Null space (or) Kernel N(T) and Range (or) Image R(T)

Null space (or) Kernel N(T) and Range (or) Image R(T)

Definition:

Null space (or) Kernel N (T)

Let V and W be vector spaces and let T: V→ W be linear, the null space (or kernel) N (T) of T to be the set of all vectors x in V such that T(x)=0.

 (ie) N(T) = { x ∈ V/T (x)=0 }.

Definition:

Range (or) Image R (T)

The Range R(T) of T be the subset of W consisting of all images (under T) of vectors in V

 (ie) R(T) = {T(x): x ∈ V} N

Definition:

Identity and zero transformation

Let V and W be vector spaces over F, I_V: V→V by I_V(x) = x for all x ∈ V.

To: V→ W by T₀ (x) = 0, for all x ∈ V,

Clearly I_V and T₀ are linear.

Theorem 1

Let V and W be the vector spaces and T: V→W be linear. Then prove that N(T) and R(T) are subspaces of V and W respectively.

Proof:

Given that V and W be vector spaces.

T: V→W be linear.

To prove: N (T) is a subspace of V.

Since T is linear

T(0_v) = 0_w ; 0_v− zero vector of V and 0_w − zero vector of W.

  T(x)=0 ; x ∈ N(T)

Similarly

 0_v ∈ N(T)

 Let x,y ∈ N(T) and c ∈ F

⇒ T(x)=0, T(y)=0

 T is linear.

1. T(x+y) = T(x)+T (y)

= 0_w+0_w=0_w

 x + y ≤ N(T)

Let x ∈ N(T), c ∈ F

T(cx) = cT (x) = C⋅ 0_w = 0_w

2. ⇒ cx ∈ N(T)

  x+y ∈ N(T) and cx ∈ N(T).

  N(T) is a subspace of V.

To prove: R (T) is a subspace of W.

Since T(0_v) = 0_w. (T is linear)

⇒ 0_w ∈ R (T)

Let x, y ∈ R(T) and c ∈ F, then there exists v and w in V such that T(v)=x and T(w) = y

 T(v+w) = T (v) + T (w)

T(v+w)=x+y

⇒ x+y ∈ R(T)

T (cv) = cT (v) = cx

⇒ cx ∈ R(T)

  R(T) is a subspace of W.

Theorem 2

Let V and W be vector spaces and let T: V→W be linear. If B = { v₁, v₂, ... v_n } is a basis for V, then R(T) = span (T (B)) = span ({T (v₁), T (v₂), ... T (v_n) }).

Proof:

Clearly T (v_i) ∈ R(T) for each i.

Since R (T) is a subspace of W,

 α₁T(v₁) + α₂ T(v₂) + ... + α_nT(v_n) ∈ R(T) where α_i ∈ F.

Span ({ T(v₁), T (v₂), ..., T (v_n) }) ∈ R(T)

span (T(B))  R(T)

                  ...(1)

To prove: R(T)  span (T(B)).

Suppose that w ∈ R(T) then there exists v ∈ V such that T (v) = w.

Since B={v₁, v₂, ... v_n} is a basis of V.

  v = a₁v₁ + a₂v₂ + ….. + a_nv_n

= a₁ T (v₁) + a₂T(v₂) + ... + a_nT(v_n)

             ∈ span {T(v₁), T(v₂), ... T(v_n) }

 ∈ span (T(B))

Hence w ∈ span (T (B)) ⇒ R (T) span (T(B))

From (1) and (2)

R(T) = span (T (B))

Definition:

Let V and W be vector spaces, and let T: V→ W be linear. If N(T) and R(T) are finite−dimensional then we define the nullity of T, denoted by nullity (T) and the rank of T, denoted by rank (T) to be the dimensions of N (T) and R (T) respectively.

 Nullity (T) = dim (N (T))

 Rank (T) = dim (R (T))

Theorem 3

Dimension Theorem: Let V and W be vector spaces, and let T: V→W be linear. If V is finite−dimensional then nullity (T) + rank (T) = dim (V)

Proof:

Suppose that dim (V) = n.

 'N (T) is a subspace of V.

dim (N (T))   n

Let dim N (T) = k ≤ n, and

 { v₁, v₂, …… v_k} is a basis of N(T).

We know that if W is a subspace of a finite dimensional vector space V then any basis for W can be extended to a basis of V.

  we may extend { v₁, v₂, v₃,..., v_k } to a basis

 B = { v₁, v₂, v₃,..., v_k, v_k+1, …… v_n } for V.

Now we claim that,

 s = {T (v_k+1), T (v_k+2), ... T(v_n)} is a basis for R (T).

First we prove that S generates R (T).

We know that if B={v₁, v₂, ..., v_n} is a basis for V, then R (T) = span (T (B)).

Since { v₁, v₂, ..., v_k } is a basis for N (T)

⇒ v_i ∈ N(T)

⇒ T(v_i) = 0

for all i = 1 to k.

we have

R(T) = span ({T (V₁), T (V₂), ... T(v_n) }) then R (T) = span (T (B))

Since { v₁, v₂, ..., v_k} is a basis for N(T)

      v_i ∈ N(T) ⇒ T (v_i) = 0 for all i = 1 to k.

R(T) = span {T (v_k+1), T (v_k+2), ... T(v_n)}

 = span (S)

Next we prove that S is linearly independent.

Let b_k+1 T (v_k+1) + b_k+2 T (v_k+2) + ... + b_n T (v_n) = 0

where     b_{k + 1}, b_{k +2},... b_n ∈ F.

Using the fact that T is linear, we have

Since { v₁, v₂, ..., v_k} is a basis for N(T)

Since { v₁, v₂, v₃,..., v_k, v_k+1, …… v_n } is a basis for V.

 (ie) { v₁, v₂, v₃,..., v_n } is linearly independent.

−c_i=0 for all i = 1 to k and

 b_i=0 for all i=k+1 to n.

In particular

 b_k+1 = b_k+2 = ... = b_n = 0

S = { T (v_k+1), T (V_{k + 2}), ... T (v_n) } is Linearly independent.

  S is a basis for R (D)

 dim (R (T)) = n−k

 rank (T) = dim V − dim (N (T))

 Nullity (T) + rank (T) = dim (V)

Theorem 4

Let V and W be vector spaces, and let T: V→W be linear. Then T is one−to−one iff N (T) = { 0 }.

Proof:

Assume: T is 1−1 (one−to−one)

Let x ∈ N(T)

T(x)=0=T(0)

T(x) = T (0)

 ⇒ x=0          (T is 1−1)

 N(T) = {0}

Conversely,

Assume: N (T) = { 0 }

Let T(x)=T(y)

T(x)−T(y)=0

 T(x−y) = 0                   ('T is linear)

  x−y ∈ N(T) = {0}

  x−y=0

  x=y

  T is 1−1          (one−to−one)

Theorem 5

Let V and W be vector spaces of equal (finite) dimension and let T:V→W be linear then the following are equivalent.

(a) T is 1−1 (one−to−one) (b) T is onto (c) rank (T) = dim (V)

Proof:

From the dimension theorem, we have nullity (T) + rank (T) = dim (V).

T is 1‒1 iff N (T) = {0}

iff Nullity (T) = 0

iff rank (T) = dim (V)

iff rank (T) = dim (W)

iff dim R(T) = dim (W)

iff R (T) = W if T is on to.