
6 / 53
Question 1(b) cont.
Solution:
W1 := {tv1 : t 2 R}.
Let u = t1 v1 , v = t2 v1 2 W1 , then
u + v = t1 v1 + t2 v1 = (t1 + t2 )v1
= t3 v1 2 W1 where t1 + t2 = t3 2 R.
Let a 2 R and u = tv1 2 W , then
0
0
au = atv1 = t v1 2 W1 , where at = t 2 R.
Hence, W1 is a subspace of R2 .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
7 / 53
Question 1(b) cont.
Solution:
W2 := {tv2 : t 2 R}.
Let u = t1 v2 , v = t2 v2 2 W2 , then
u + v = t1 v2 + t2 v2 = (t1 + t2 )v2
= t3 v2 2 W2 where t1 + t2 = t3 2 R.
Let a 2 R and u = tv2 2 W2 , then
0
0
au = atv2 = t v2 2 W2 , where at = t 2 R.
Hence, W2 is a subspace of R2 .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
8 / 53
Solution contd.
W3 := {sv1 + tv2 : s, t 2 R}.
Let u = s1 v1 + t1 v2 , v = s2 v1 + t2 v2 2 W3 , then
u + v = s 1 v 1 + t1 v 2 + s 2 v 1 + t 2 v 2
= s3 v1 + t3 v2 2 W3 where s1 + s2 = s3 , t1 + t2 = t3 2 R.
Let a 2 R and u = sv1 + tv2 2 W2 , then
0
0
0
0
au = a(sv1 + tv2 ) = s v1 + t v2 2 W3 , where as = s , at = t 2 R.
Hence, W3 is a subspace of R2 .
W4 := {sv1 + tv2 : 0  s, t  1}.
Let v1 2 W4 and 2 2 R. Then 2v1 = (2, 4) 2
/ W4 . Hence, W4 is NOT
a subspace of R2 .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
9 / 53
Question 1(c)
Question 1(c)
Show that {v1 , v2 } is a linearly independent subset of R2 .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
10 / 53
Question 1(c)
Question 1(c)
Show that {v1 , v2 } is a linearly independent subset of R2 .
Recall:
A subset X = {v1 , v2 , ..., vm } of a vector space V over F is said to be
linearly dependent if there exist scalars a1 , a2 , ..., am 2 F such that at
least one of these scalars is nonzero and a1 v1 + a2 v2 + + am vm = 0.
A finite subset X = {v1 , v2 , ..., vm } is linearly independent if and only
if a1 v1 + a2 v2 + · · · + am vm = 0 =) a1 = a2 = = am = 0.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
10 / 53
Question 1(c) cont.
Solution:
Consider a1 v1 + a2 v2 = (0, 0) for a1 , a2 2 R.
We have
av1 + bv2 = a(1, 2) + b(0, 1) = (0, 0)
) (a, 2a + b) = (0, 0)
)a=0 &
)a=0 &
2a + b = 0
b = 0.
Hence {v1 , v2 } is linearly independent.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
11 / 53
Question 1(d)
Question 1(d)
Suppose v3 = (2, 3). Is {v1 , v2 , v3 } linearly independent?
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
12 / 53
Question 1(d)
Question 1(d)
Suppose v3 = (2, 3). Is {v1 , v2 , v3 } linearly independent?
Solution: Observe that
2v1
v2
v3 = (0, 0)
Hence, the set {v1 , v2 , v3 } is linearly dependent.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
12 / 53
Question 2
Question 2
Suppose V := C2 is the complex vector space (over C) under
component-wise addition.
(a) Show that {(1 + i, 2), (2, 1)} is linearly independent.
(b) Show that {(1, 2), (0, i), (i, 1
i)} is linearly dependent.
(c) Show that every ordered pair can be written as a linear combination
of v1 = (1 + i, 2) and v2 = (2, 1). Also show that up to change of
order (of v1 and v2 ) such a linear combination is unique (for each
ordered pair).
(d) Show that every ordered pair can be written as a linear combination
of v1 = (1, 2), v2 = (0, i), v3 = (i, 1 i) in more than one ways.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
13 / 53
Question 2(a)
Question 2(a)
Show that {(1 + i, 2), (2, 1)} is linearly independent.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
14 / 53
Question 2(a)
Question 2(a)
Show that {(1 + i, 2), (2, 1)} is linearly independent.
Solution:
Consider a1 (1 + i, 2) + a2 (2, 1) = (0, 0), for a1 , a2 2 C.
