Nullity And Rank Of A Linear Transformation


Linear Algebra / Saturday, November 10th, 2018
(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 Im(T) of W.

If V be a finite dimensional vector space then both ker T and Im(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, …, αk} be a basis of ker T. There are vectors αk+1, αk+2, …, αn in V such that {α1, α2, …, αn} is a basis for V. We shall now prove that {T(αk+1), T(αk+2), …, T(αn)} is a basis for Im(T). We have noted that the vectors T(α1), T(α2), …, T(αn) certainly span Im(T), and since α1, α2, …, αk ∈ ker T, we have T(α1) = T(α2) = … = T(αk) = 0’. Hence it follows that the vectors T(αk+1), T(αk+2), …, T(αn) span Im(T). We shall show that T(αk+1), T(αk+2), …, T(αn) are linearly independent.

Suppose that there are ck+1, ck+2, …, cn ∈ F such that

ck+1 T(αk+1) + ck+2 T(αk+2) + … + cn T(αn) = 0’

Since T is a linear transformation, it follows that

T(ck+1 αk+1 + ck+2 αk+2 + … + cn αn) = 0’

Hence α = ck+1 αk+1 + ck+2 αk+2 + … + cn αn ∈ ker T.

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

α = b1 α1 + b2 α2 + … + bk αk , b1, b2, …, bk ∈ F.

Thus, α = b1 α1 + b2 α2 + … + bk αk = ck+1 αk+1 + ck+2 αk+2 + … + cn αn

or, b1 α1 + b2 α2 + … + bk αk – ck+1 αk+1 – ck+2 αk+2 – … – cn αn = 0

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

b1 = b2 = … = bk = ck+1 = ck+2 = … = cn = 0

Hence T(αk+1), T(αk+2), …, T(αn) are linearly independent and as they span Im(T), {T(αk+1), T(αk+2), …, T(αn)} is a basis of Im(T).

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

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

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

Hence the transformation T is onto.

Conversely, suppose that T is onto.

Then Im(T) = W

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

Since dim ker T + dim Im(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 Im(T) = W, therefore dim Im(T) = dim W.

Now, dim ker T + dim Im(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

rank ex

 

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