Null space (or) Kernel N(T) and Range (or) Image R(T): Example Solved Problems - Part 2

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

Worked Example Problems – Part 2

 

Example 10

Let T: R³ → R². Define T (a₁, a₂, a₃) = (a₁−a₂2a₃). Find the bases for N(T) and evaluate the nullity of T.

Solution:

Given that T: R³ → R

We know that N (T) = { x ∈ V/T (x) = 0} & R(T) = { x ∈ V/T (x) }

To find the bases for N (T):

Let x ∈ R³ and let x= (a₁, a₂, α₃)

T(x)=0

T(a₁, a₂, a₃)=0

(.N(T)={x ∈ V,T(x)=0 })

 (a₁−a₂, 2a₃) = T (a₁, a₂, α₃)

.. T (a₁, a₂, a₃) = (a₁−a₂, 2a₃)=0

                     ….(1)

(ie) a₁−a₂=0 & 2a₃=0

 T(a₁, a₂, a₃)=(0,0) if a₁ = a₂ & a₃=0.

.Any element in the null space is of the form (a₁, a₂, 0)

 N (T) = { (a, a, 0) ∈ R³/T (a, a, 0) = 0}

N (T) = 1 with basis (1, 1, 0)

 

Example 11

Let T: M_2×3(F) → M_2×2(F).

Define 

Find the basis for N(T) and calculate N(T).

Solution:

⇒ 2a₁₁−a₁₂=0, a₁₃+2a₁₂=0

 a₁₁= (1/2)a₁₂ & a₁₃ = ‒2a₁₂

The number of basis element of N(T) = 4.

  N (T) = 4.

 

Example 12

Let T: R² → R³. Define T (a₁, a₂)=(a₁+a₂, 0, 2 a₁−α₂). Find the basis for N (T) and compute N (T).

Solution:

Let N.(T) = {x ∈ V,T(x)=0}

Let x ∈ R² such that x= (a₁, a₂)

Then T(x)=0

T (a₁, a₂) = 0  ⇒ (a₁ + a₂, 0, 2α₁−α₂) = 0

⇒ a₁+a₂=0, 2a₁−a₂=0 ⇒ a₁ = a₂ &

                    …….. (1)

  2a₁ = a₂

                     …....(2)

From equations (1) & (2) we get a₁ =0 and a₂ = 0

The kernal consists of all that is transformed to zero.

Any element in the null space will be of the form (0,0)

 N(T) = { (0,0) ∈ R²/T (0, 0) = 0 }

The Null space of T is zero dimensional.

 N (T) = 0.

 

Example 13

Suppose that T: P₂(R) → P₂(R). Let us define T[f(x)]=xf(x)+f ' (x). Find the basis for N(T) and compute the value.

Solution:

For the domain P₂(R), the standard basis is { 1, x, x² ́}

For the codomain (Range) P₃(R), the standard basis is {1, x, x², x³ }

Let T[f(x)] = 0(x)

⇒ x f(x) + ƒ′ (x)=0

⇒ f(x) = ‒ f '(x) / x

If f is an n^th degree polynomial, then f ' is an (n−1)^th degree polynomial and ‒f '(x)/x is an (n−2)^th degree polynomial.

:. f(x) = ‒f '(x)/x means n^th degree polynomial = (n−2)^th degree polynomial.

  This is not possible in {1, x, x²} except at f(x)=0.

  N (T) = 0

 

Example 14

Let T: R³ → R². Define T (a₁, a₂, α₃) = (a₁−a₂, 2a₃). Find the basis for R(T) and find the value of R (T).

Solution:

Given that T: R³ → R² and T (a₁, a₂, a₃) = (a₁ ‒ a₂, 2a₃)

To find the basis for R (T):

For every (a, b, c) ∈ R³ of T, there exists a T(a, b, c) = (a − b, 2c)

Any vector in R² can be written using the transformed elements in R³.

Suppose that, (x, y) ∈ R² then we write it as (x, y) = (a−b, 2c) where a, b & c are real numbers.

⇒  a−b=x, 2c=y.

 a−b=x has many solutions and assume that ½ y=c

 T (a, b, c) = (a−b, 2c) = (x, y)

Since {(1, 0), (0, 1)} is a standard basis for R², it is also a standard basis for R (T).

Hence the basis of R (T) is {(1, 0), (0, 1)} and the rank is 2.

 

Example 15

Let T: R² → R³. Define T (a₁, a₂) = (a₁+a₂, 0, 2α₁ − a₂). Find the basis for R(T) and find the value.

