MA 106 : Linear Algebra

Tutorial 4: Linear Transformations

Vrinda Jindal
Indian Institute of Technology Bombay
 Problem 1

Problem 1:Define f : R5 −→ R3 by
f ((x1 , x2 , x3 , x4 , x5 )t ) =
(2x3 − 2x4 + x5 , 2x2 − 8x3 + 14x4 − 5x5 , x2 + 3x3 + x5 )t .
Find bases for the null-space and the range of f , using GJEM.

Solution: We first find the 3 × 5 matrix A that is associated with the

linear transformation f with respect to standard basis of R5 and R3
 
0 0 2 −2 1
A = 0 2 −8 14 −5
 
0 1 3 0 1

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 Problem 1 (Cont.)

Now,
 perform EROs on Ato bring it to Row
 Echelon Form. 
0 0 2 −2 1 0 2 −8 14 −5
 0 2 −8 14 −5  R1 ∼ R2 −→  0 0 2 −2 1 
   
0 1 3 0 1 0 1 3 0 1
 
) 0 1 −4 7 −5/2
R1 /2
−→  0 0 2 −2 1 
 
R3 − R1
0 0 7 −7 7/2
 
) 0 1 −4 7 −5/2
R2 /2
−→  0 0 1 −1 1/2 
 
R3 − 7R2
0 0 0 0 0

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 Problem 1 (cont.)
 
0 1 −4 7 −5/2
A is transformed into G(A) =  0 0 1 −1 1/2 .
 
0 0 0 0 0
Range / Image of f is the column-space of A.
From the Row Echelon Form of A, G(A), it is clear that Column 2 and
Column 3 are the pivotal columns.
So, the columns (0, 2, 1)t and (2, −8, 3)t of A form a basis for the
range of f .

The null-space of f is same as the null-space of A.

i.e. the set of solutions to the homogeneous equation, Ax = 0

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 Problem 1 (cont.)
The above homogeneous equation:
 
  x1
0 1 −4 7 −5/2  x2 

0 0 1 −1 1/2   3 = 0
x
  

 
0 0 0 0 0 x4 
x5
Gives us:

x5
x3 − x4 + =0 (1)
2
5x5
x2 − 4x3 + 7x4 − =0 (2)
2
The nullity of A is 3, and columns 1, 4 and 5 are non-pivotal columns.
Use three sets of special values of x1 , x4 , x5 to find the basis of the null-space.
Put x1 = 1, x4 = 0, x5 = 0 to get (1, 0, 0, 0, 0)t
Put x1 = 0, x4 = 1, x5 = 0 to get (0, −3, 1, 1, 0)t
Put x1 = 0, x4 = 0, x5 = 1 to get (0, 1/2, −1/2, 0, 1)t
These three vectors are a basis of the null-space of A and hence of f .

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 Problem 2 (i)
Problem 2: Find the range and null-space of the following linear
transformations. Also find the rank and nullity wherever applicable.
(i) T : R2 −→ R2 defined by T(x1 , x2 )t = (x1 + x2 , x1 )t .
" #
1 1
Solution: The 2 × 2 associated matrix A =
1 0
Perform EROs to get Row Echelon Form, G(A)
) " #
R2 − R1 1 1
−→
(−1)R2 0 1
There are no pivot free columns, so nullity of A (and hence of T) is 0.
The null-space is singular as it contains only the zero-vector, (0).
Both the columns of A are pivot-columns, so the rank of A (and hence
of T) is 2. The dimension of the range is 2 and the range is R2 .

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 Problem 2 (ii)

Problem 2: Find the range and null-space of the following linear

transformations. Also find the rank and nullity wherever applicable.
(ii) T : C1 (0, 1) −→ C(0, 1) defined by T(f )(x) = f 0 (x)ex
Solution: C1 (0, 1) is the set of all functions that are derivable on
(0, 1) and this derivative is continuous on (0,1).
C(0, 1) is the set of all functions that are continuous over the interval
(0, 1).
For finding the null-space of T, we want f 0 (x)ex = 0 for x ∈ (0, 1) ,
which implies f 0 (x) = 0, as ex is never zero in (0, 1).
So the null-space of T is the set of all constant functions. i.e.
{f (x) = c|c ∈ R}.

