Null space (or) Kernel N(T) and Range (or) Image R(T): Example Solved Problems

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

Worked Example Problems

Example 1

Define the linear transformation T: P₂(R)−M_2×2(R) by T (f(x)) = . Find dim [R (T)].

Solution:

Since B= { 1, x, x² } is a basis for P₂(R)

We know that R (T) = span (T (B))

   = span ({ T(1), T(x), T(x²) }

{} is linearly independent and also generates R (T).

{} is a basis for R (T).

  dim (R (T)) = 2.

Example 2

Let T : R³ – R² by T (a₁, a₂, a₃) = (α₁ − a₂, 2a₃). Verify it is 1–1 and onto.

Solution:

 If T: V→W a linear transformation and, N (T) and R(T) are null space and range of T then (i) T is 1 ‒ 1 iff N(T) = 0

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

   N(T) = {(a, a, 0)/a ∈ R}

Here N (T) = 1, so T is not 1 ‒ 1.

 R(T)={ T(x) : x ∈V }

R(T) = 2, so T is on to.

dim (R (T)) = rank (T) = 2

nullity (T) = dim (V) – rank (T)

 = 3−2 = 1

T is not 1−1, and T is onto.

Example 3

Let T: R²→R³ defined by T (a₁, a₂)=(α₁ +α₂, 0, 2α₁−a₂). Find nullity (T).

Solution:

Let N(T) = {x ∈ V, T(x)=0 }. Let x ∈ R²

 x(a₁, a₂). T(x) = 0, T (a₁, a₂) = 0

⇒ (a₁ + a₂, 0, 2a₁ − a₂) = (0, 0, 0)

 a₁+ a₂ = 0 and 2a₁−a₂ =0

⇒ a₁=0 and a₂=0. N (T) = { (0, 0) ∈ R²; T (0,0) = 0 }

N(T) = {0}

dim (R (T)) = rank (T) = 2

nullity (T) = dim (V) − rank (T) = 2−2 = 0.

Example 4

Suppose that T: R²→R² is linear, and T (1, 0) = (1, 4) and T (1, 1) = (2,5) then what is T (2,3).

Solution:

Let (2, 3) = a (1, 0) + b (1, 1) = (a, 0) + (b, b) = (a + b, b)              …(1)

Comparing on both sides we get

 a+b=2; a= − 1 then b=3.

 T (2, 3) = T (−1 (1, 0) + 3 (1, 1))

                              (from equation (1))

= − 1T (1, 0)+3T (1, 1) = −1 (1, 4)+3 (2, 5)

= (−1,−4)+(6, 15) = (−1+6, −4+15)

 T (2, 3) = (5, 11)

Example 5

Let T: R²→ R³ such that T (1, 1) = (1, 0, 2) ; T (2, 3) = (1, − 1, 4) find T (8, 11).

Solution:

 Let (8, 11) = a (1, 1) + b(2, 3) = (a, a) + (2b, 3b)

⇒   (8, 11) = (a + 2b, a + 3b). Comparing on both sides,

 a+2b=8

and a+3b= 11. Solving these two equations, we get

 a=2 and b=3.

T (8, 11)=T(2 (1, 1) +3 (2, 3))

 = 2T(1, 1)+3T (2, 3)

=2(1,0,2)+3(1,−1, 4)

= (2, 0, 4) + (3, −3, 12)

= (2 + 3, 0 − 3, 4 + 12).

T (8, 11) = (5, −3, 16)

Example 6

Is there a linear transformation T: R³→R² such that T (1,0,3)=(1, 1) and T(−2, 0, −6)=(2, 1)

Solution:

T (− 2, 0, − 6) = T (− 2 (1, 0, 3))

= −2T (1, 0, 3)

= −2 (1, 1)

= (−2,−2)

(2, 1) ≠ (−2,−2)

It is not a linear transformation.

Example 7

Let T: P₂(R) → P₃(R) be the linear transformation defined by T (f(x)) = 2ƒ' (x) + 3^x∫₀ f(t) dt. Check T is 1−1 or not.

Solution:

We know that R (T) = span (T (B))

(.' from Theorem: 2)

B is a basis of V = P₂(R)

= span [T (1), T (x), T (x²)]

T(1)=2(0)+ 3₀ʃ^xdt = 3[1]₀^x = 3x.

T(x)=2(1)+ 3₀ʃ^xtdt = 2 + 3(t²/2)₀^x = 2 + 3(x²/2)

T(x²)=2(2x)+ 3₀ʃ^xt²dt = 4x + [t³/3]₀^x

T(x²)=4x+x³

R(T) = span { 3x, 2+ 3x²/2, 4x + x³ }

To prove that { 3x, 2+3x²/2, 4x + x³ } is linearly independent,

 α(3x) + β( 2+ 3x²/2 ) + γ(4x+x³)=0

3αx + 2β + 3/2βx² + 4γx + γx³ = 0

 γx³ + (3/2)βx² + (3α+4γ)x + 2β = 0x³ +0x²+0x+0

Comparing the corresponding coefficients on both sides

we get,

 γ=0; 3β/2 = 0 3α+4γ=0

 γ=0; β=0; 3α=0. (α = 0).

 {3x, 2 + 3x²/2, 4x + x³) is linearly independent and basis of R (T).

 rank (T) = dim R (T) = 3 ≠ 4 = dim (P₃(R)).

  T is not onto.

By dimension theorem, we know

nullity (T)+ rank (T) = dim (V).

nullity (T)=3−3=0.

 T is 1−1.

Example 8

 Let T: R²→ R² be linear transformation defined by T (a₁, a₂) = (a₁+a₂, a₁). Verify whether it is onto or not?

Solution:

We know that N (T) = { 0 }. (ie) nullity (T) = 0.

.. T is 1−1

nullity (T) + rank (T) = dim (V)

rank (T) = 2

 T is onto.

Example 9

 Let T: P₂ (R) → R³ be the linear transformation defined by T(a₀ + a₁x + a₂x²) = (a₀, a₁, a₂). Verify it is 1–1.

Solution:

Clearly T is linear and one to one. Let S = {2−x+3x², x+x², 1−2x²}. Then S is linearly independent in P₂(R) because T (S) = { (2, − 1, 3), (0, 1, 1), (1, 0, − 2)}.

T is 1−1 (one−to−one).