We have,
a1 (1 + i, 2) + a2 (2, 1) = (0, 0)
)((1 + i)a1 + 2a2 , 2a1 + a2 ) = (0, 0)
)(1 + i)a1 + 2a2 = 0 &
Solving for a1
)(1 + i)a1 + 2a2 = 0 &
(1 + i)a1
2a1 + a2 = 0
a2 =
2a1
4a1 = ( 3 + i)a1 = 0.
)a1 = 0, a2 = 0.
Hence, {(1 + i, 2), (2, 1)} is linearly independent.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
14 / 53
Question 2(b)
Question 2(b)
Show that {(1, 2), (0, i), (i, 1
(Maths Dept., IIT Delhi)
i)} is linearly dependent.
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
15 / 53
Question 2(b)
Question 2(b)
Show that {(1, 2), (0, i), (i, 1
i)} is linearly dependent.
Solution:
Let a1 , a2 , a3 2 C and a1 (1, 2) + a2 (0, i) + a3 (i, 1
i) = (0, 0).
Consider
a1 (1, 2) + a2 (0, i) + a3 (i, 1
i) = (0, 0)
=) a1 + ia3 = 0,
2a1 + ia2 + (1
i)a3 = 0.
We get the system of two linear equations in three unknowns, gives,
more than one non-zero solution i.e., not all scalars are zero. One
such solution is a1 = i + 3, a2 = 10i, a3 = 1 + 3i.
Hence, {(1, 2), (0, i), (i, 1
(Maths Dept., IIT Delhi)
i)} is linearly dependent.
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
15 / 53
Question 2(c)
Question 2(c)
Show that every ordered pair can be written as a linear combination of
v1 = (1 + i, 2) and v2 = (2, 1). Also show that up to change of order (of
v1 and v2 ) such a linear combination is unique (for each ordered pair).
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
16 / 53
Question 2(c)
Question 2(c)
Show that every ordered pair can be written as a linear combination of
v1 = (1 + i, 2) and v2 = (2, 1). Also show that up to change of order (of
v1 and v2 ) such a linear combination is unique (for each ordered pair).
Solution:
Let (x, y ) 2 C2 .
Suppose, (x, y ) = ↵(1 + i, 2) + (2, 1) for some ↵,
Solving above equation for ↵ and , we get
2y
↵ = x 3+i
, = 2x+(1+i)y
.
3+i
2 C.
Hence, every ordered pair (x, y ) can be written as a linear
combination of v1 = (1 + i, 2) and v2 = (2, 1).
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
16 / 53
Question 2(c) cont.
Uniqueness:
Suppose there exists ↵,
and ↵0 ,
0 2 C such that
(x, y ) = ↵(1 + i, 2) + (2, 1) = ↵0 (1 + i, 2) +
(↵
)↵
↵0 )(1 + i, 2) + (
↵0 =
) ↵ = ↵0 ,
(Maths Dept., IIT Delhi)
0
=
0
0
0
(2, 1),
)(2, 1) = (0, 0)
= 0, since {(1 + i, 2), (2, 1)} is linearly independent
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
17 / 53
Question 2(d)
Question 2(d)
Show that every ordered pair can be written as a linear combination of
v1 = (1, 2), v2 = (0, i), v3 = (i, 1 i) in more than one ways.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
18 / 53
Question 2(d)
Question 2(d)
Show that every ordered pair can be written as a linear combination of
v1 = (1, 2), v2 = (0, i), v3 = (i, 1 i) in more than one ways.
Solution:
Let (x, y ) 2 C2 .
Suppose (x, y ) = a1 (1, 2) + a2 (0, i) + a3 (i, 1
i).
This implies
a1 + ia3 = x,
2a1 + ia2 + (1
i)a3 = y
Which is the system of two linear equations in three unknowns
(a1 , a2 , a3 ), hence, there exists more than one solution.
Two such solutions are, (x, y ) = x(1, 2) + i(2x y )(0, i) + 0(i, 1
(x, y ) = (x + 1)(1, 2) + ( 1 + i(2x y + 3))(0, i) + i(i, 1 i).
i),
(Maths Dept., IIT Delhi)
18 / 53
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
Question 3
Show that X = {(1 + i, 1 i), (1 i, 1 + i), (2, i), (3, 2i)} is linearly
independent in C2 (R). Express (a + ib, c + id) as an R-linear combination
of vectors belonging to X .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
19 / 53
Question 3
Show that X = {(1 + i, 1 i), (1 i, 1 + i), (2, i), (3, 2i)} is linearly
independent in C2 (R). Express (a + ib, c + id) as an R-linear combination
of vectors belonging to X .