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 Problem 2 (ii) Cont.

Given any g ∈ C(0, 1), take f (x) = g(x)e−x .

R

Now, f (x) ∈ C1 (0, 1), and T(f )(x) = g(x). Therefore the range of T
is the entire space C(0, 1).

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 Problem 3 (i)

Problem 3 (i): Find a linear transformation T : R3 −→ R3 such that

the set of all vectors satisfying 4x1 − 3x2 + x3 = 0 is – (i) the
null-space of T
Solution: Notice, that the vectors that satisfy the given equation
4x1 − 3x2 + x3 = 0 form a 2-dimensional subspace of R3 (say C).
To be precise, C is a plane in R3 .
So we pick up 2 independent vectors in the given plane to find a basis
of this subspace C, v1 = (3, 4, 0)t and v2 = (0, 1, 3)t
We then extend this set by adding v3 to it, such that (v1 , v2 , v3 ) form a
basis for R3 .
In order to do so, we pick v3 that is perpendicular to the plane of v1
and v2 . Picking v3 = (4, −3, 1)t .

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 Problem 3 (i) (Cont.)

Now, we want x ∈ C ⇐⇒ T(x) = 0.

Take T(v1 ) = 0, T(v2 ) = 0, T(v3 ) = 1.
For any x ∈ C, x = a1 v1 + a2 v2 for some unique a1 , a2 ∈ R.
T(x) = T(a1 v1 + a2 v2 ) = a1 T(v1 ) + a2 T(v2 ) = 0, So null-space of T
is indeed the plane given above.
Similarly, if for x ∈ R3 , say x = a1 v1 + a2 v2 + a3 v3 for some unique
reals a1 , a2 , a3 , T(x) = 0, implies a3 = 0, which is the same as x ∈ C
For any general y ∈ R3 , y = a1 v1 + a2 v2 + a3 v3 for some unique
a1 , a2 , a3 ∈ R, So,
T(y) = T(a1 v1 + a2 v2 + a3 v3 ) = a1 T(v1 ) + a2 T(v2 ) + a3 T(v3 ) = a3 1.
So T is a valid linear transformation that satisfies the given condition.

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 Problem 3 (i) Cont.

Note that such a T is not unique.

Another solution is the linear transformation, whose matrix w.r.t the
standard
 basis is 
4 −3 1
A = 0 0 0
 
0 0 0

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 Problem 3 (ii)
Problem 3 (ii): Find a linear transformation T : R3 −→ R3 such that
the set of all vectors satisfying 4x1 − 3x2 + x3 = 0 is – (ii) the range
of T.
Solution: Let C, v1 , v2 be same as described in the previous part.
The set (e1 , e2 , e3 ) form a basis of R3 .
Recall, Range(T) = {T(x)|x ∈ R3 } and we want Range(T) = C.
Take T(e1 ) = v1 , T(e2 ) = v2 , T(e3 ) = 0.
Now, for any x ∈ R3 , x = a1 e1 + a2 e2 + a3 e3 , for some uniquely
determined a1 , a2 , a3 ∈ R
T(x) = T(a1 e1 + a2 e2 + a3 e3 ) = a1 T(e1 ) + a2 T(e2 ) + a3 T(e3 ) =
a1 v1 + a2 v2 ∈ C. (Similarly prove other way round).
Hence the range of T is indeed C.
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 Problem 4

Problem 4: Let P[x] denote the space of all real polynomials in one
variable. Let V = {p(x) ∈ P[x] : p(0) = 0}.
Prove that taking the derivative defines a one-to-one linear
Z x
transformation from D : V −→ P and D−1 (p)(x) = p(t) dt .
0
Solution: Proving that D is a linear transformations:
Let f , g be arbitrary one-variable real polynomials ∈ V and α, β ∈ R
be some scalars.
D(αf + βg) = (αf + βg)0 = (αf )0 + (βg)0 = αf 0 + βg0 =
αD(f ) + βD(g).
Hence, proved.

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 Problem 4 Cont.

Z x
Proving that D−1 (p)(x) = p(t) dt .
Z x 0

Let F = p(t) dt.

0

If we prove, D(F(p(x)) = p(x), we are done.

