1 Linear Transformations

We will study mainly finite-dimensional vector spaces over an arbitrary field

F —i.e. vector spaces with a basis. (Recall that the dimension of a vector
space V (dim V ) is the number of elements in a basis of V .)

DEFINITION 1.1
(Linear transformation)
Given vector spaces U and V , T : U 7→ V is a linear transformation (LT)
if
T (λu + µv) = λT (u) + µT (v)
for all λ, µ ∈ F , and u, v ∈ U . Then T (u+v) = T (u)+T (v), T (λu) = λT (u)
and
n n
!
X X
T λk uk = λk T (uk ).
k=1 k=1

EXAMPLES 1.1
Consider the linear transformation

T = TA : Vn (F ) 7→ Vm (F )

where A = [aij ] is m × n, defined by TA (X) = AX.  

x1
Note that Vn (F ) = the set of all n-dimensional column vectors  ...  of
 

xn
F —sometimes written F n .
Note that if T : Vn (F ) 7→ Vm (F ) is a linear transformation, then T = TA ,
where A = [T (E1 )| · · · |T (En )] and
   
1 0
 0   .. 
E1 =  .  , . . . , E n =  . 
   
.
 .   0 
0 1

Note:
 
x1
v ∈ Vn (F ), v =  ...  = x1 E1 + · · · + xn En
 

xn

1
If V is a vector space of all infinitely differentiable functions on R, then

T (f ) = a0 Dn f + a1 Dn−1 f + · · · + an−1 Df + an f

defines a linear transformation T : V 7→ V .

The set of f such that T (f ) = 0 (i.e. the kernel of T ) is important.

Let T : U 7→ V be a linear transformation. Then we have the following

definition:

DEFINITIONS 1.1
(Kernel of a linear transformation)

Ker T = {u ∈ U | T (u) = 0}

(Image of T )

Im T = {v ∈ V | ∃u ∈ U such that T (u) = v}

Note: Ker T is a subspace of U . Recall that W is a subspace of U if

1. 0 ∈ W ,
2. W is closed under addition, and
3. W is closed under scalar multiplication.
PROOF. that Ker T is a subspace of U :

1. T (0) + 0 = T (0) = T (0 + 0) = T (0) + T (0). Thus T (0) = 0, so

0 ∈ Ker T .
2. Let u, v ∈ Ker T ; then T (u) = 0 and T (v) = 0. So T (u + v) =
T (u) + T (v) = 0 + 0 = 0 and u + v ∈ Ker T .
3. Let u ∈ Ker T and λ ∈ F . Then T (λu) = λT (u) = λ0 = 0. So
λu ∈ Ker T .

EXAMPLE 1.1

Ker TA = N (A), the null space of A

= {X ∈ Vn (F ) | AX = 0}
and Im TA = C(A), the column space of A
= hA∗1 , . . . , A∗n i

2
 Generally, if U = hu1 , . . . , un i, then Im T = hT (u1 ), . . . , T (un )i.
Note: Even if u1 , . . . , un form a basis for U , T (u1 ), . . . , T (un ) may not
form a basis for Im T . I.e. it may happen that T (u1 ), . . . , T (un ) are linearly
dependent.

1.1 Rank + Nullity Theorems (for Linear Maps)

THEOREM 1.1 (General rank + nullity theorem)
If T : U 7→ V is a linear transformation then

rank T + nullity T = dim U.

PROOF.

1. Ker T = {0}.
Then nullity T = 0.
We first show that the vectors T (u1 ), . . . , T (un ), where u1 , . . . , un are
a basis for U , are LI (linearly independent):
Suppose x1 T (u1 ) + · · · + xn T (un ) = 0 where x1 , . . . , xn ∈ F .
Then

T (x1 u1 + · · · + xn un ) = 0 (by linearity)

x1 u1 + · · · + xn un = 0 (since Ker T = {0})
x1 = 0, . . . , xn = 0 (since ui are LI)

Hence Im T = hT (u1 ), . . . , T (un )i so

rank T + nullity T = dim Im T + 0 = n = dim V.

2. Ker T = U .
So nullity T = dim U .
Hence Im T = {0} ⇒ rank T = 0

⇒ rank T + nullity T = 0 + dim U

= dim U.

3. 0 < nullity T < dim U .

