Basis, N(T), R(T), Span dimension theorem

(b) Basis, N(T), R(T), Span dimension theorem

Definition :

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

We define the range (or image) R(T) of T to be the subset of W consisting of all images (under T) of vectors in V; that is, R(T) = {T(x) : x ЄV}.

Example 1.

Let V and W be vector spaces, and let I : V → V and T₀: V → W be the identity and zero transformations, respectively.

THEOREM 1.

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

Proof :

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 nullity (T), and the rank of T, denoted rank (T), to be the dimensions of N(T) and R(T), respectively.

THEOREM 2

(Dimension Theorem) [Rank nullity 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 :

First we prove that S generates R(T). The fact that T(v_i) = 0 for 1 ≤ i ≤ k, we have

Now, we show that S is linearly independent.

Using the fact that T is linear, we have

Hence, there exist c₁, c₂, ..., c_k Є F such that

Since, β is a basis for V, we have b_i = 0 for all i.

Hence, S is linearly independent.

Note: This argument also show that  are distinct; therefore rank(T) = n - k

THEOREM 3.

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

Proof:

Let T is one-to-one and x Є N(T).

Then T(x) = 0 = T(0). Since T is one-to-one, we have x = 0.

Hence, N(T) = {0}.

Now, assume that N(T) = {0}, and let T(x) = T(y).

Then 0 = T(x) - T(y) = T(x - y) by property.

Therefore, x - y Є N(T) = {0}. So x - y = 0, or x = y.

This means that T is one-to-one.

THEOREM 4.

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 one-to-one, (b) T is onto, (c) rank(T) = dim(V)

THEOREM 5.

Let V and W be vector spaces, and let T : V → W be linear.

If β = {v₁, v₂, ..., v_n} is a basis for V, then

Proof :

We know that T(v_i) Є R(T) for each i. Since R(T) is a subspace, R(T) contains span 

Example

Thus we have found a basis for R(T), and so dim (R(T)) = 2

(b)(i) Basis for N(T) and compute N(T)

Problem 1.

 Find the bases for N(T) and compute the nullity of T.

Solution :

To find : N(T) and R(T)

Find basis of N(T)

So, any element in the null space will be of the form (a₁, a₂, 0)

Problem 2.

Find the basis for N(T) and compute N(T)

Solution :

.'. N(T) is shaped by four matrices

The number of basis elements of N(T) is 4.

.'. N(T) = 4

Problem 3.

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

Solution :

Domain is P₂(R). So, its standard basis is {1, x, x²}

Co-domain is P₃(R). Its standard bassis is {1, x, x², x³}

To find N(T), suppose T(f(x)) = 0(x)

We know that, if f is an n^th degree polynomial, then f ' is an (n-1)^th degree polynomial and consequently, -f '(x) / x is an (n-2)^th degree polynomial.

Problem 4.

Find the basis for N(T) and compute N(T)

Solution :

To find the null space of T :

i.e., N(T) = n² - 1

Range space of (T) = 

Dimension of range space of (T) = 1

i.e., R(T) = 1 with basis 1.

Problem 5.

Let T: R² → R³ defined by T (a₁, a₂) = (a₁ + a₂, 0, 2a₁-a₂)

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

Solution :

Let N(T) = {x Є V, T(x) = 0}

To find the basis for N(T)

Let x Є R², such that x = (a₁, a₂)

Then

Clearly, the kernel (null space) consists of everything that is transformed to zero.

Here, T(a₁, a₂) = (0, 0, 0) if a₁ = 0, a₂ = 0

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

In general,

N(T) = {(0, 0) Є R²: T(0, 0) = 0}

Then, the null space of T is zero dimensional.

Therefore, N(T) = 0 with basis φ

(b)(ii) Basis for R(T) and R(T)

Problem 6.

Let T: R³ → R² defined by

T (a₁, a₂, a₃) = (a₁ - a₂, 2a₃)

Find the basis for R(T) and compute R(T)

Solution :

Given: T: R³ → R²

T (a₁, a₂, a₃) = (a₁-a₂, 2a₃)

To find a basis of range space R(T):

Clearly, for every (a, b, c) Є R³ of T, there exists a T(a, b, c) in R², such that,

T(a, b, c) = (a - b, 2c)

That is, any vector in R² can be written using the transformed elements in R³.

Suppose, (x, y) Є R², then write (x, y) = (a - b, 2c), where a, b and c are any real numbers.

Clearly, a - b = x have infinitely many solutions, and assume (1/2)y = c

So,

T (a, b, c) = (a - b, 2c) = (x, y)

Then, R(T) is all of R² or R(T) = R²

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

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

Problem 7.

Let T: R² → R³ defined by T(a₁, a₂)

= (a₁ + a₂, 0, 2a₁ - a₂)

Find the basis for R(T) and compute R(T)

Solution :

Given T: R² → R³

To find a basis of range space R(T):

In general, if T from V to W is a linear transformation, and β = {v₁, v₂, ... v_n} is a basis for V, then

Here, β = {(1, 0), (0, 1)} is a standard basis for R³, so,

R(T) = span (T(β))

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

Then,

Also, clearly (1, 0, 2) and (1, 0, -1) are linearly independent.