Solution:
Let ↵, , , 2 R such that
↵(1 + i, 1
)(↵ +
i) + (1
i, 1 + i) + (2, i) + (3, 2i) = (0, 0),
+ 2 + 3 + i(↵
), ↵ +
+ i( ↵ +
+
+ 2 )) = (0, 0)
Comparing the real and imaginary part, we get,
↵+
+ 2 + 3 = 0,
↵
=0
↵+
= 0,
↵+
+
=) ↵ =
(Maths Dept., IIT Delhi)
+2 =0
=
=
MTL100 - Calculus, Tutorial-2
= 0.
Semester-I, 2020-21
19 / 53
Question 3 Cont.
Suppose,
↵(1 + i, 1
)(↵ +
i) + (1
i, 1 + i) + (2, i) + (3, 2i) = (a + ib, c + id),
+ 2 + 3 + i(↵
), ↵ +
+ i( ↵ +
+
+ 2 ))
= (a + ib, c + id)
Comparing the real and imaginary part, we get,
↵+
+ 2 + 3 = a,
↵
=b
↵+
= c,
↵+
On solving, we get,
c b
↵ = b+c
,
=
2
2 , = 2a
(Maths Dept., IIT Delhi)
3b
+
+2 =d
2c
3d, = 2b + c + 2d
MTL100 - Calculus, Tutorial-2
a.
Semester-I, 2020-21
20 / 53
Question 4
Question 4:
Let V be a vector space over F. Show that u, v , w 2 V are linearly
independent if and only if u + v , v + w , w + u are linearly independent.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
21 / 53
Question 4
Question 4:
Let V be a vector space over F. Show that u, v , w 2 V are linearly
independent if and only if u + v , v + w , w + u are linearly independent.
Solution:
Let ↵, ,
2 F such that
↵(u + v ) + (v + w ) + (w + u) = 0
(↵ + )u + (↵ + )v + ( + )w = 0.
Since u, v , w are linearly independent, hence,
↵+
=0 &
=) ↵ =
(Maths Dept., IIT Delhi)
=
↵+
=0 &
+
=0
= 0.
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
21 / 53
Question 4 Cont.
Conversely, Let ↵, ,
2 F such that
↵u + v + w = 0.
(1)
Then,
↵u + v + w = ↵(u + v ) + (v + w ) + (w + u)
= ↵(u + v ) + (v + w ) + (w + u)
h
↵(v + w ) + (w + u) + (u + v )
= (↵
)(u + v ) + (
⇥
⇤
+ ↵w + u + v
⇥
↵)(v + w ) + (
⇥
↵v + w + u
↵w + u + v
)(w + u)
= (↵
)(u + v ) + (
↵)(v + w ) + (
)(w + u)
⇥
⇤
+ ↵(w + u) + (u + v ) + (v + w ) ↵u + v + w
= (↵ +
)(u + v ) + ( +
⇥
⇤
↵u + v + w .
(Maths Dept., IIT Delhi)
↵)(v + w ) + ( + ↵
MTL100 - Calculus, Tutorial-2
⇤i
⇤
)(w + u)
Semester-I, 2020-21
22 / 53
Question 4 Cont.
Since u + v , v + w , w + u are linearly independent. Hence,
↵+
= 0,
↵+
↵
+
+
= 0,
= 0,
(2)
=) ↵ =
=
= 0.
Hence u, v , w are linearly independent.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
23 / 53
Question 5
Question 5
(a) Find the coordinates of (a, b, c) 2 R 3 relative to ordered basis
{(1, 0, 0), (1, 1, 0), (1, 1, 1)}.
(b) Find the coordinates of a + bx + cx 2 relative to ordered basis
{1, 1 + x, 1 + x 2 } in the space P3 of polynomials of degree at most 2
with coefficients from R.
(c) Find the coordinate vector of an element 2 R 3 with respect to
following ordered bases B1 = {(1, 2, 1), (1, 2, 3), (0, 1, 1)} and
B2 = {(1, 0, 0), (1, 1, 0), (1, 1, 1). Also write the change of coordinate
matrix.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
24 / 53
Question 5
Recall:
Suppose B = {v1 , v2 , ..., vn } be a basis of a vector space V over the field
F. Fix the ordering of elements in B as they are listed. Then we know that
v = a1 v1 + a2 v2 + + an vn (recall that the coefficients are uniquely
determined). We write [v ]B = (a1 , a2 , · · · , an )T which is in F and called
the coordinate vector of v with respect to the basis B.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
25 / 53
Question 5
Question 5(a)
Find the coordinates of (a, b, c) 2 R 3 relative to ordered basis
{(1, 0, 0), (1, 1, 0), (1, 1, 1)}.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
26 / 53
Question 5
Question 5(a)
Find the coordinates of (a, b, c) 2 R 3 relative to ordered basis
{(1, 0, 0), (1, 1, 0), (1, 1, 1)}.
