(Last Updated On: November 10, 2018)

In this article we will learn about nullity and rank of a linear transformation.

Definition

Let V and W be vector spaces over the field F and let T be a linear transformation from V to W. The nullity of T is the dimension of the subspace ker T of V and the rank of T is the dimension of the subspace I_m(T) of W.

If V be a finite dimensional vector space then both ker T and I_m(T) are finite dimensional.

Theorem

Let V and W be vector spaces over the field F and V is finite dimensional. If T: V→W is a linear transformation then the nullity of T + rank of T = dim V.

Proof:

Let dim V  = n.

Let {α₁, α₂, …, α_k} be a basis of ker T. There are vectors α_k+1, α_k+2, …, α_n in V such that {α₁, α₂, …, α_n} is a basis for V. We shall now prove that {T(α_k+1), T(α_k+2), …, T(α_n)} is a basis for I_m(T). We have noted that the vectors T(α₁), T(α₂), …, T(α_n) certainly span I_m(T), and since α₁, α₂, …, α_k ∈ ker T, we have T(α₁) = T(α₂) = … = T(α_k) = 0’. Hence it follows that the vectors T(α_k+1), T(α_k+2), …, T(α_n) span I_m(T). We shall show that T(α_k+1), T(α_k+2), …, T(α_n) are linearly independent.

Suppose that there are c_k+1, c_k+2, …, c_n ∈ F such that

c_k+1 T(α_k+1) + c_k+2 T(α_k+2) + … + c_n T(α_n) = 0’

Since T is a linear transformation, it follows that

T(c_k+1 α_k+1 + c_k+2 α_k+2 + … + c_n α_n) = 0’

Hence α = c_k+1 α_k+1 + c_k+2 α_k+2 + … + c_n α_n ∈ ker T.

Again, by assumption, {α₁, α₂, …, α_k} be a basis of ker T so

α = b₁ α₁ + b₂ α₂ + … + b_k α_k , b₁, b₂, …, b_k ∈ F.

Thus, α = b₁ α₁ + b₂ α₂ + … + b_k α_k = c_k+1 α_k+1 + c_k+2 α_k+2 + … + c_n α_n

or, b₁ α₁ + b₂ α₂ + … + b_k α_k – c_k+1 α_k+1 – c_k+2 α_k+2 – … – c_n α_n = 0

But {α₁, α₂, …, α_n} is a basis of V, so we must have,

b₁ = b₂ = … = b_k = c_k+1 = c_k+2 = … = c_n = 0

Hence T(α_k+1), T(α_k+2), …, T(α_n) are linearly independent and as they span I_m(T), {T(α_k+1), T(α_k+2), …, T(α_n)} is a basis of I_m(T).

Clearly, dim (I_m(T)) = n – k = rank of T.

Also dim(ker T) = k = nullity of T

Hence rank of T + nullity of T = n = dim V.

Theorem

Let V and W are both finite dimensional vector spaces of the same dimension and T is a linear transformation from V to W, then T is one-one if and only if T is onto.

Proof:

Let T be one-one then clearly ker T = {0} and so dim ker T = 0.

Now, ker T is a subspace of V and I_m(T) is a subspace of W and also dim ker T + dim I_m(T) = dim V.

Since dim ker T = 0, we have dim I_m(T) = dim V = dim W (given)

Therefore, dim I_m(T) = dim W ⇒ I_m(T) = W

Hence the transformation T is onto.

Conversely, suppose that T is onto.

Then I_m(T) = W

Therefore dim I_m(T) = dim W = dim V.

Since dim ker T + dim I_m(T) = dim V, we have dim ker T = 0

Hence ker T = {0} and T is one-one.

Theorem

If V and W are finite dimensional vector spaces over a field F and if there exists a linear transformation T: V → W which is both one-one and onto, then dim V = dim W.

Proof:

Since by assumption T is one-one, we have ker T = {0}, therefore dim ker T = 0.

Again, since T is onto, we have I_m(T) = W, therefore dim I_m(T) = dim W.

Now, dim ker T + dim I_m(T) = dim V

⇒ 0 + dim W = dim V

i,e., dim V = dim W.

Example 01

Let V be the vector space of n-square matrices over the field F. Let M be an arbitrary matrix in V. Let T: V → W be defined by T(A) = AM + MA where A ∈ V. Show that T is linear.

Solution:

For any A, B ∈ V and c ∈ F, we have

T(A + B) = (A + B)M + M(A + B)

=AM + BM + MA + MB

=(AM + MA) + (BM + MB)

=T(A) + T(B)

And T(cA) = (cA)M + M(cA)

=c(AM) + c(MA) = c(AM +MA) = cT(A)

Accordingly, T is linear.

Example 02

 

<<Previous   Next>>