Vector Space & Linear Transformation- 3 | Mathematics for Competitive Exams PDF Download

Linear Transformation

We begin with a definition.

Definition: Let V and U vector spaces over the same field K. A mapping F: V → U is called a linear mapping or linear transformation if it satisfies the following two conditions:

1. For any vectors v, w ∈ V, F(v + w) = F(v) + F(w).
2. For any scalar k and vector v ∈ V, F(kv) = kF(v).

Namely, F: V → U is linear if it “preserves” the two basic operations of a vector space, that of vector addition and that of scalar multiplication.
Substituting k = 0 into condition (2), we obtain F(0) = 0. Thus, every linear mapping takes the zero vector into the zero vector.

Now for any scalars a, b ∈ K and any vector v, w ∈ V,
we obtain F(av + bw) = F(av) + F(bw) = aF(v) + bF(w)
More generally, for any scalars a_i ∈ K and any vectors v_i ∈ V, we obtain the following basic property of linear mappings:

Remark: A linear mapping F: V→ U is completely characterized by the condition
F(av + bw) = aF(v) + bF(w)     (*)
and so this condition is sometimes used as its definition.
Remark: The term linear transformation rather than linear mapping is frequently used for linear mappings of the form F: ℝⁿ -> ℝ^m.

Some definitions

1. Linear Functional: If t is a linear transformation from V(F) to F(F) i.e. t: V → F, then t is called a Linear Functional on V.
2. Linear Operator: If t is a linear transformation from V to V i.e., t: V → V, then t is called a Linear Operator of V.
3. Homomorphic Image: If t: V→V’ is an onto homomorphism, then V’ is called the Homomorphic image of V. 

Examples of Linear Transformation:
Example: Let V(F) and V’(F) be two vector spaces. Then
 
is a linear transformation from V to V'
Since for any a, b ∈ F and α, β ∈ V
t(aα + bα) = 0

= a0 + b0
= at(α) + bt(α)
This is called Zero linear mapping.

Example: Let V(F) be a vector space, then
 
is a linear transformation from V to V (i.e. linear operation of V).
Since for any a, b ∈ F and α, β ∈ V
t(aα + bβ) = aα + bβ
= at(α) + bt(β)

Example: Let V₃(F) and V₂(F) be two vector spaces over the field F, then

is a linear transformation from V₃ to V₂ since for any a, b ∈ F and (a₁, a₂, a₃), (b₁, b₂, b₃) ∈ V₃

Example: t : V₂(ℝ) → ℝ(ℝ), t(a₁, a₂) = a₁ + a₂ is a linear transformation from V₂(ℝ) to ℝ(ℝ) i.e. a linear functional on V₁ 
Since for any a, b ∈ R and (a₁, a₂); (b₁, b₂) ∈ V₂

Example: t:V₂(F) → F(F), t(a₁, a₂) = a₁ is a linear functional on V₂(F) since for any a, b ∈ F and (a₁, a₂); (b₁, b₂) ∈ V₂.

Properties of Linear mapping

Theorem: If t is a linear mapping from the vector space V(F) to V'(F), then:
(a) t(0) = 0’ where 0 and O’ are zero vectors of V and V’ respectively.
(b) t(-α) = -t(α),    ∀ α ∈ V
(c) t(α - β) = t(α) - t(β),    ∀ α, β ∈ V

Proof:
(a) Let α ∈ V. Since 0 is zero element in V, therefore
α + 0 = α
⇒ t(α + 0) = t(α)
⇒ t(α) + t(0) = t(α)    [∵ t is a linear transformation]
⇒ t(α) + t(0) = t(α) + O'    [∵ O’ is zero element in V’]
⇒ t(0) = 0’    [by cancellation law, (V’, +) being an abelian group]

(b) ∵ α ∈ V ⇒ -a e V is a additive inverse of a, therefore
α + (-α) = 0
⇒ t[α + (-α)] = t(0) = 0’    [by (a)]
⇒ t(α) + t(-α) = 0’    [∵ t is a linear transformation]
⇒ t(-α) is additive inverse of t(α) in V’
⇒ t(-α) = -t(α)