Solution:
Let
(a, b, c) = ↵(1, 0, 0) + (1, 1, 0) + (1, 1, 1),
=) a = ↵ +
b=
+ ,
+ ,
c= .
After solving above set of equations, we get,
↵ = (a b), = (b c), = c.
Hence, coordinates are given by, (a, b, c)B = (a
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
b, b
c, c)T .
Semester-I, 2020-21
26 / 53
Question 5
Question 5(b)
Find the coordinates of a + bx + cx 2 relative to ordered basis
{1, 1 + x, 1 + x 2 } in the space P3 of polynomials of degree at most 2 with
coefficients from R.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
27 / 53
Question 5
Question 5(b)
Find the coordinates of a + bx + cx 2 relative to ordered basis
{1, 1 + x, 1 + x 2 } in the space P3 of polynomials of degree at most 2 with
coefficients from R.
Solution:
Let
a + bx + cx 2 = ↵(1) + (1 + x) + (1 + x 2 ),
=) a = ↵ +
+ ,
b= ,
c= .
Hence, ↵ = (a
b
c),
= b,
= c.
Hence, coordinates are given by, (a + bx + cx 2 )B = (a
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
b
c, b, c)T .
Semester-I, 2020-21
27 / 53
Question 5
Question 5(c)
Find the coordinate vector of an element 2 R 3 with respect to following
ordered bases B1 = {(1, 2, 1), (1, 2, 3), (0, 1, 1)} and
B2 = {(1, 0, 0), (1, 1, 0), (1, 1, 1). Also write the change of coordinate
matrix.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
28 / 53
Question 5
Question 5(c)
Find the coordinate vector of an element 2 R 3 with respect to following
ordered bases B1 = {(1, 2, 1), (1, 2, 3), (0, 1, 1)} and
B2 = {(1, 0, 0), (1, 1, 0), (1, 1, 1). Also write the change of coordinate
matrix.
Solution:
Let (a, b, c) 2 R3 .
Also,
(a, b, c) = ↵(1, 2, 1) + (1, 2, 3) + (0, 1, 1),
=) a = ↵ + ,
b = 2↵ + 2 + ,
c =↵+3 +
a+b c
=) ↵ =
,
2
(Maths Dept., IIT Delhi)
=
a
b+c
,
2
MTL100 - Calculus, Tutorial-2
= (b
2a).
Semester-I, 2020-21
28 / 53
Question 5(c) cont.
Hence, the coordinates with respect to B1 are given by
T.
( a+b2 c , a b+c
,
(b
2a))
2
Similarly, for the coordinate with respect to ordered basis B2 ,
(a, b, c) = ↵(1, 0, 0) + (1, 1, 0) + (1, 1, 1).
After solving, we get, the coordinates with respect to B2
(a b, b c, c)T .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
29 / 53
Question 5(c) cont.
Change of coordinate matrix,
First express the vectors of B1 in linear combinations of vectors of B2
(1, 2, 1) = a11 (1, 0, 0) + a12 (1, 1, 0) + a13 (1, 1, 1),
(1, 2, 3) = a21 (1, 0, 0) + a22 (1, 1, 0) + a23 (1, 1, 1),
(0, 1, 1) = a31 (1, 0, 0) + a32 (1, 1, 0) + a33 (1, 1, 1)
Solve the obtained system of equation for the scalars
a11 =
1, a12 = 1, a13 = 1,
a21 =
1, a22 =
a31 =
1, a32 = 0, a33 = 1
1, a23 = 3,
Arrange the scalars column wise to get the change of coordinate
matrix,
2
3
1
1
1
41
1 05
1
3
1
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
30 / 53
Question 6
Question 6
(a) Show that if v 2 V then Fv := { v :
vector space V over F.
2 F} is a subspace of any
(b) Show that if W1 , W2 are subspaces of V , then W1 \ W2 is a subspace
of V .
(c) Show that the intersection of any collection of subspaces of a vector
space is a subspace.
(d) Suppose W1 and W2 are subspaces of a vector space V . Show that
W1 [ W2 is a subspace of V if and only if either W1 ⇢ W2 or
W2 ⇢ W1 .
(e) Let X be a non
Pnempty subset of a vector space V over F. Let
span(X ) := { i=1 ai vi : n 2 N, ai 2 F, vi 2 X } and let < X > be the
intersection of all the subspace of V which contain X . Show that
span(X ) and < X > are subspaces of V . Also show that
span(X ) =< X >.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
31 / 53
Question 6
Question 6(a)
Show that if v 2 V then Fv := { v :
space V over F.