So, ((1, 0, 2), (1, 0, -1)} is a basis for R(T) and the rank of T is R(T) = 2

(b) (iii) Verify the dimension theorem

Problem 8.

Verify the dimension theorem for T: R³ → R² defined by

T(a₁, a₂, a₃) = (a₁ - a₂, 2a₃)

Solution :

The dimension theorem is

If V is finite dimensional, then N(T) + R(T) = dim(V) ... (1)

Here, R³, R² are two vector spaces and T : R³ → R² is linear.

dimension of R³ is 3 dim(V) = 3

N(T) + R(T) = 1 + 2 by problem 1 Page No. 26 and problem 6 Page No. 32.

.'. N(T) + R (T) = dim(V)

Hence, the dimension theorem verified.

Problem 9.

Verify the dimension theorem for

Solution :

The dimension theorem is

If V is finite dimensional, then N (T) + R (T) = dim(V) ......(1)

Here, R², R³ are two vector spaces and T : R² → R³ is linear

dimension of R² is 2 => dim(V) = 2

N(T) + R(T) = 0 + 2 by problem 5 page no.31 and problem 7 page no. 33.

.'. N(T) + R(T) = dim(V)

Hence, the dimension theorem verified.

Problem 10.

Verify the dimension theorem for

Solution :

The dimension theorem is

If V is finite dimensional, then N (T) + R (T) = dim(V) ... (1)

Here, M_{2 x 3}(F) and M_{2 x 2}(F) are two vector spaces and

N(T) + R(T) = 4 + 2 by problem 2 and problem 3

.'. N(T) + R(T) = dim(V)

Hence, the dimension theorem verified.

Problem 11

Verify the dimension theorem for

Solution :

The dimension theorem is

If V is finite dimensional, then N(T) + R(T) = dim(V) ...... (1)

Here, P₂(R), P₃(R) are two vector spaces and animals

T : P₂(R) → P₃(R) is linear.

dim [P₂(R)] = 3

N(T) + R(T) = 0 + 3 = 3 by problem 3 & Problem 9

.'. N(T) + R(T) = dim(V)

Hence, the dimension theorem verified.

(b) (iv) Determine T is one-to-one or onto V

Problem 12

Determine T is one-to-one or onto for T : R³ → R² defined by

T(a₁, a₂, a₃) = (a₁ - a₂, 2a₃)

Solution :

To determine whether the given transformation is one-to-one or onto.

In general, if T : V → W a linear transformation, and N(T), and R(T) are null space and range of T, then

1. T is one-to-one if and only if N(T) = 0

2. T is onto if and only if R(T) = W

Here, N(T) = 1, so T is not one-to-one by problem 1

And R(T) = 2, so T is onto by problem 6

Therefore, the given transformation is onto, but not one-to-one

Problem 13.

Determine T is one-to-one or onto for T: R² → R³ defined by

T(a₁, a₂) = (a₁ + a₂, 0, 2a₁-a₂)

Solution :

To determine whether the given transformation is one-to-one or onto.

In general, if T : V → W a linear transformation, and N(T), and R(T) are null space and range of T, then,

1. T is one-to-one if and only if N(T) = 0

2. T is onto if and only if R(T) = W

Here, N(T) = 0, so T is not one-to-one by problem 5 page no.31

Clearly, T is not onto, because, T will never map all of R³.

For example T will never map (1, 1, 1).

Also,

rank (T) ≠ dim (R3)

2 ≠ 3

Therefore, the given transformation T is one-to-one, but not onto

Problem 14.

Determine T is one-to-one or onto

T: M_{2 x 3}(F) → M_{2 x 2}(F) defined by

Solution :

We know that, a linear mapping T : V → W is one-to-one if and only if N(T) = {0}

Since N(T) ≠ {0}, therefore T is not one-to-one.

Since, rank(T) = 2 and dim (M_{2 x 3}(F)) = 4, that is

rank(T) ≠ dim (M_{2 × 3}(V)}

therefore T is not onto by problem 2 & 8.

Hence, T is neither one-to-one nor onto.

Problem 15.

Determine T is one-to-one or onto T: P₂(R) → P₃(R) defined by

T(f(x)) = x f(x) + f'(x)

Solution :

We know that the dimension of domain P₂(R) < dimension of the co-domain P₃(R)

From this, the linear transformation T is not onto.

Theorem :

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

Here, we have N(T) = {0}

So, the above theorem confirms that T(f(x)) = x f(x) + ƒ'(x) is a one-to-one function.

Problem 16.

Give an example of a linear transformation T: R² → R² such that N(T) = R(T)

Solution :

Consider the linear transformation T : R² → R²

Construct an example, such that N(T) = R(T)

Take T(x, y) = (y, 0)

Problem 17.

Give an example of distinct linear transformations T and U such that N(T) = N(U) and R(T) = R(U)

Solution :

Let the linear transformations; T and U

Let T : R → R, defined by T(x) = 2x

And also U : R → R is defined by U(x) = x

The nullity of T is the set of all vectors x in R.

Such that T(x) = 0

Similarly, for the transformation U.

So, that N(T) = {0} and N(U) = {0}

⇒ N(T) = N(U) = {0}

The range of T or U is a subset of co-domain consisting of all images