Let u1 , . . . , ur be a basis for Ker T and n = dim U , so r = nullity T
and r < n.
Extend the basis u1 , . . . , ur to form a basis u1 , . . . , ur , ur+1 , . . . , un of

3
 U (refer to last year’s notes to show that this can be done).
Then T (ur+1 ), . . . , T (un ) span Im T . For

Im T = hT (u1 ), . . . , T (ur ), T (ur+1 ), . . . , T (un )i

= h0, . . . , 0, T (ur+1 ), . . . , T (un )i
= hT (ur+1 ), . . . , T (un )i

So assume

x1 T (ur+1 ) + · · · + xn−r T (un ) = 0

⇒ T (x1 ur+1 + · · · + xn−r un ) = 0
⇒ x1 ur+1 + · · · + xn−r un ∈ Ker T
⇒ x1 ur+1 + · · · + xn−r un = y1 u1 + · · · + yr ur
for some y1 , . . . , yr
⇒ (−y1 )u1 + · · · + (−yr )ur + x1 ur+1 + · · · + xn−r un = 0

and since u1 , . . . , un is a basis for U , all coefficients vanish.

Thus

rank T + nullity T = (n − r) + r
= n
= dim U.

We now apply this theorem to prove the following result:

THEOREM 1.2 (Dimension theorem for subspaces)

dim(U ∩ V ) + dim(U + V ) = dim U + dim V

where U and V are subspaces of a vector space W .

(Recall that U + V = {u + v | u ∈ U, v ∈ V }.)

For the proof we need the following definition:

DEFINITION 1.2
If U and V are any two vector spaces, then the direct sum is

U ⊕ V = {(u, v) | u ∈ U, v ∈ V }

(i.e. the cartesian product of U and V ) made into a vector space by the
component-wise definitions:

4
 1. (u1 , v1 ) + (u2 , v2 ) = (u1 + u2 , v1 + v2 ),
2. λ(u, v) = (λu, λv), and
3. (0, 0) is an identity for U ⊕ V and (−u, −v) is an additive inverse for
(u, v).
We need the following result:
THEOREM 1.3

dim(U ⊕ V ) = dim U + dim V

PROOF.
Case 1: U = {0}
Case 2: V = {0}
Proof of cases 1 and 2 are left as an exercise.
Case 3: U 6= {0} and V 6= {0}
Let u1 , . . . , um be a basis for U , and
v1 , . . . , vn be a basis for V .

We assert that (u1 , 0), . . . , (um , 0), (0, v1 ), . . . , (0, vn ) form a basis for U ⊕ V .
Firstly, spanning:
Let (u, v) ∈ U ⊕ V , say u = x1 u1 + · · · + xm um and v = y1 v1 + · · · + yn vn .
Then
(u, v) = (u, 0) + (0, v)
= (x1 u1 + · · · + xm um , 0) + (0, y1 v1 + · · · + yn vn )
= x1 (u1 , 0) + · · · + xm (um , 0) + y1 (0, v1 ) + · · · + yn (0, vn )
So U ⊕ V = h(u1 , 0), . . . , (um , 0), (0, v1 ), . . . , (0, vn )i
Secondly, independence: assume x1 (u1 , 0) + · · · + xm (um , 0) + y1 (0, v1 ) +
· · · + yn (0, vn ) = (0, 0). Then
(x1 u1 + · · · + xm um , y1 v1 + · · · + yn vn ) = 0
⇒ x1 u1 + · · · + xm um = 0
and y1 v1 + · · · + yn vn = 0
⇒ xi = 0, ∀i
and yi = 0, ∀i

5
 Hence the assertion is true and the result follows.
PROOF.
Let T : U ⊕ V 7→ U + V where U and V are subspaces of some W , such
that T (u, v) = u + v.
Thus Im T = U + V , and
Ker T = {(u, v) | u ∈ U, v ∈ V, and u + v = 0}
= {(t, −t) | t ∈ U ∩ V }
Clearly then, dim Ker T = dim(U ∩ V )1 and so
rank T + nullity T = dim(U ⊕ V )
⇒ dim(U + V ) + dim(U ∩ V ) = dim U + dim V.