(Maths Dept., IIT Delhi)
2 F} is a subspace of any vector
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
32 / 53
Question 6
Question 6(a)
Show that if v 2 V then Fv := { v :
space V over F.
2 F} is a subspace of any vector
Solution:
Let
= 0, then v = 0 2 W ) W 6= .
Let x =
1v , y =
2 v 2 W and ↵,
↵x + y = (↵ 1 +
2 )v =
2 F, then
0
v 2 W for some
0
2F
Hence, W is a subspace.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
32 / 53
Question 6
Question 6(b)
Show that if W1 , W2 are subspaces of V , then W1 \ W2 is a subspace of
V.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
33 / 53
Question 6
Question 6(b)
Show that if W1 , W2 are subspaces of V , then W1 \ W2 is a subspace of
V.
Solution:
Since 0 2 W1 \ W2 , we have W1 \ W2 6=
Let x, y 2 W1 \ W2 and ↵,
Then
2 F.
x, y 2 W1 & x, y 2 W2
) ↵x + y 2 W1 & ↵x + y 2 W2
) ↵x + y 2 W1 \ W2 .
Hence, W1 \ W2 is a subspace of V .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
33 / 53
Question 6
Question 6(c)
Show that the intersection of any collection of subspaces of a vector space
is a subspace.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
34 / 53
Question 6
Question 6(c)
Show that the intersection of any collection of subspaces of a vector space
is a subspace.
Solution: Let V be a vector space over a field F, and W = \ 2^ W ,
where W is a subspace of V .
Since 0 2 W , W 6= .
Let x, y 2 W and ↵, 2 F.
Then
x, y 2 W 8
2^
) ↵x + y 2 W 8
) ↵x + y 2 W .
2^
Hence intersection of any collection of subspaces of a vector space is
a subspace.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
34 / 53
Question 6
Question 6(d)
Suppose W1 and W2 are subspaces of a vector space V . Show that
W1 [ W2 is a subspace of V if and only if either W1 ⇢ W2 or W2 ⇢ W1 .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
35 / 53
Question 6
Question 6(d)
Suppose W1 and W2 are subspaces of a vector space V . Show that
W1 [ W2 is a subspace of V if and only if either W1 ⇢ W2 or W2 ⇢ W1 .
Solution:
Suppose W1 [ W2 is a subspace of V . Let x 2 W1 and y 2 W2 .
Then
x + y 2 W1 [ W2
) x + y 2 W1 or x + y 2 W2
)x +y
x = y 2 W1 or x + y
) W1 ⇢ W2 or W2 ⇢ W1 .
y = x 2 W2
Conversely, suppose W1 ⇢ W2 or W2 ⇢ W1 then W1 [ W2 ⇢ W2 or
⇢ W1 , Hence, it is a subspace of V .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
35 / 53
Question 6
Question 6(e)
Let X be a non
Pnempty subset of a vector space V over F. Let
span(X ) := { i=1 ai vi : n 2 N, ai 2 F, vi 2 X } and let < X > be the
intersection of all the subspace of V which contain X . Show that span(X )
and < X > are subspaces of V . Also show that span(X ) =< X >.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
36 / 53
Question 6
Question 6(e)
Let X be a non
Pnempty subset of a vector space V over F. Let
span(X ) := { i=1 ai vi : n 2 N, ai 2 F, vi 2 X } and let < X > be the
intersection of all the subspace of V which contain X . Show that span(X )
and < X > are subspaces of V . Also show that span(X ) =< X >.
Solution: First prove that span(X ) and < X > are subspaces of V .
Pn
Let ai = 0 8i, then i=1 ai vi = 0 2 span(X) ) span(X) 6= .
Pn
Pn
Let x, y 2 span(X), then x = i=1 ci vi and y = i=1 di vi and let
↵, 2 F
) ↵x + y =
n
X
i=1
(↵bi + ci )vi 2 span(X ).
Hence, span(X) is a subspace of V .
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
36 / 53
Question 6(e) cont.
Since, the intersection of subspaces of V is a subspace of V, Hence,
< X > is also a subspace of V .
To show: span(X)=< X >.
Clearly, X ⇢ span(X), hence < X > ⇢ span(X)
Pn
Let x = i=1 ai vi 2 span(X).
Since, vi 2< X >,
n
X
i=1
ai vi 2< X >
) x 2< X >
) span(X ) ⇢< X >
Hence span(X)=< X >.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
37 / 53
Question 7
Question 7
In each case show that W1 + W2 = V (directly) and find dim(W1 \ W2 ).
Verify the formula dim(W1 + W2 ) = dimW1 + dimW2 dim(W1 \ W2 ).