Z x Z x
d
Now, D(F(p(x)) = D( p(t) dt) = dx ( p(t) dt) =
0 0
d d
p(x) dx x + p(0) dx 0 = p(x)
Hence, proved.

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 Problem 5 (a)
Problem 5: Let f : V −→ W be a linear transformation.
(a) Suppose f is injective and S ⊂ V is linearly independent. Then
show that f (S) is linearly independent.
Solution: Let S = {v1 , v2 , . . . , vk }, then,
f (S) = {f (v1 ), f (v2 ), . . . , f (vk )}.
We need to prove that f (S) is linearly independent.
In other words, we need to prove that ki=1 αi f (vi ) = 0 only if
P

α1 = · · · = αk = 0
Now, by virtue of f being a linear transformation, ki=1 αi f (vi ) = 0
P
P
means f ( i αi vi ) = 0.
Recall, a linear-map f of vector spaces is injective if and only if
N (f ) = {0}. (See lecture notes for proof)
P P
Since, f in injective,f ( i αi vi ) = 0 implies i αi vi = 0. And since S
is a linearly independent set, this means that α1 = · · · = αk = 0.
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 Problem 5(b)

Problem 5: Let f : V −→ W be a linear transformation.

(b) Suppose f is onto and S spans V. Then show that f (S) spans W.
Solution:
f is onto, this means that ∀w ∈ W, ∃v ∈ V s.t. f (v) = w.
Let S = {v1 , v2 , . . . , vk }, (S spans V).
For w ∈ W choose v ∈ V s.t. f (v) = w.
P
Since, v ∈ V, we can write v = i αi vi for scalars αi .
P P
So, f (v) = f ( i αi vi ) = w = i αi f (vi ) ∈ L(f (S)).
And, since f is onto, the choice of w can be anything in W, So,
L(f (S)) = W

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 Problem 5(c)

Problem 5: Let f : V −→ W be a linear transformation.

(c) Suppose S is a basis for V and f is an isomorphism then show that
f (S) is a basis for W.
Solution:
Combining the previous two parts,
From (a) we know that f (S) is linearly independent.
From (b) we know that f (S) spans W.
So, f (S) in a linearly independent set that spans W, this means that
f (S) must be a basis of W.
Hence, proved.

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 Problem 6
Problem 6: Let V be a finite dimensional vector space and
f : V −→ V be a linear map. Prove that the following are equivalent:
(i) f is an isomorphism.
(ii) f is injective.
(iii) f is surjective.
Solution:
Clearly (i) implies all the other statements. So it remains to show that
each one of the other statements implies (i).
Let S be a basis for V. Clearly, S has n elements where n = dim V.
(ii) =⇒ (i)
Since f if one-one, N (f ) = {0}., i.e. nullity of f if 0.
So, by rank-nullity theorem, the rank of f if n.
This means, that the range of f has n dimensions, but it is also a subset
of V. And V satisfies this, so f if surjective.
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 Problem 6 Cont.

(iii) =⇒ (i) f is surjective. This means that range of f if V. The rank

of f is equal to dim(v) = n.
By rank-nullity theorem, nullity of f is 0, i.e. N (f ) = {0}. By the
property proved in class, this means that f in one-one.

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 Problem 7
Problem 7: Consider the linear transformations T1 : U −→ V and
T2 : V −→ W. If T2 is one-one then show that
rank(T2 ◦ T1 ) = rank(T1 ).
Solution:
Recall that rank of a linear is the dimension of its image.
Now R(T2 ◦ T1 ) = T2 (R(T1 )).
Let B = {v1 , . . . , vk } be a basis for R(T1 ).
Then from Q5, it follows that T2 (B) = {T2 (v1 ), . . . , T2 (vk )} is L.I.
P
Any x ∈ R(T1 ) can be written as i αi vi for some unique reals αi s.
P
So, T2 (x) = i αi T2 (vi ) i.e., T2 (B) spans the range of T2 ◦ T1 .
Hence it is a basis for T2 (T1 (U)). So, the dimension of the image of
T2 ◦ T1 is equal to k = cardinality of B = Rank of T1 .