(c) t(α - β) = t[a + (-P)l

= t(α) + t(-β)]    [by linear mapping]

= t(α) - t(β)    [by (b)]

Kernel of a Homomorphism : Definition:

Let V(F) and V’(F) be two vector spaces and t be a homomorphism from V to V'. Then the set K of all those elements of V whose t-images are the identity O’ of V’ is called the kernel of t.
It is generally denoted by Ker t or ker t or Ker(t).
i.e. Ker t = K = {α ∈ V I t(α) = O’ ∈ V’}

Remark: The kernel of the linear transformation of t is called Null space of t and is denoted by N(t).
Theorem: The kernel of a linear transformation is a subspace.

Proof: Let V(F) and V’(F) be the two vector spaces and 0 and O’ be their zero vectors respectively. Let t be a linear transformation from V to V’ and K be the kernel of t i.e. Ker t = K = {α ∈ V I t(α) = O’ ∈ V’}

To show that K is a subspace of V:

Therefore let α, β ∈ K, then by the definition of kernel,
t(α) = 0’ and t(β) = 0’    ...(1)
Now if a and b be any two elements of the field F, then
t(aα + bβ) = at(α) + bt(β)    [∵ t is a linear transformation]
= a(0') + b(0’)    [by (1)]
= 0’ + 0’
or t(aα + bβ) = 0’    [by definition]

⇒ aα + bβ ∈ K
Therefore a, b ∈ F and a, b ∈ F ⇒ aα + bβ ∈ K
∴ K is a subspace of V.

Isomorphism of Vector Spaces : Definition
Let V(F) and V'(F) be the two vector spaces, then a map
t : V → V' is called an isomorphism, if t is :
(i) one-one

(ii) onto and

(iii) a linear mapping i.e.

t(aα + bβ) = at(α) + bt(β), ∀ a, b ∈ F, V a, b ∈ V
Under these conditions, the vector spaces V and V’ are called isomorphic images any symbolically it is expressed as
V ≅ V’
Again vector space V’ is also called the isomorphic image of V.

Example 1: Prove that the mapping t : V₂(ℝ) → V₃(ℝ) which is defined by t(a, b) = (a, b, 0) is a linear transformation from V₂ to V₃.

Let α = (a, b,) and β = (a₂, b₂) be any two elements of V2(ℝ) and a, b ∈ ℝ, then

Therefore t is a linear transformation from V₂ to V₃.

Example 2: Show that the mapping t: V₂(ℝ) → V₃(ℝ) where t(a, b, c) = (c, a + b) is a linear mapping.

Let α = (a₁, b₁, c,) and β = (a₂, b₂, c₂) be any two elements of the space V₃(ℝ) and a, b ∈ R, then



Therefore t is a linear transformation from V₃ to V₂.

Example 3: Show that the mapping f : V₃(F) → V₂(F) which is defined by f(a₁, a₂, a₃) = (a₁, a₂) is a homomorphism from V₃ onto V₂. Also find its kernel.

Let α = (a₁, a₂, a₃) and β = (b₁, b₂, b₃) be any two elements of the space V₃(F) and a, b ∈ F, then



∴ f is a homomorphism from space V₃ to V₂.

Onto: Let (a₁, a₂) ∈ V₂, then corresponding to (a₁, a₂), there exist (a₁, a₂, 0) ∈ V₃ for which f(a₁, a₂, 0) = (a₁, a₂)

⇒ there exist f-pre image of each element of V₂ in V₃ 
∴ f is onto homomorphism.

Kernel: Let Ker f = K, then K will be the set of all the those vectors of V₃ which map on the zero vector (0, 0) of V₂ i.e.

Example 4: If P[x] denotes the vector space over the field of real numbers ℝ  of all polynomials d in x of degree ≤ n, then prove that the mapping f : P[x] → P[x] where
 is a linear transformation.

and  be any two elements of P[x] where a₀, a₁, .... a_n; b₀, b₁,....., b_n are all real numbers. Then




Again, for any a ∈ ℝ,



Therefore f is a linear transformation.