Solution:

Given that T: R² → R³ & T (a₁, a₂) = (a₁ + a₂, 0, 2α₁ − a₂).

We know that R(T) = { T(x)/x ∈ V}

To find the basis of R (T)

We know that, if T: V→W is a linear transformation and B = { v₁, v₂, v₃ ... v_n} is a basis for V.

Then R (T) = span [T (B)] = span [T (v₁), T (v₂), T (v₃), ... T (V_n)]

Here B = {(1, 0), (0, 1)} is a standard basis for R³.

  R(T) = span [T (B)] = span [T (1, 0), T (0, 1)]

 T (1, 0) = (1+0, 0, 2−0) since T (a₁, a₂) = (a + a₂, 0, 2a₁−a₂)

= (1,0,2)

T (0, 1) = (0 + 1, 0, 2 (0) − 1) = (1, 0, − 1)

R(T) = span [T (B)]

= span [T (1, 0), T (0, 1)]

= span [T (1, 0), T (0, 1)]

= span [(1, 0, 2), (1, 0, −1)]

 (1, 0, 2) & (1, 0, ‒1) are linearly independent.

{ (1, 0, 2), (1, 0, −1)} is a basis for R (T) & R(T) = 2.

 

Example 16

Let T: P₂(R) → P₃(R). Define T[f (x)]=xf(x)+f '(x) Find the basis for R(T) and find the value of R (T).

Solution:

 Given that T: P₂(R) → P₃(R).

 T[f(x)] = xf(x) +ƒ'(x)

 R(T) = span [T (1), T(x), T(x²)]

 = span [x.1 + 0, x . x + 1, x. x² + 2x]

= span [x, x² + 1, x³ + 2x]

Let us consider ax + b (x²+1) + c(x² + 2x) = 0

⇒ ax + bx²+b+cx³ +2cx=0

⇒ cx³ + bx²+x (a + 2c) + b = 0x³+0x²+0x +0

Equating the corresponding coefficients on both sides we get

 c=0,b=0, a+2c=0

 a=b=c

⇒ x,x²+1,x³+2x are linearly independent.

 { x, x² + 1, x³ + 2x } form a basis for R (T).

  R(T) = 3

 

Example 17

Let T: M_2×3(F) → M_2×2(F).

Define:  

Find the basis for R (T) and also find the value of R (T).

Solution:

Given that 

We know that the standard basis for the vector M_2×3 (F) is

Since B is a basis for M2x3 (F), we have

R (T) = span [T (B)]

  R(T) is spanned by two matrices 

Since these two matrices are linearly independent, the basis of R(T) is

R(T) = 2

 

Example 18

Let T: R²→R³. Define T (a₁, a₂)=(a₁+a₂, 0, 2α₁−a₂). Verify whether T is one−to−one or on−to.

Solution:

We know that if T: V→W is a linear transformation and N (T), R (T) are nullspace and range of T then,

(i) T is 1−1 if and only if N (T) = 0 and

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

Here N (T) = 0, So T is 1−1 and T is not on to because T never maps all the values of R³.

  T is 1−1 but not on−to.

 

Example 19

Let T: M_2×3(F) → M_2×2(F).

Define 

Verify whether T is 1−1 or on−to.

Solution:

We know that if T: V→W is a linear transformation and N (T), R (T) are null space & range of T then

(i) T is 1−1 iff N (T)=0 & (ii) T is onto if and only if R(T) = W.

Since N (T) ≠ 0, T is not 1−1.

Since R (T) = 2 and dim [ M_2×3(F) ] = 4,

(ie) R (T) ≠ dim [M_2×3 (F)], T is not onto.

  Hence T is neither 1−1 nor on to.

 

Example 20

Let T: R² → R² be linear and T (1, 0) = (1, 4) and T (1, 1) =  (2, 5). Then find the value of T (2, 3). Verify whether it is one−to−one or not?

Solution:

 Given that T: R² → R².

T (1, 0) = (1, 4) & T (1, 1) = (2,5)

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

Comparing on both sides we get a+b=2 & b=3.

Solving these two equations we get a=−1 & b=3

T (2,3) = T(−1 (1, 0)) + 3 (1, 1))

= −1 T (1, 0) + 3 T (1, 1)

= −1(1, 4) + 3(2,5)

= (−1,−4)+(6, 15)

= (−1+6, −4+15)

T (2, 3) = (5, 11)

We know that a linear transformation T is one−to−one if and only if N(T) = {(0,0) }

 (ie) If (a, b) ∈ T then T (a, b) = (0, 0) = T (0, 0).

Let us consider T (a, b) = (0, 0).