(a) V = R2 , W1 is the X-axis, W2 Y-axis.
(b) V = R2 , W1 and W2 are distinct line passing through the origin.
(c) V = R3 , W1 is XY plane and W2 is YZ plane.
(d) V = Mn (R), W1 = {A 2 Mn (R) : A is upper triangular } and
W2 = {A 2 Mn (R) : A is lower triangular }.
(e) V = Mn (R), W1 = {A 2 Mn (R) : A is a symmetric } and
W2 = {A 2 Mn (R) : A is skew-symmetric }.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
38 / 53
Question 7(a)
Question 7(a)
V = R2 , W1 is the X-axis, W2 Y-axis.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
39 / 53
Question 7(a)
Question 7(a)
V = R2 , W1 is the X-axis, W2 Y-axis.
Solution:
Let (x, y ) 2 R2 , then (x, y ) = (x, 0) + (0, y ) ) V = W1 + W2
dimW1 = dimW2 = 1
Clearly, W1 \ W2 = (0, 0) ) dim(W1 \ W2 ) = 0.
Hence, dim(W1 + W2 ) = dimW1 + dimW2
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
dim(W1 \ W2 ).
Semester-I, 2020-21
39 / 53
Question 7
Question 7(b)
V = R2 , W1 and W2 are distinct line passing through the origin.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
40 / 53
Question 7
Question 7(b)
V = R2 , W1 and W2 are distinct line passing through the origin.
Solution:
Since, (x, y ) 2 R2 can be expressed as sum of the points lies on two
distinct lines(Using parallelogram law). So, ) V = W1 + W2
dimV = 2, dimW1 = dimW2 = 1
Since lines are distinct and passing through origin
W1 \ W2 = (0, 0) ) dim(W1 \ W2 ) = 0.
Hence, dim(W1 + W2 ) = dimW1 + dimW2
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
dim(W1 \ W2 ).
Semester-I, 2020-21
40 / 53
Question 7
Question 7(c)
V = R3 , W1 is XY plane and W2 is YZ plane.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
41 / 53
Question 7
Question 7(c)
V = R3 , W1 is XY plane and W2 is YZ plane.
Solution:
V = R3 =) dimV = 3.
W1 = {(x, y , 0); x, y 2 R} =) dimW1 = 2.
W2 = {(0, y , z); y , z 2 R} =) dimW2 = 2.
W1 \ W2 = {(0, y , 0); y 2 R} =) dimW1 \ W2 = 1.
Let (x, y , z) 2 R3 , then
(x, y , z) = (x, y , 0) + (0, 0, z) = (x, 0, 0) + (0, y , z) ) V = W1 + W2 .
Hence, dim(W1 + W2 ) = dimW1 + dimW2 - dim(W1 \ W2 ).
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
41 / 53
Question 7
Question 7(d)
V = Mn (R), W1 = {A 2 Mn (R) : A is upper triangular } and
W2 = {A 2 Mn (R) : A is lower triangular }.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
42 / 53
Question 7
Question 7(d)
V = Mn (R), W1 = {A 2 Mn (R) : A is upper triangular } and
W2 = {A 2 Mn (R) : A is lower triangular }.
Solution:
Since, any matrix can be written as sum of an upper triangular matrix
and a lower triangular matrix, so ) V = W1 + W2 .
n(n+1)
dimV = n2 , dimW1 = n(n+1)
and
dimW
=
. (How?)
2
2
2
Since, W1 \ W2 = {A 2 Mn (R) : A is a diagonal matrix } so,
dim(W1 \ W2 ) = n.
Hence, dim(W1 + W2 ) = dimW1 + dimW2 - dim(W1 \ W2 ).
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
42 / 53
Question 7
Question 7(e)
V = Mn (R), W1 = {A 2 Mn (R) : A is a symmetric } and
W2 = {A 2 Mn (R) : A is skew-symmetric }.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
43 / 53
Question 7
Question 7(e)
V = Mn (R), W1 = {A 2 Mn (R) : A is a symmetric } and
W2 = {A 2 Mn (R) : A is skew-symmetric }.
Solution:
A+A0
A A0
A+A0
Let A 2 Mn (R), A = 2 + 2 , where 2 is symmetric and
A A0
is skew-symmetric, so ) V = W1 + W2
2
n(n 1)
dimV = n2 , dimW1 = n(n+1)
and
dimW
=
2
2
2
Since, only zero matrice is both symmetric and skew-symmetric, so
W1 \ W2 = { zero matrix } so, dim(W1 \ W2 ) = 0.
Hence, dim(W1 + W2 ) = dimW1 + dimW2 - dim(W1 \ W2 ).
