# 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’ ∈ I_{m}(T), it follows that I_{m}(T) is non-empty.

Let I_{m}(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’ ∈ I_{m}(T).

Again, let α’, β’ ∈ I_{m}(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’ ∈ I_{m}(T).

Hence α’, β’ ∈ I_{m}(T) ⇒ α’ + β’ ∈ I_{m}(T).

And cα’ ∈ I_{m}(T), c ∈ F

Hence I_{m}(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 {α_{1}, α_{2}, …, α_{n}} be a linearly independent set in V, we have to prove that {T(α_{1}), T(α_{2}), …, T(α_{n})} is a linearly independent set of vectors in W.

Let c_{1}, c_{2}, …, c_{n} ∈ F, such that

c_{1} T(α_{1}) + c_{2} T(α_{2}) + … + c_{n} T(α_{n}) = 0’

Since T is a linear transformation, we have

T(c_{1}α_{1} + c_{2}α_{2} + … + c_{n}α_{n}) = c_{1} T(α_{1}) + c_{2} T(α_{2}) + … + c_{n} T(α_{n})

Therefore, T(c_{1}α_{1} + c_{2}α_{2} + … + c_{n}α_{n}) = 0’

This implies that c_{1}α_{1} + c_{2}α_{2} + … + c_{n}α_{n} ∈ ker T.

But ket T = {0} is given.

Hence c_{1}α_{1} + c_{2}α_{2} + … + c_{n}α_{n} = 0

But since {α_{1}, α_{2}, …, α_{n}} is a linearly independent set of vectors in V, we must have

c_{1} = c_{2} = … = c_{n} = 0

Thus it follows that, c_{1} T(α_{1}) + c_{2} T(α_{2}) + … + c_{n} T(α_{n}) = 0’

Implies that c_{1} = c_{2} = … = c_{n} = 0

Hence {T(α_{1}), T(α_{2}), …, T(α_{n})} is a linearly independent set of vectors in W.

Theorem |

**Let T: V → W be a linear transformation and {α _{1}, α_{2}, …, α_{n}} be a basis of V. Then {T(α_{1}), T(α_{2}), …, T(α_{n})} generates I_{m}(T).**

**Proof:**

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

Since {α_{1}, α_{2}, …, α_{n}} be a basis of V, there exists c_{1}, c_{2},…, c_{n} ∈ F such that

α = c_{1}α_{1} + c_{2}α_{2} + … + c_{n}α_{n}

Therefore, T(α) = T(c_{1}α_{1} + c_{2}α_{2} + … + c_{n}α_{n}) = c_{1} T(α_{1}) + c_{2} T(α_{2}) + … + c_{n} T(α_{n}), since T is linear.

Thus, α’ = c_{1} T(α_{1}) + c_{2} T(α_{2}) + … + c_{n} T(α_{n})

Since eact T(α_{i}) ∈ I_{m}(T), it follows that I_{m}(T) is generated by {T(α_{1}), T(α_{2}), …, T(α_{n})}.