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 Selected Practice Problems

Problem 10: Find the range and null-space of the following linear
transformations. Also find the rank and nullity wherever applicable.
(i) T : R2 −→ R3 defined by T(x1 , x2 )t = (x1 , x1 + x2 , x2 )t .
Solution:
  The associated matrix (w.r.t the standard basis) is
1 0
 1 1 .
 
0 1
No non-pivotal columns, so nullity is 0, and rank is 2.
The range is given by y1 − y2 + y3 = 0. The null space is (0).

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 Select Practice Problems

Problem 15: Let A be a m × n matrix where m ≤ n. Use the

Rank-Nulity theorem to show that the space of solutions of Ax = 0
has dimension atleast n − m.
Solution:
m ≤ n =⇒rank(A) ≤ m.
Dimension of space of solution of Ax = 0 is nulity of A.
By rank-nulity theorem, rank(A) + null(A) = n.
This means, null(A) ≥ n − m.
Hence, proved.

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 Selected Practice Problems
Problem 9: Show that each of the following linear transformations is
1-1 and onto and find its inverse.
(i) T1 : R2 −→ R2 defined by T(x1 , x2 ) = (5x1 , 3x2 )
Solution:
(i) Find the 2 × 2 matrix A, associated with f with respect to the
standard
" basis.
#
5 0
A=
0 3
A is in the REF, and has rank 2 , This means that nullity of A is 0, and
hence f "is one-one.
#
1
0
A−1 = 5 1
0 3
−1
A is the matrix associated with the inverse of T1 .
So, T −1 (x1 , x2 ) = (x1 /5, x2 /5)
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 Selected Practice Problems

Problem 9: Show that each of the following linear transformations is

1-1 and onto and find its inverse.
(ii) T2 : R3 −→ R3 defined by
T(x1 , x2 , x3 ) = (x1 + x2 + x3 , x3 + x2 , x3 ).
Solution:
 the 3 ×3 matrix 
(i) Find A, associated with
 f.
1 1 1 1 −1 0
A = 0 1 1, A−1 = 0 1 −1
   
0 0 1 0 0 1
A has rank 3 , This means that nullity of A is 0, and hence f is one-one.
A−1 is the matrix associated with the inverse of T1 .
So, T −1 (x1 , x2 , x3 ) = (x1 − x2 , x2 − x3 , x3 ).

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 Selected Practice Problems
Problem 12: Let T : U −→ V be a linear transformation. Prove that:
(i) T is one-one if and only if N (T) = {0}.
(ii) if L({u1 , u2 , . . . , un }) = U then
R(T) = L({T(u1 ), T(u2 ), . . . , T(un )})
Solution:
(i) Done in class.
P
(ii) For any u ∈ U, we can write u as i αi ui for some scalars αi ’s.
P P
T(u) = T( i αi ui ) = i αi T(ui ) ∈ L({T(u1 ), T(u2 ), . . . , T(un )}).
Hence, R(T) ⊆ L({T(u1 ), T(u2 ), . . . , T(un )})
Similarly, let x ∈ L({T(u1 ), T(u2 ), . . . , T(un )}).
P P
So x = i βi T(ui ) for some scalars βi ’s. See that T( i βi ui ) = x.
This implies L({T(u1 ), T(u2 ), . . . , T(un )}) ⊆ R(T).
Hence proved both ways.
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 Selected Practice Problems
Problem 8: Let f , g : V −→ V be two linear maps which commute
with each other, i.e., f ◦ g = g ◦ f . Show that f (Image(g)) ⊆ Image(g)
and f (Null(g)) ⊆ Null(g).
Solution:
Image(g ◦ f ) = Image(f ◦ g)
Image(g ◦ f ) = g(Image(f ))
Image(f ) ⊆ V =⇒ g(Image(f )) ⊆ g(V) = Image(g)
So, Image(g ◦ f ) ⊆ Image(g)
=⇒ Image(f ◦ g) ⊆ Image(g) =⇒ f (Image(g)) ⊆ Image(g)
Let a ∈ f (Null(g))
This means ∃x ∈ Null(g) s.t. f (x) = a and g(x) = 0
g(f (x)) = g(a) = f (g(x)) = f (0) = 0
So, g(a) = 0 =⇒ a ∈ Null(g)
Hence proved.
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