1.2 Matrix of a Linear Transformation

DEFINITION 1.3
Let T : U 7→ V be a LT with bases β : u1 , . . . , un and γ : v1 , . . . , vm for
U and V respectively.
Then
a1j v1
+
a2j v2 a1j
T (uj ) = + for some .. ∈ F.
.
..
. a mj
+
amj vm
The m × n matrix
A = [aij ]
is called the matrix of T relative to the bases β and γ and is also written
A = [T ]γβ
Note: The j-th column of A is the co-ordinate vector of T (uj ), where
uj is the j-th vector of the basis β.  
x1
 ..
Also if u = x1 u1 + · · · + xn un , the co-ordinate vector  .  is denoted by


xn
[u]β .
1
True if U ∩ V = {0}; if not, let S = Ker T and u1 , . . . , ur be a basis for U ∩ V . Then
(u1 , −u1 ), . . . , (ur , −ur ) form a basis for S and hence dim Ker T = dim S.

6
EXAMPLE 1.2 
a b
Let A = ∈ M2×2 (F ) and let T : M2×2 (F ) 7→ M2×2 (F ) be
c d
defined by
T (X) = AX − XA.
Then T is linear2 , and Ker T consists of all 2 × 2 matrices A where AX =
XA.
Take β to be the basis E11 , E12 , E21 , and E22 , defined by
       
1 0 0 1 0 0 0 0
E11 = , E12 = , E21 = , E22 =
0 0 0 0 1 0 0 1

(so we can define a matrix for the transformation, consider these henceforth
to be column vectors of four elements).
Calculate [T ]ββ = B :

T (E11 ) = AE11 − E11 A

     
a b 1 0 1 0 a b
= −
c d 0 0 0 0 c d
 
0 −b
=
c 0
= 0E11 − bE12 + cE21 + 0E22

and similar calculations for the image of other basis vectors show that
 
0 −c b 0
 −b a − d 0 b 
B= 
 c 0 d − a −c 
0 c b 0

Exercise: Prove that rank B = 2 if A is not a scalar matrix (i.e. if

A 6= tIn ).
Later, we will show that rank B = rank T . Hence

nullity T = 4 − 2 = 2
2

T (λX + µY ) = A(λX + µY ) − (λX + µY )A

= λ(AX − XA) + µ(AY − Y A)
= λT (X) + µT (Y )

7
 Note: I2 , A ∈ Ker T which has dimension 2. Hence if A is not a scalar
matrix, since I2 and A are LI they form a basis for Ker T . Hence

AX = XA ⇒ X = αI2 + βA.

DEFINITIONS 1.2
Let T1 and T2 be LT’s mapping U to V.
Then T1 + T2 : U 7→ V is defined by

(T1 + T2 )(x) = T1 (x) + T2 (x) ; ∀x ∈ U

For T a LT and λ ∈ F , define λT : U 7→ V by

(λT )(x) = λT (x) ∀x ∈ U

Now . . .

[T1 + T2 ]γβ = [T1 ]γβ + [T2 ]γβ

[λT ]γβ = λ[T ]γβ

DEFINITION 1.4

Hom (U, V ) = {T |T : U 7→ V is a LT}.

Hom (U, V ) is sometimes written L(U, V ).

The zero transformation 0 : U 7→ V is such that 0(x) = 0, ∀x.

If T ∈ Hom (U, V ), then (−T ) ∈ Hom (U, V ) is defined by

(−T )(x) = −(T (x)) ∀x ∈ U.

Clearly, Hom (U, V ) is a vector space.

Also

[0]γβ = 0
and [−T ]γβ = −[T ]γβ

The following result reduces the computation of T (u) to matrix multi-

plication:

THEOREM 1.4

[T (u)]γ = [T ]γβ [u]β

8
PROOF.
Let A = [T ]γβ , where β is the basis u1 , . . . , un , γ is the basis v1 , . . . , vm ,
and
m
X
T (uj ) = aij vi .
i=1
 
x1
 .. 
Also let [u]β =  . .
xn
Then u = nj=1 xj uj , so
P

n
X
T (u) = xj T (uj )
j=1
Xn m
X
= xj aij vi
j=1 i=1
 
m
X Xn
=  aij xj  vi
i=1 j=1
 
a11 x1 + · · · + a1n xn
⇒ [T (u)]γ = 
 .. 
. 
am1 x1 + · · · + amn xn
= A[u]β

DEFINITION 1.5
(Composition of LTs)
If T1 : U 7→ V and T2 : V 7→ W are LTs, then T2 T1 : U 7→ W defined by

(T2 T1 )(x) = T2 (T1 (x)) ∀x ∈ U

is a LT.