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
43 / 53
Question 8
Question 8
Which of the following is a linear transformation? Justify.
(a) T1 : R2 ! R2 , define by T1 (x, y ) = (x 2 + y 2 , x
(b) T2 : R2 ! R2 , define by T2 (x, y ) = (x + y + 1, x
y ) over R.
y ) over R.
(c) T3 : R2 ! R2 , define by T3 (x, y ) = (ax + by , cx + dy ) over R.
(d) T4 : C ! C define by T4 (z) = z̄ over C.
(e) T : R2 ! R2 the rotation about the origin by an angle ✓.(Write the
expression for rotation)
(f) T5 : Mm⇥n (F) ! Mn⇥m (F) define by T5 (A) = A0 (A0 transpose of A)
(g) T6 : Mn (F) ! (F) define by T6 (A) =tr(A)
(h) T7 : Pn ! Pn such that T7 (p)(x) = p(x
1).
(i) T8 : Pn ! Pn+1 such that T8 (p)(x) = xp(x) + p(1).
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
44 / 53
Question 8
Recall:
Suppose V and W are vector spaces over the same field F. A map
T : V ! W is called a linear transformation if
T (au + bv ) = aT (u) + bT (v )
for any a, b 2 F and any u, v 2 V .
In particular,
T (0V ) = 0W where 0V is the zero vector of vector space V and 0W
is the zero vector space of W .
T (au) = aT (u).
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
45 / 53
Question 8 cont.
Question 8(a)
T1 : R2 ! R2 , define by T1 (x, y ) = (x 2 + y 2 , x
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
y ) over R.
Semester-I, 2020-21
46 / 53
Question 8 cont.
Question 8(a)
T1 : R2 ! R2 , define by T1 (x, y ) = (x 2 + y 2 , x
y ) over R.
Solution:
Since,
T1 (a(x, y )) = T1 (ax, ay ) = (a2 (x 2 + y 2 ), a(x
Hence, T1 is not a linear transformation.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
y )) 6= aT1 (x, y ),
Semester-I, 2020-21
46 / 53
Question 8 cont.
Question 8(a)
T1 : R2 ! R2 , define by T1 (x, y ) = (x 2 + y 2 , x
y ) over R.
Solution:
Since,
T1 (a(x, y )) = T1 (ax, ay ) = (a2 (x 2 + y 2 ), a(x
Hence, T1 is not a linear transformation.
y )) 6= aT1 (x, y ),
Question 8(b)
T2 : R2 ! R2 , define by T2 (x, y ) = (x + y + 1, x
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
y ) over R.
Semester-I, 2020-21
46 / 53
Question 8 cont.
Question 8(a)
T1 : R2 ! R2 , define by T1 (x, y ) = (x 2 + y 2 , x
y ) over R.
Solution:
Since,
T1 (a(x, y )) = T1 (ax, ay ) = (a2 (x 2 + y 2 ), a(x
Hence, T1 is not a linear transformation.
y )) 6= aT1 (x, y ),
Question 8(b)
T2 : R2 ! R2 , define by T2 (x, y ) = (x + y + 1, x
y ) over R.
Solution:
Since, T2 (0, 0) 6= (0, 0),
Hence, T1 is not a linear transformation.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
46 / 53
Question 8 cont.
Question 8(c)
T3 : R2 ! R2 , define by T3 (x, y ) = (ax + by , cx + dy ) over R.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
47 / 53
Question 8 cont.
Question 8(c)
T3 : R2 ! R2 , define by T3 (x, y ) = (ax + by , cx + dy ) over R.
Solution:
Let u = (x1 , y1 ), v = (x2 , y2 ) 2 R2 and ↵,
2 R, then
T3 (↵u + v ) = T3 (↵(x1 , y1 ) + (x2 , y2 ))
= T3 (↵x1 + x2 , ↵y1 + y2 )
✓
◆
= a(↵x1 + x2 ) + b(↵y1 + y2 ), c(↵x1 + x2 ) + d(↵y1 + y2 )
✓
◆
= ↵(ax1 + by1 ) + (ax2 + by2 ), ↵(cx1 + dy1 ) + (cx2 + dy2 )
✓
◆
= ↵(ax1 + by1 , cx1 + dy1 ) + (ax2 + by2 , cx2 + dy2 )
= ↵T3 (u) + T3 (v ).
Hence, T3 is a linear transformation.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
47 / 53
Question 8 cont.
Question 8(d)
T4 : C ! C define by T4 (z) = z̄ over C.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
48 / 53
Question 8 cont.
Question 8(d)
T4 : C ! C define by T4 (z) = z̄ over C.