Example 5: Show that the mapping f: V₃(ℝ) → V₃(ℝ) which is defined by f(a₁, a_2, a₃) = (a₁,a₂, aa₃) where a ≠0 is fixed in ℝ is an isomorphism on V₃(ℝ).

One-one : Let α = (a₁, a₂, a₃) and p = (b₁, b₂, b₃) be any two elements of the space V₃(ℝ). Then


  [∵ a ≠ 0]

∴ f is one-one.
Onto: For every a = (a₁, a₂, a₃) ∈ V₃(ℝ) there exist  for which

∴ f is onto.
f is a linear transformation: For any a’, b’ ∈ ℝ



∴ f is a linear transformation.
Consequently, f is an isomorphism on V₃(ℝ)

Example 6: Show that the mapping f : V₂(ℝ) → V₂(ℝ), where
(x, y) = (x cos θ - y sin θ, x sin θ + y cos θ) is an isomorphism on V₂(ℝ) 

One-one: Let α = (a₁,a₂) and β(b₁,b₂) be any two elements of the vector space V₂(ℝ), then

∴ f is one.
Onto: Since for every (x cos θ - y sineθ, x sinθ + y cosθ) ∈ V₂(ℝ) there exist (x, y) el such that f(x, y) = (x cosθ - y sinθ, x sinθ + y cosθ)
∴ f is onto.
f is a linear transformation : For any a, b ∈ ℝ




∴ f is a linear transformation.

Consequently, f is an isomorphism on V₂(ℝ).

Example 7: If V(F) be the vector space of all n x n matrices over the field F and M e V be a given matrix, then prove that the mapping

is a linear transformation. 

Let A, B ∈ V, then
f(A + B) = (A + B)M + M(A + B)
= AM + BM + MA + MA    [By distributivity of Matrices]
= (AM + MA) + (BM + MB)    [By associativity for +]
= f(A) + f(B)    ...(1)
Again, for any a ∈ F
f(aA) = (aA)M + M(aA)
= a(AM) + a(MA)
= a(AM + MA)    By scalar multiplication of Matrices]
= af(A)    ...(2)
From (1) and (2), f is a linear transformation on V(F).

Example 8: If the mapping t : V(F) → V(F’) is one-one onto linear map; then show that t^-1: V’ → V will also be a linear map.

Let a₁, b₁ ∈ V and a₂, b₂ ∈ V' such that
    ...(1)
 .... (2)
∵ t is a linear, therefore
    [by (1)]

 [by (2)]
    ...(3)
and for any


    ...(4)
From (3) and (4), it is clear than t^-1 is also a linear transformation.

Example: (a) Let F ; ℝ³->:ℝ³ be the “projection” mapping into the xy-plane; that is, F is the mapping defined by F(x, y, z) = (x, y, 0). We show that F is linear. Let v = (a, b, c) and w = (a’, b’, c’). Then

and, for any scalar k,
F(kv) = K(ka, kb, kc) = (ka, kb, 0) = k(a, b, 0) = kF(v)
Thus, F is linear.

Matrices as Linear Mappings 

Let A be any real m * n matrix. Recall that A determines a mapping  (where the vectors in Kⁿ and K^m are written as columns). We show F_A is linear. By matrix multiplication,


In other words, using A to represent the mapping, we have
A(v + w) = Av + Aw and A(kv) = k(Av)
Thus, the matrix mapping A is linear.

Vector Space Isomorphism

Definition: Two vector spaces V and U over K are isomorphic, written  if there exists a bijective (one- to-one and onto) linear mapping F : V → U. The mapping F is then called an isomorphism between V and U.
Consider any vector space V of dimension n and let S be any basis of V. Then the mapping

which maps each vector v ∈V into its coordinate vector [v]_s, is an isomorphism between V and Kⁿ.

Some Theorem on Space Morphism 

Theorem: If W be a subspace of a vector space V(F), then the quotient space V/W, is a homomorphic image of V with kernel W.

Proof. Let a mapping f from V to V/W be defined as follows :

To prove (a) f is an onto linear transformation and (b) ker f = W
(a) If α, β ∈ V and α, β ∈ F, then
f(aα + bβ) = W + (aα + bβ)
= (W + aα) + (W + bβ)
= a(W + α) + b(W + β)
= af(α) + bf(β)
∴ f is a linear transformation.