THEOREM 1.5
If β, γ and δ are bases for U , V and W , then

[T2 T1 ]δβ = [T2 ]δγ [T1 ]γβ

9
PROOF. Let u ∈ U . Then

[T2 T1 (u)]δ = [T2 T1 ]δβ [u]β

and = [T2 (T1 (u))]δ
= [T2 ]δγ [T1 (u)]γ

Hence
[T2 T1 ]δβ [u]β = [T2 ]δγ [T1 ]γβ [u]β (1)
(note that we can’t just “cancel off” the [u]β to obtain the desired result!)
Finally, if β is u1 , . . . , un , note that [uj ]β = Ej (since uj = 0u1 + · · · +
0uj−1 + 1uj + 0uj+1 + · · · + 0un ) then for an appropriately sized matrix B,

BEj = B∗j , the jth column of B.

Then (1) shows that the matrices

[T2 T1 ]δβ and [T2 ]δγ [T1 ]γβ

have their first, second, . . . , nth columns respectively equal.

EXAMPLE 1.3
If A is m × n and B is n × p, then

TA TB = TAB .

DEFINITION 1.6
(the identity transformation)
Let U be a vector space. Then the identity transformation IU : U 7→ U
defined by
IU (x) = x ∀x ∈ U
is a linear transformation, and

[IU ]ββ = In if n = dim U .

Also note that IVn (F ) = TIn .

THEOREM 1.6
Let T : U 7→ V be a LT. Then

IV T = T IU = T.

10
 Then
TIm TA = TIm A = TA = TA TAIn = TAIn
and consequently we have the familiar result

Im A = A = AIn .

DEFINITION 1.7
(Invertible LTs)
Let T : U 7→ V be a LT.
If ∃S : V 7→ U such that S is linear and satisfies

ST = IU and T S = IV

then we say that T is invertible and that S is an inverse of T .

Such inverses are unique and we thus denote S by T −1 .
Explicitly,

S(T (x)) = x ∀x ∈ U and T (S(y)) = y ∀y ∈ V

There is a corresponding definition of an invertible matrix: A ∈ Mm×n (F )

is called invertible if ∃B ∈ Mn×m (F ) such that

AB = Im and BA = In

Evidently

THEOREM 1.7
TA is invertible iff A is invertible (i.e. if A−1 exists). Then,

(TA )−1 = TA−1

THEOREM 1.8
If u1 , . . . , un is a basis for U and v1 , . . . , vn are vectors in V , then there
is one and only one linear transformation T : U → V satisfying

T (u1 ) = v1 , . . . , T (un ) = vn ,

namely T (x1 u1 + · · · + xn un ) = x1 v1 + · · · + xn vn .

(In words, a linear transformation is determined by its action on a basis.)

11
1.3 Isomorphisms
DEFINITION 1.8
A linear map T : U 7→ V is called an isomorphism if T is 1-1 and onto,
i.e.

1. T (x) = T (y) ⇒ x = y ∀x, y ∈ U , and

2. Im T = V , that is, if v ∈ V , ∃u ∈ U such that T (u) = v.

Lemma: A linear map T is 1-1 iff Ker T = {0}.

Proof:

1. (⇒) Suppose T is 1-1 and let x ∈ Ker T .

We have T (x) = 0 = T (0), and so x = 0.

2. (⇐) Assume Ker T = {0} and T (x) = T (y) for some x, y ∈ U .

Then

T (x − y) = T (x) − T (y) = 0
⇒x−y ∈ Ker T
⇒x−y = 0⇒x=y

THEOREM 1.9
Let A ∈ Mm×n (F ). Then TA : Vn (F ) → Vm (F ) is

(a) onto: ⇔ dim C(A) = m ⇔ the rows of A are LI;

(b) 1–1: ⇔ dim N (A) = 0 ⇔ rank A = n ⇔ the columns of A are LI.

EXAMPLE 1.4
Let TA : Vn (F ) 7→ Vn (F ) with A invertible; so TA (X) = AX.
We will show this to be an isomorphism.