Solution:
Since, T4 (az) = az
¯ = āz̄ 6= aT4 (z)
Hence, T4 is not a linear transformation.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
48 / 53
Question 8
Question 8(e)
T : R2 ! R2 the rotation about the origin by an angle ✓.(Write the expression for
rotation)
Solution:
T (x, y ) = (x cos ✓
tutsheet 1).
y sin ✓, x sin ✓ + y cos ✓) 8 (x, y ) 2 R (from question 6
Let u = (x1 , y1 ), v = (x2 , y2 ) 2 R2 and ↵,
2 R, then
T (↵u + v ) = T (↵(x1 , y1 ) + (x2 , y2 ))
= T (↵x1 + x2 , ↵y1 + y2 )
✓
= (↵x1 + x2 ) cos ✓ (↵y1 + y2 ) sin ✓
◆
, (↵x1 + x2 ) sin ✓ + (↵y1 + y2 ) cos ✓
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
49 / 53
Question 8(e)cont.
=
✓
↵(x1 cos ✓
y1 sin ✓) + (x2 cos ✓
y2 sin ✓)
◆
, ↵(x1 sin ✓ + y1 cos ✓) + (x2 sin ✓ + y2 cos ✓)
✓
= ↵(x1 cos ✓ y1 sin ✓, x1 sin ✓ + y1 cos ✓)
◆
+ (x2 cos ✓ y2 sin ✓, x2 sin ✓ + y2 cos ✓)
= ↵T (u) + T (v ).
Hence, T is a linear transformation.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
50 / 53
Question 8
Question 8(f)
T5 : Mm⇥n (F) ! Mn⇥m (F) define by T5 (A) = A0 (A0 transpose of A)
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
51 / 53
Question 8
Question 8(f)
T5 : Mm⇥n (F) ! Mn⇥m (F) define by T5 (A) = A0 (A0 transpose of A)
Solution:
Since, T5 (aA + bB) = (aA + bB)0 = aA0 + bB 0 = aT5 (A) + bT5 (B)
for all A, B 2 Mm⇥n and a, b 2 F.
Hence, T5 is a linear transformation.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
51 / 53
Question 8
Question 8(f)
T5 : Mm⇥n (F) ! Mn⇥m (F) define by T5 (A) = A0 (A0 transpose of A)
Solution:
Since, T5 (aA + bB) = (aA + bB)0 = aA0 + bB 0 = aT5 (A) + bT5 (B)
for all A, B 2 Mm⇥n and a, b 2 F.
Hence, T5 is a linear transformation.
Question 8(g)
T6 : Mn (F) ! (F) define by T6 (A) =tr(A)
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
51 / 53
Question 8
Question 8(f)
T5 : Mm⇥n (F) ! Mn⇥m (F) define by T5 (A) = A0 (A0 transpose of A)
Solution:
Since, T5 (aA + bB) = (aA + bB)0 = aA0 + bB 0 = aT5 (A) + bT5 (B)
for all A, B 2 Mm⇥n and a, b 2 F.
Hence, T5 is a linear transformation.
Question 8(g)
T6 : Mn (F) ! (F) define by T6 (A) =tr(A)
Solution:
Since, T6 (aA + bB) =
tr(aA + bB) = a trA + b trB = aT6 (A) + bT6 (B) for all A, B 2 Mn
and a, b 2 F.
Hence, T6 is a linear transformation.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
51 / 53
Question 8
Question 8(h)
T7 : Pn ! Pn such that T7 (p)(x) = p(x
(Maths Dept., IIT Delhi)
1).
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
52 / 53
Question 8
Question 8(h)
T7 : Pn ! Pn such that T7 (p)(x) = p(x
1).
Solution:
Since,
T7 (ap + bq)(x) = (ap + bq)(x
= ap(x
1)
1) + bq(x
1)
= aT7 (p)(x) + bT7 (q)(x) 8p, q 2 Pn , a, b 2 F.
Hence, T7 is a linear transformation.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
52 / 53
Question 8
Question 8(i)
T8 : Pn ! Pn+1 such that T8 (p)(x) = xp(x) + p(1).
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
53 / 53
Question 8
Question 8(i)
T8 : Pn ! Pn+1 such that T8 (p)(x) = xp(x) + p(1).
Solution:
Since,
T8 (ap + bq)(x) = x(ap + bq)(x) + (ap + bq)(1)
= a(xp(x) + p(1)) + b(xq(x) + q(1))
= aT8 (p)(x) + bT8 (q)(x) 8p, q 2 Pn , a, b 2 F.
Hence, T8 is a linear transformation.
(Maths Dept., IIT Delhi)
MTL100 - Calculus, Tutorial-2
Semester-I, 2020-21
53 / 53