Again for every W + α ∈ V/W, there exist α ∈ V such that

f(α) = W + α
∴ f is onto i.e. f(V) = V/W
Therefore f : V → V/W is onto homomorphism
⇒ V/W is homomorphic image of V.
(b) ker f = {α ∈ V I f(α) = W}    [∵ W is zero element of V/W]

= α ∈V I W + α = W}

= {α ∈ V I α ∈ W} = W

Theorem : [Fundamental theorem on Space Homomorphism] :
If f be a linear transformation from a vector space V(F) onto V’(F), then

Proof: Let Ker f = K.

Now define a linear transformation ϕ from V/K to V’ as follows :
 ....(1)
ϕ is well defined: Here it will be shown that for any two elements K + α and K + β of V/K

 [∵ f is a linear transformation]

∴ ϕ is well defined.

ϕ is one-one: Let K + α, K + β ∈ V/K, then
 ....([by (1)])

ϕ is one-one.
ϕ is onto: ∵ f is onto, therefore for every α’ ∈ V', there exist α ∈ V such that
   [by (1)]
Therefore for α’ ∈ V there exist K + α ∈ V/K such that

∴ ϕ is onto

ϕ is a linear transformation : For any a, b ∈ F
 

∴ ϕ is a linear transformation.

Hence ϕ is an isomorphism from V/K to V'
Consequently,

Theorem : [Isomorphism theorem for Vector Spaces]
Two FDVS V and V’ over the same field F are isomorphic iff they are of the same dimensions
i.e V ≌ V’ dim V = dimV’

Proof: Let V(F) and V’(F) be the two vector spaces and dim V = n.

Firstly, let V ≌ V', then there exist a map f from V to V' which is one-one onto linear transformation.
Let  be the basis of the space V.
if
then to show that B’ is the basis of V’.

B’ is LI: Let a₁, a₂, ..., a_n ∈ F are such that

 [ ∵ f is linear transformation]
 where 0 ∈ V is zero element
[∵ f is one-one and f(0) = O’]

Therefore

L (B’) = V’ : Let α' ∈ V ’ since f is onto,
therefore for every α' ∈ V there exist α ∈ V such that f(α) = α’

⇒ α' ∈ V' is a LC of vectors of B’
⇒ L(B’) = V’

Hence B’ is a basis of V’. Consequently dim V’ = n
∴ dim V = n = dim V’

Therefore V = V’ => dimV = dimV'

Conversely: Let dimV = dimV’. Let  be the bases of V and V’ respectively.

If α ∈ V, then there exist  
such that

Define a map f from V to V’ as follows :
f : V → V’ such that

To prove that f is an isomorphism:
f is one-one: Let α, β ∈ V where

Then f(a) = f(b)



∴ f is one-one.
f is onto: Let  be any element of V’, then there exist such that

∴ f is onto.
f is a linear transformation: For any a, b ∈ F


   [by (1)]

∴ f is a linear transformation.

Therefore  f is an isomorphism from V to V ’ .

Consequently, V≅ V ’
Hence

Theorem: Every n-dimensional vector space V(F) is isomorphic to V_n(F) i.e.
V(F) ≅ V_n(F)

Proof: dimV(F) = n, therefore let  be any basis of the space V_n(F). Then every vector of V can be expressed as a LC of elements of B. Therefore for any α ∈ V there exist a₁, a₂, .... a_n ∈ F such that

Now define a map f from V(F) to V_n(F) as follows:
 

To prove that f is an isomorphism: 
f is one-one: Let α, β ∈ V where

then f(α) = f(β)


∴ f is one-one.

f is onto: Let  then there exist an element,    in V such that

∴ f is onto.

f is a linear transformation: Let α, β ∈ V and a, b ∈ F, then



∴ f is a linear transformation.
Hence f is an isomorphism from V(F) to V_n(F).
Consequently, V(F) ≅ V_n(F).

Range space of a linear transformation: Definition:

Let V(F) and V'(F) be two vector space and t: V → V’ be a linear transformation. Then the image set of V under t is called the range space of t.
This is expressed by R(t) or l_m(t).