⇒ a=0 & b=0

  N(T) = {(0,0)}

Hence T is one−to−one.

 

Example 21

Let T: P (R) → P (R). Define T [f(x)] = ₀ʃ^x f (t) dt. Prove that T is linear, one−to−one but not onto, for the above operation.

Solution:

Given that T: P (R) → P (R) and T [f(x)] = ₀ʃ^x f(t) dt

To prove T is linear

To prove T is linear we have to show that

T (cx+y)= c T (x) + T (y) for all x, y ∈ V & c ∈ F.

Let C ∈ F & f, g ∈ P(R).

 T[cf(x)+ g(x)] = ₀ʃ^x [cf (t) + g (t)] dt

 = ₀ʃ^x cf (t) dt + ₀ʃ^x g(t) dt

= c ₀ʃ^x f (t) dt + ₀ʃ^x g(t) dt

= cT.[f(x)]+ T[g (x)]

Hence T is linear.

To prove T is one−to−one

We know that the linear transformation T is 1−1 if and only if N(T) = 0.

The standard basis for P (R) is {1, x, x², x³,... xⁿ}

The range of T is R(T) = span [(T(1), T (x), T (x²) ... T(xⁿ)]

is linearly independent.

Then, the dimension of the range of T is n+1

(ie) rank (T) =n+1.

The dimension theorem states that if V and W are the vector spaces

and if T: V→W is linear.

Let V be the finite dimensional.

N (T) + R (T) = dim (V)

Since dim [P (R)] = n + 1 & rank (T) = n + 1

N (T) + R (T) = dim [P (R)]

N(T) + (n + 1) = (n+1)

 N(T)=0

Since N (T) = 0, T is one−to−one.

To prove it is not on−to:

The basis of P (R) (domain) is span [1, x, x², x³,... xⁿ] and the basis for the codomain P (R) (range) is

Span [ x, x²/2, x³/3, x⁴/4, x²/2, …. Xⁿ⁺¹/n+1,]

 T is not on to, because no integral is equal to 1.

 

Example 22

Let V and W be finite dimensional vector spaces. Define T: V→W be linear. Prove that if dim (V) < dim (W), then T cannot be onto.

Solution:

By the dimension theorem, we know that

N(T)+R(T) = dim (V)

Let dim (V) < dim (V)

To prove T is not on to:

Suppose that T is onto. Then R(T) = dim (W)

                 ……………(1)

And also by the dimension theorem N (T) + R (T) = dim (V).

Since the dimension is a non−negative (always positive) number.

  N(T)=0

R(T) = dim (V)

                ………………(2)

From equations (1) & (2): dim (W) = dim (V)

But this a contradiction to our hypothesis dim (V) < dim (W).

Hence T is not onto.

 

Example 23

Let V and W be finite dimensional vector spaces.

Define: T: V→W be linear. Prove that if dim (V) > dim (W), then T cannot be one−to−one.

Solution:

Let us consider dim (V) > dim (W).

To prove T is not one−to−one:

Suppose T is one−to−one. Then N (T) = 0

Also since V is finite dimensional T is onto.

 Now by dimension theorem we know that N (T) + R (T) = dim (V)

Since N (T) = 0    ⇒ R (T) = dim (V)

 Also T is one−to−one R (T) = dim (W)

 dim (V) = dim (W)

This is a contradiction to the assumption

dim (V) > dim (W)

Hence T is not one−to−one.

 

Example 24

Let T: P (R) → P (R) be linear. Define T [f(x)] = f '(x). Prove that T is onto but not one−to−one.

Solution:

Given that T: P (R) → P (R), T [ƒ (x)] = ƒ′(x)

To prove T is onto but not one−to−one:

  Let g(x)=5, & h(x) = 6 both are in P (R).

⇒ g(x) ≠ h(x))

Now T [g(x)] = g′ (x) = d/dx [g(x)] = d/dx [5] = 0

and T [h (x)] = h′ (x) = d/dx [h(x)] = d/dx [6] = 0

 T(5)=0 & T (6)=0. Null space (T) ≠ 0

Hence T is not one−to−one.

To prove T is onto:

For any element g(x) in P(R), we can find f'(x) = ʃ g(x) dx in

P (R) which satisfies T[f(x)]=f '(x) = g(x)

 g(x) is in the range (T) ⇒ P(R)  R (T)

Given that T: P (R) → P (R) we have R(T)  P(R)

P (R) = R(T)

Hence T is onto.

 

Example 25

Verify that T: R³ → R and T (u) = || u || is a linear transformation or not.

Solution:

T(u + v) = || u+v || ≤ || u || + || u ||

Here T (u+v) ≠ T (u) +T (v) for all u, v ∈ R³.

