# Kernel Of Linear Transformation

Linear Algebra / Tuesday, November 6th, 2018
(Last Updated On: November 6, 2018)

# Kernel Definition

Let V and W be vector spaces over a the field F and T: V → W be a linear transformation, The kernel of a linear transformation T, written ker T, is the set of elements in V which map into α’ ∈ W such that

0’ is the zero or null element in W.

 Theorem

Let T: V → W be a linear transformation. Then ker T is subspace of V.

Proof:

We have,

Since T(0) = 0’, 0 ∈ ker T and so ker T is non-empty. If ker T contains 0 alone then ker T is a subspace of V.

Let us suppose that ker T contains a non-zero element α of V, then T(α) = 0’.

Then for any c ∈ F (the field on which V is defined) we have T(cα) = cT(α) = c0’ = 0’

This implies that cα ∈ ker T.

Let α, β ∈ ker T, then T(α) = 0’, T(β) = 0’

Now since T is linear transformation,

T(α + β) = T(α) + T(β) = 0’ + 0’ = 0’

So, α, β ∈ ker T ⇒ α + β ∈ ker T

And α ∈ ker T ⇒ cα ∈ ker T, c ∈ F

This shows that ker T is a subspace of V.

 Theorem

Let T: V → W be a linear transformation. Then ker T is subspace of W.

Proof:

Since T(0) = 0’ ∈ Im(T), it follows that Im(T) is non-empty.

Let Im(T) contain a non-zero element say α’. Then there exists an element α ∈ V such that T(α) = α’.

This implies that T(cα) = c T(α) = cα’ for all c in F.

Therefore, cα’ ∈ Im(T).

Again, let α’, β’ ∈ Im(T), then there are elements α, β ∈ V such that

T(α) = α’, T(β) = β’

Now, T(α + β) = T(α) + T(β) = α’ + β’

⇒ α’ + β’ ∈ Im(T).

Hence α’, β’ ∈ Im(T) ⇒ α’ + β’ ∈ Im(T).

And cα’ ∈ Im(T), c ∈ F

Hence Im(T) is a subspace of W.

 Theorem

Let T: V → W be a linear transformation such that ker T = {0}. Then the images of a linearly independent set of vectors in V are linearly independent in W.

Proof:

Let {α1, α2, …, αn} be a linearly independent set in V, we have to prove that {T(α1), T(α2), …, T(αn)} is a linearly independent set of vectors in W.

Let c1, c2, …, cn ∈ F, such that

c1 T(α1) + c2 T(α2) + … + cn T(αn) = 0’

Since T is a linear transformation, we have

T(c1α1 + c2α2 + … + cnαn) = c1 T(α1) + c2 T(α2) + … + cn T(αn)

Therefore, T(c1α1 + c2α2 + … + cnαn) = 0’

This implies that c1α1 + c2α2 + … + cnαn ∈ ker T.

But ket T = {0} is given.

Hence c1α1 + c2α2 + … + cnαn = 0

But since {α1, α2, …, αn} is a linearly independent set of vectors in V, we must have

c1 = c2 = … = cn = 0

Thus it follows that, c1 T(α1) + c2 T(α2) + … + cn T(αn) = 0’

Implies that c1 = c2 = … = cn = 0

Hence {T(α1), T(α2), …, T(αn)} is a linearly independent set of vectors in W.

 Theorem

Let T: V → W be a linear transformation and {α1, α2, …, αn} be a basis of V. Then {T(α1), T(α2), …, T(αn)} generates Im(T).

Proof:

Let α’ ∈ Im(T). Then exists at least one element, say α in V such that T(α) = α’.

Since {α1, α2, …, αn} be a basis of V, there exists c1, c2,…, cn ∈ F such that

α = c1α1 + c2α2 + … + cnαn

Therefore, T(α) = T(c1α1 + c2α2 + … + cnαn) = c1 T(α1) + c2 T(α2) + … + cn T(αn), since T is linear.

Thus, α’ = c1 T(α1) + c2 T(α2) + … + cn T(αn)

Since eact T(αi) ∈ Im(T), it follows that Im(T) is generated by {T(α1), T(α2), …, T(αn)}.