1. Let X ∈ Ker TA , i.e. AX = 0. Then

A−1 (AX) = A−1 0

⇒ In X = 0
⇒X = 0
⇒ Ker T = {0}
⇔ T is 1-1.

12
 2. Let Y ∈ Vn (F ) : then,

T (A−1 Y ) = A(A−1 Y )
= In Y = Y
so Im TA = Vn (F )

THEOREM 1.10
If T is an isomorphism between U and V , then

dim U = dim V

PROOF.
Let u1 , . . . , un be a basis for U . Then

T (u1 ), . . . , T (un )

is a basis for V (i.e. hui i = U and hT (ui )i = V , with ui , vi independent

families), so
dim U = n = dim V

THEOREM 1.11

Φ : Hom (U, V ) 7→ Mm×n (F ) defined by Φ(T ) = [T ]γβ

is an isomorphism.
Here dim U = n, dim V = m, and β and γ are bases for U and V , re-
spectively.

THEOREM 1.12

T : U 7→ V is invertible
⇔ T is an isomorphism between U and V .

PROOF.
⇒ Assume T is invertible. Then

T −1 T = IU
−1
and T T = IV
−1
⇒T (T (x)) = x ∀x ∈ U
and T (T −1 (y)) = y ∀y ∈ V

13
 1. We prove Ker T = {0}.
Let T (x) = 0. Then

T −1 (T (x)) = T −1 (0) = 0 = x

So T is 1-1.
2. We show Im T = V .
Let y ∈ V . Now T (T −1 (y)) = y, so taking x = T −1 (y) gives

T (x) = y.

Hence Im T = V .
⇐ Assume T is an isomorphism, and let S be the inverse map of T

S : V 7→ U

Then ST = IU and T S = IV . It remains to show that S is linear.

We note that
x = S(y) ⇔ y = T (x)
And thus, using linearity of T only, for any y1 , y2 ∈ V , x1 = S(y1 ), and
x2 = S(y2 ) we obtain

S(λy1 + µy2 ) = S(λT (x1 ) + µT (x2 ))

= S(T (λx1 + µx2 ))
= λx1 + µx2
= λS(y1 ) + µS(y2 )

COROLLARY 1.1
If A ∈ Mm×n (F ) is invertible, then m = n.

PROOF.
Suppose A is invertible. Then TA is invertible and thus an isomorphism
between Vn (F ) and Vm (F ).
Hence dim Vn (F ) = dim Vm (F ) and hence m = n.

THEOREM 1.13
If dim U = dim V and T : U 7→ V is a LT, then

T is 1-1 (injective) ⇔ T is onto (surjective)

( ⇔ T is an isomorphism )

14
PROOF.

⇒ Suppose T is 1-1.
Then Ker T = {0} and we have to show that Im T = V .

rank T + nullity T = dim U

⇒ rank T + 0 = dim V
i.e. dim( Im T ) = dim V
⇒ Im T = V as T ⊆ V .

⇐ Suppose T is onto.
Then Im T = V and we must show that Ker T = {0}. The above
argument is reversible:

Im T = V
rank T = dim V
= dim U
= rank T + nullity T
⇒ nullity T = 0
or Ker T = {0}

COROLLARY 1.2
Let A, B ∈ Mn×n (F ). Then

AB = In ⇒ BA = In .

PROOF Suppose AB = In . Then Ker TB = {0}. For

BX = 0 ⇒ A(BX) = A0 = 0
⇒ In X = 0 ⇒ X = 0.

But dim U = dim V = n, so TB is an isomorphism and hence invertible.

Thus ∃C ∈ Mn×n (F ) such that

TB TC = IVn (F ) = TC TB
⇒ BC = In = CB,

noting that IVn (F ) = TIn .