Theorem: If t is a linear transformation from a vector space V(F) to V’(F), then the range of t is a subspace of V’.
Proof: Let the range of t be R(t), then

Clearly, R(t) ⊂ V’ and R(t) ≠ ϕ
Let β_1, β₂ ∈ R(t), then there exist α₁, α_2,  ∈ V such that

But V(F) is a vector space, therefore


Therefore

∴ R(t) is a vector subspace of V’.

Definition: A linear transformation t: U→V is called non-singular if t(u) = 0 implies u = 0.
NOTE: Moreover, U and V must be of same dimension otherwise t: U→V may not be surjective.
Example 1: A linear transformation t : U → V is an isomorphism iff it is non singular. 

To show that if t is an isomorphism ⇒ t is non singular i.e. t is one-one, therefore 0 ∈ U is the only vector whose image 0 ∈ V or, only zero vector is in the zero space of t. t is non singular.
Conversely: If t is non singular ⇒ t is an isomorphism
Let t(v₁) = t(v₂)

Hence t is one-one.

t is nonsingular and one-one, therefore this is one-one also, 

∴ t is an isomorphism.

Example 2: If t : U → V is a linear transformation; then t is non singular if and only if the range of LI set by t in U is LI set in V.
First let t be nonsingular i.e. only zero vector is in the empty space of t.

Let B = {v₁, v₂, ....., v_n} be a LI subset of U, then its image set under t is

   [t is linear]
 [t is nonsingular]
  [B is LI]
Hence the image B, of B under t is LI.

Conversely: Let t-image of any LI set of U be LI in V.
Consider a nonzero vector v of U, then the set
S = {v} is a LI set in V.

Now t-image S, of S contains only an element t(v)

[Note]

S, = {t(v)},

S, is LI in V, therefore for v ≠ 0, t(v) ≠ 0,
because the set containing zero element alone is LI.
Consequently, the null space of t is null subspace, i.e. t is nonsingular.

Kernel and Image of a Mapping

We begin by defining two concepts.
Definition Let F : V → U be a linear mapping. The kernel of F, written Ker F, is the set of elements in V that map into zero vector 0 in U; that is,
Ker F = {v ∈ V : F(v) = 0}
The image (or range) of F, written Im F, is the set of image points in U; that is,
Im F = (u ∈ U : there exists v ∈ V for which F(v) = u}

Theorem: Let F : V → U be a linear mapping. Then the kernel of F is a subspace of V and the image of F is a subspace of U.
Now suppose that v₁, v₂, ..., v_m span a vector space V and that F : V → U is linear. We show that F(v₁), F(v₂), ..., F(v_m) span Im F. Let u ∈ Im F. Then there exists v ∈ V such that F(v) = u. Because the v_i’s span V and v ∈ V, there exist scalars a₁, a₂, ..., a_m for which

Therefore,

Thus, the vectors F(v₁), F(v₂),....., F(v_m) span Im F.

Proposition: Supposev₁, v₂, ..., v_m span a vector space V, and suppose F : V → U is linear. Then F(v₁), F(v₂),....., F(v_m) space Im F.

Example: (a) Let F : ℝ³ → ℝ³ be the projection of a vector v into the xy-plane, i.e.
F(x, y, z) = (x, y, 0)
Clearly the image of F is the entire xy-plane—that is, points of the form (x, y, 0). Moreover, the kernel of F is the z-axis—that is, points of the form (0, 0, c). That is,
Im F = {(a, b, c) : c = 0} = xy-plane and Ker F = {(a, b, c) : a = 0, b = 0} = z-axis

Kernel and Image of Matrix Mappings

Consider, say, a 3 * 4 matrix A and the usual basis {e₁, e₂, e₃, e₄} of K⁴ (written as columns):

Recall that A may be viewed as a linear mapping A : K⁴ → K³, where the vectors in K⁴ and K³ are viewed as column vectors. Now the usual basis vectors span K⁴ so their images Ae₁, Ae₂, Ae₃, Ae₄, span the image of A. But the vectors Ae₁, Ae₂, Ae₃, Ae₄ are precisely the columns of A:

Thus, the image of A is precisely the column space of A.
On the other hand, the kernel of A consists of all vectors v for which Av = 0. This means that the kernel of A is the solution space of the homogeneous system AX = 0, called the null space of A. 