Hence it is not linear transformation.

 

Example 26

Let T: R² → R³ be defined by T (x, y) = (x + 3y, 0, 2x−4y). Compute the matrix of the transformation with respect to the standard bases of R² and R³. Find N (T) & R (T).

Is T one−to−one? Is T onto?. Justify your answer.

Solution:

Let T (1, 0) = (1, 0, 2) = 1 (1, 0, 0) + 0 (0, 1, 0)+2 (0, 0, 1)

T (0, 1) = (3, 0, −4) = 3 (1, 0, 0) + 0 (0, 1, 0) + (− 4) (0, 0, 1).

                   ………. (1) 

A = [T]= Matrix of linear transformation.

A=

N(T) = {(x, y) / T (x, y) = (0, 0, 0) } = {(x, y)/x+3y=0, 2x−4y=0}=(0, 0).

R(T) = span { T (1, 0), T (0, 1) } = span { (1, 0, 2), (3, 0, −4) } = xz plane

T is one to one because N (T) = { (0, 0) }.

T is not onto because R (T) is in xz - plane

 (0,1,0) does not have pre image.

 

Example 27

Let T: R³→R², be defined by T (x, y, z) = (2x−y, 3z). Verify whether T is linear or not. Find N (T) and R (T) and hence verify the dimension theorem.

Solution:

 Let x₁, x₂, y₁, y₂, z₁, z₂ ∈ R³

T(x₁+x₂, y₁+y₂, z₁+z₂) = [2x₁+2x₂−y₁−y₂, 3z₁+3z₂]

= [2x₁−y₁+ 2x₂ ‒ y₂, 3z₁ + 3z₂]

= (2x₁ ‒ y₁, 3z₁) + (2x₂ − y₂, 3z₂)

= T(x₁, y₁, z₁) + T(x₂, y₂, z₂)

Now T[α(x₁,y₁, z₁)] = T (α x₁, α y₁, α z₁)

= (2αx₁ − αу₁, 3αz₁)

= α (2x₁ − y₁, 3z₁)

= α T(x₁,y₁, z₁)

T is linear.

N(T) = {(x, y, z) / T (x, y, z) = (0, 0)}

(ie) N (T) = {(x, y, z)/2x−y=0 & 3z=0}

= {(x, y, 0)/y = 2x }

The line y=2x lies on the xy plane

  dim N (T)=1

R(T) = span {T (1, 0, 0), T (0, 1, 0), T (0, 0, 1)}

= span { (2, 0), (−1, 0), (0, 3)}

= span { (2, 0), (0, 3)}

dim R (T) = 2

Dimension theorem states that dim N (T) + dim R (T) = dim R³

LHS = 1+2 = 3 & RHS = 3

Hence dimension theorem is verified

 

Example 28

Consider the basis S={v₁, v₂, v₃} for R³, where v₁ = (1, 1, 1), v₂ = (1, 1, 0) & v₃ = (1, 0, 0). Let T: R³ → R² be the linear transformation, such that T(v₁) = (1, 0), T (v₂) = (2,−1) and T (v₃)=(4,3). Find the formula for T (x₁,x₂, x₃), then use this formula to compute T (2,−3, 5).

Solution:

Let us express X = (x₁, x₂, x₃) as a linear combination of  v₁ = (1, 1, 1), v₂ = (1, 1, 0) and v₃ = (1, 0, 0).

If we write

(x₁, x₂, x₃) = a (1, 1, 1) + b (1, 1, 0) + c (1, 0, 0)

(x₁, x₂, x₃) = (a, a, a) + (b, b, 0) + (c, 0, 0)

⇒ a+b+c=x₁, a+b=x₂, a = x₃

⇒ a=x₃, b=x₂−x₃ & c=x₁= x₂

 (x₁, x₂, x₃) = x₃(1, 1, 1) + (x₂−x₃)(1, 1, 0) + (x₁ ‒ x₂)(1, 0, 0)

= x₃v₁+ (x₂−x₃)v₂ + (x₁−x₃)v₃

 T (x₁, x₂, x₃) = x₃T(v₁) + (x₂ − x₃) T (v₂) + (x₁ − x₂)T(v₃)

= x³(1, 0) + (x₂ − x₃) (2, − 1) + (x₁ − x₂) (4, 3)

T (x₁, x₂, x₃) = (4x₁ − 2x₂ − x₃, 3x₁ − 4x₂ + x₃)

Put x₁ =2, x₂ = −3 & x₃=5 in the above equation, we get

T (2, 3, 5)=(9, 23)