15
 Now, knowing AB = In ,

⇒ A(BC) = A
(AB)C = A
In C = A
⇒C = A
⇒ BA = In

DEFINITION 1.9
Another standard isomorphism: Let dim V = m, with basis γ = v1 , . . . , vm .
Then φγ : V 7→ Vm (F ) is the isomorphism defined by

φγ (v) = [v]γ

THEOREM 1.14

rank T = rank [T ]γβ

PROOF
T
U −→ V
φβ ↓ ↓ φγ
−→
Vn (F ) TA Vm (F )
With
β : u1 , . . . , u n U
a basis for
γ : v1 , . . . , vm V,
let A = [T ]γβ . Then the commutative diagram is an abbreviation for the
equation
φγ T = TA φβ . (2)
Equivalently
φγ T (u) = TA φβ (u) ∀u ∈ U
or
[T (u)]γ = A[u]β
which we saw in Theorem 1.4.
But rank (ST ) = rank T if S is invertible and rank (T R) = rank T if R
is invertible. Hence, since φβ and φγ are both invertible,

(2) ⇒ rank T = rank TA = rank A

16
and the result is proven.
Note:
Observe that φγ (T (uj )) = A∗j , the jth column of A. So Im T is mapped
under φγ into C(A). Also Ker T is mapped by φβ into N (A). Consequently
we get bases for Im T and Ker T from bases for C(A) and N (A), respectively.

(u ∈ Ker T ⇔ T (u) = 0 ⇔ φγ (T (u)) = 0

⇔ TA φβ (u) = 0
⇔ φβ (u) ∈ N (A).)

THEOREM 1.15
Let β and γ be bases for some vector space V . Then, with n = dim V ,

[IV ]γβ

is non-singular and its inverse

n o−1
[IV ]γβ = [IV ]βγ .

PROOF

IV IV = IV
⇒ [IV IV ]ββ = [IV ]ββ = In
= [IV ]βγ [IV ]γβ .

The matrix P = [IV ]γβ = [pij ] is called the change of basis matrix. For if
β : u1 , . . . , un and γ : v1 , . . . , vn then

uj = IV (uj )
= p1j v1 + · · · + pnj vn for j = 1, . . . , n.

It is also called the change of co-ordinate matrix, since

[v]γ = [IV (v)]γβ [v]β

i.e. if

v = x1 u1 + · · · + xn un
= y1 v1 + · · · + yn vn

17
then    
y1 x1
 ..  .. ,
 .  = P

. 
yn xn
or, more explicitly,
y1 = p11 x1 + · · · + p1n xn
..
.
yn = pn1 x1 + · · · + p1n xn .

THEOREM 1.16 (Effect of changing basis on matrices of LTs)

Let T : V 7→ V be a LT with bases β and γ. Then
[T ]ββ = P −1 [T ]γγ P
where
P = [IV ]γβ
as above.

PROOF
IV T = T = T IV
⇒ [IV T ]γβ = [T IV ]γβ
⇒ [IV ]γβ [T ]ββ = [T ]γγ [IV ]γβ

DEFINITION 1.10
(Similar matrices)
If A and B are two matrices in Mm×n (F ), then if there exists a non-
singular matrix P such that
B = P −1 AP
we say that A and B are similar over F .

1.4 Change of Basis Theorem for TA

In the MP274 course we are often proving results about linear transforma-
tions T : V 7→ V which state that a basis β can be found for V so that
[T ]ββ = B, where B has some special property. If we apply the result to
the linear transformation TA : Vn (F ) 7→ Vn (F ), the change of basis theorem
applied to TA tells us that A is similar to B. More explicitly, we have the
following:

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THEOREM 1.17
Let A ∈ Mn×n (F ) and suppose that v1 , . . . , vn ∈ Vn (F ) form a basis β
for Vn (F ). Then if P = [v1 | · · · |vn ] we have

P −1 AP = [TA ]ββ .

PROOF. Let γ be the standard basis for Vn (F ) consisting of the unit vectors
E1 , . . . , En and let β : v1 , . . . , vn be a basis for Vn (F ). Then the change of
basis theorem applied to T = TA gives

[TA ]ββ = P −1 [TA ]γγ P,

where P = [IV ]γβ is the change of coordinate matrix.

Now the definition of P gives

v1 = IV (v1 ) = p11 E1 + · · · + pn1 En

..
.
vn = IV (vn ) = p1n E1 + · · · + pnn En ,

or, more explicitly,

   
p11 p1n
v1 =  ...  ,
 .. 
...,  . .
 

pn1 pnn

In other words, P = [v1 | · · · |vn ], the matrix whose columns are v1 , . . . , vn

respectively.
Finally, we observe that [TA ]γγ = A.

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