Proposition: Let A be any m * n matrix over a field K viewed as a linear map A : Kⁿ → K^m. Then
Ker A = nullsp(A) and Im A = colsp(A)
Here colsp(A) denotes the column space of A, and nullsp(A) denotes the null space of A.

Rank and Nullity of a Linear Mapping

Let F : V → U be a linear mapping. The rank of F is defined to be the dimension of its image, and the nullity of F is defined to be the dimension of its kernel; namely,
rank(F) = dim(lm F) and nullity(F) = dim(Ker F)

Theorem: Let V be of finite dimension, and let F : V → U be linear. Then
dim V = dim(Ker F) + dim(lm F) = nullity(F) + rank(F)

Matrix Representation of a Linear Operator

Let T be a linear operator (transformation) from a vector space V into itself, and suppose S = {u₁, u₂, ...., u_n} is a basis of V. Now T(u₁), T(u₂),..., T(u_n) are vectors in V, and so each is a linear combination of the vectors in the basis S; say,

Definition
The transpose of the above matrix of coefficients, denoted by m_s(T) or [T]_s, is called the matrix representation of T relative to the basis S, or simply the matrix of T in the basis S.
Example: Let F: ℝ² → ℝ² be the linear operator defined by F(x, y) = (2x + 3y, 4x - 5y).

(a) Find the matrix representation of F relative to the basis S = {u₁, u₂} = {(1, 2), (2, 5)}.

1. First find F(u₁), and then write it as a linear combination of the basis vectors u₁ and u₂. (For notational convenience, we use column vectors.) We have
   
   Solve the system to obtain x = 52, y = -22. Hence, F(ut) = 52u₁ - 22u₂.
2. Next find F(u₂), and then write it as a linear combination of u₁ and u₂:
   Solve the system to get x = 129, y = -55. Thus F(u₂) = 129u₁ - 55u₂.
   Now write the coordinates of F(u,) and F(u2) as columns to obtain the matrix
   

Example: Let F : ℝ⁴ → ℝ³ be the linear mapping defined by
F(x, y, z, t) = (x - y + z + t, 2x - 2y + 3z + 4t, 3x - 3y + 4z + 5t)
(a) Find a basis of the dimension of the image of F.
Find first the image of the usual basis vectors of R4.
F(1, 0, 0, 0) = (1. 2, 3),    F(0, 0, 1, 0) = (1, 3, 4)
F(0, 1, 0, 0) = (-1, -2, -3), F(0, 0, 0, 1) = (1, 4, 5)
By Proposition the image vectors span Im F. Hence, form the matrix M whose rows are these image vectors and row reduce to echelon form:

Thus, (1, 2, 3) and (0, 1, 1) form a basis of Im F. Hence, dim(lm F) = 2 and rank(F) = 2.

Change of Basis
Let V be an n-dimensional vector space over a field K. We have shown that once we have selected a basis S of V, every vector v ∈ V can be represented by means of an n-tuple [v]_s in Kⁿ, and every linear operator T in A(V) can be represented by an n * n matrix over K.

Definition
Let S = {u₁, u₂, ...., u_n} be a basis of a vector space V, and let S’ = {v₁, v₂, ..., v_n} be another basis. (For reference, we will call S the “old” basis and S’ the “new" basis.) Because S is a basis, each vector in the “new” basis S' can be written uniquely as a linear combination of the vectors in S; say, 

Let P be transpose of the above matrix of coefficients; that is, let P = [p_ij], where p_ij = a_ji, Then P is called the change-of-basis matrix (or transition matrix) from the “old” basis S to the “new” basis S’.
The following remarks are in order.
Remark: The above change-of-basis matrix P may also be viewed as the matrix whose columns are, respectively, the coordinate column vectors of the “new” basis vectors v. relative to the “old" basis S; namely,
P = [[V₁]_s, [V₂]_s,......, [V_n]_S]

Remark: Analogously, there is a change-of-basis matrix Q from the “new” basis S' to the “old” basis S. Similarly, Q may be viewed as the matrix whose columns are, respectively, the coordinate column vectors of the “old” basis vectors u. relative to the “new” basis S’; namely,
Q = [[u₁]_s, [u₂]_s,......, [u_n]_S]

Remark : Because the vectors v₁, v₂, ..., v_n in the new basis S’ are linearly independent, the matrix P is invertible. Similarly, Q is invertible. In fact, we have the following proposition.

Proposition: Let P and Q be the above change-of-basis matrices. Then Q = P^-1.
Now suppose S = {u₁, u₂,.....,u_n} is a basis of a vector space V, and suppose P = [p_ij] is any nonsingular matrix. Then the n vectors

corresponding to the columns of P, are linearly independent. Thus they form another basis S’ of V. Moreover, P will be the change-of-basis matrix from S to the new basis S’.

Example: Consider the following two bases of ℝ²:

(a) Find the change-of-basis matrix P from S to the “new” basis S”. Write each of the new basis vectors of S’ as a linear combination of the original basis vectors u, and u2 of S. We have


Thus

Note that the coordinates of v₁ and v₂ are the columns, not rows, of the change-of-basis matrix P.

Rank-Nullity Theorem 

If t is a linear transformation defined from a vector space V(F) to V'(F) where V(F) is a finite dimensional, then :

Rank (t) + Nullity (t) = Dim V.

Proof. Let V(F) be a n-dimensional vector space and R and K be their range and kernel respectively.

Define a mapping ϕ from R to V/K as follows:
 


∴ ϕ is one-one.
ϕ is onto : Let K + α ∈ V/E, then there exist t(α) ∈ ℝ such that
ϕ[t(α)] = K + α
Therefore preimage of each element of V/K exist in ℝ.

∴ ϕ is onto.
ϕ is a linear transformation: For any a, b ∈ F and t(α), t(β) ∈ ℝ f[at(α) + bt(β)] = ϕ[t(aα + bβ)] [∵ t is a linear transformation]

∴ ϕ is a linear transformation.

Hence ϕ is an isomorphism from R to V/K.
⇒ ℝ ≅ V/K
⇒ dim ℝ = dim(V/K)
⇒ dim ℝ = dim V - dim K
⇒ dim ℝ + dim K = dim V
⇒ Rank(t) + Nullity(t) = dimV.

Example 1: If

find a basis for nullspace (A) and verify Theorem.

We must find all solutions to Ax = 0. Reducing the augmented matrix of this system yields

Consequently, there are two free variables, x₃ = t, and x₄ = t₂, so that
x₂ = 7t, + 7t₂, x₁ = —9t₁ - 10t₂.

Hence,
nullspace (A)

Since the two vectors in this spanning set are not proportional, they are linearly independent. Consequently, a basis for nullspace(A) is {(-9, 7, 1, 0), (-10, 7, 0, 1)}, so that nullity(A) = 2. In this problem, A is 3 * 4 matrix, and so, in the Rank-Nullity Theorem, n = 4. Further, from the foregoing row- echelon form of the augmented matrix of the system Ax = 0, we see that rank(A) = 2. Hence, rank(A) + nullity(A) = 2 + 2 = 4 = Dim(A) and the Rank-Nullity Theorem is verified.

Example: Find rank (A) and nullity (A) for A

• rank (A). It is enough to put A in row-echelon form and count the number of leading ones.
  The reader will verify that a row-echelon form of A = is  . There are three leading ones, therefore rank (A) = 3.
• nullity (A). For this, we need to find a basis for the solution set of Ax = 0. Its reduced row-echelon form is:  Then corresponding system is  The leading variables are x₁, x₂, and x₄. Hence, the free variables are X₃ and x₅. We can write the solution in parametric form as follows:  
  Thus, nullity (A) = 2 .

Example : Find the rank and nullity of
Note this matrix has 5 columns. The row-echelon form is
We see that rank(A) = 2 (2 leading 1’s). Therefore nullity (A) = 5 - 2 = 3.